Electric Potential

J = C-V

Reason — The electric potential at a point is the work done in carrying a unit positive test charge from infinity to that point,

$$\begin{aligned}\text V = \frac{\text W}{\text q_0} \\[1em] \text W = \text q_0 \text V\end{aligned}$$

Hence work (joule) = charge (coulomb) × potential (volt), that is 1 J = 1 C × 1 V or J = C-V. Equivalently, 1 volt = 1 joule/coulomb.

10 V

Reason —

Given,

• Charge, q = 3 C
• Force, F = 3000 N
• Distance, d = 1 cm = 10^-2 m

The intensity of the uniform electric field is

$$\begin{aligned}\text E = \frac{\text F}{\text q} \\[1em] \text E = \frac{3000\ \text N}{3\ \text C} \\[1em] \text E = 1000\ \text N\text C^{-1}\end{aligned}$$

In a uniform electric field the potential difference between two points separated by a distance d along the field is

$$\begin{aligned}\text V = \text E\text d \\[1em] \text V = 1000 \times 10^{-2} \\[1em] \text V = 10\ \text V\end{aligned}$$

1 × 10^-1 V

Reason —

Given,

• Work done, W = 2 J
• Charge, q = 20 C

By the definition of potential difference,

$$\begin{aligned}\text V_\text A - \text V_\text B = \frac{\text W}{\text q_0} \\[1em] \text V_\text A - \text V_\text B = \frac{2\ \text J}{20\ \text C} \\[1em] \text V_\text A - \text V_\text B = 0.1\ \text V \\[1em] \text V_\text A - \text V_\text B = 1 \times 10^{-1}\ \text V\end{aligned}$$

The distance between the points is not required, since the potential difference is fixed by the work done per unit charge alone.

does not depend on r

Reason — The electric field intensity at a point is equal to the negative potential gradient at that point,

$\text E = -\frac{\text{dV}}{\text{dr}}$

Inside a hollow charged sphere E = 0 everywhere, so $\frac{\text{dV}}{\text{dr}} = 0$. Hence the potential does not change with r; it is constant throughout the interior and equal to the potential at the surface, $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R}$.

1000 J

Reason —

Given,

• Charge, q = 0.5 C
• Potential difference, V = 2000 V

The kinetic energy gained by a charge q accelerated from rest through a potential difference V is

$$\begin{aligned}\text K = \text q\text V \\[1em] \text K = 0.5 \times 2000 \\[1em] \text K = 1000\ \text J\end{aligned}$$

zero

Reason — The perpendicular bisector of a dipole is its equatorial line. At every point on this line the point is equidistant from the charges + q and − q, so the potentials due to the two charges are equal and opposite,

$$\begin{aligned}\text V = \frac{\text q}{4\pi\epsilon_0}\left(\frac{1}{\text{BP}} - \frac{1}{\text{AP}}\right) \\[1em] \text V = 0 \qquad (\because\ \text{BP} = \text{AP})\end{aligned}$$

Since the potential is zero everywhere on this line and also zero at infinity, the work done W = q₀(V_final − V_initial) = 0.

zero

Reason — The potential energy of an electric dipole placed in a uniform electric field at an angle θ to the field is

$\text U_\theta = -\text p\text E\cos\theta$

U is most negative when cos θ has its greatest value, that is cos θ = 1 or θ = 0. Thus U₀ = − pE, which is the least (maximum negative) value. In this position the dipole moment is along the field and the dipole is in stable equilibrium.

everywhere same as on the surface

Reason — The whole charge given to a hollow metal sphere resides on its outer surface, so the electric field inside the sphere is zero at every point. Since

$$\begin{aligned}\text E = -\frac{\text{dV}}{\text{dr}} \\[1em] \text E = 0\end{aligned}$$

there is no potential gradient inside, and hence no work is done in moving a charge from one interior point to another. The potential is therefore uniform throughout the interior and equal to its value at the surface, $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R}$.

8 J

Reason —

Given,

• Radius of the ring, R = 0.45 m
• Charge on the ring, Q = 2 × 10^-10 C
• Charge to be brought, q = 2 C

Every element of the ring is at the same distance R from the centre, so the potential at the centre is

$$\begin{aligned}\text V_\text C = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text R} \\[1em] \text V_\text C = \frac{(9.0\times10^9)(2\times10^{-10})}{0.45} \\[1em] \text V_\text C = \frac{1.8}{0.45} \\[1em] \text V_\text C = 4\ \text V\end{aligned}$$

The work done in bringing the charge q from infinity (where V = 0) to the centre is

$$\begin{aligned}\text W = \text q(\text V_\text C - \text V_\infty) \\[1em] \text W = 2 \times 4 \\[1em] \text W = 8\ \text J\end{aligned}$$

decreases because the charge moves along the electric field.

Reason — A free positive charge experiences a force $\vec{\text F} = \text q\vec{\text E}$in the direction of the field, so on release it moves along the electric field, that is from a region of higher potential to a region of lower potential. Its potential energy U = qV therefore falls, the loss of potential energy appearing as kinetic energy.

zero

Reason — The potential at a general point (r, θ) due to a short dipole is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p\cos\theta}{\text r^2}$

For an axial point θ = 0° and $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2}$, while for a point on the equatorial line θ = 90°, so cos θ = 0 and V = 0. Physically, an equatorial point is equidistant from + q and − q, so their potentials cancel.

Intensity of electric field due to an electric dipole at a distant point.

Reason — For a point charge $\text E \propto \frac{1}{\text r^2}$, and a charged sphere behaves at an external point as if its whole charge were placed at its centre, so there also $\text E \propto \frac{1}{\text r^2}$. For a short dipole the potential varies as $\text V \propto \frac{1}{\text r^2}$, which is an inverse-square dependence. But the intensity of the electric field of a dipole at a distant point varies as $\text E \propto \frac{1}{\text r^3}$, and hence it does not obey the inverse-square law.

S₁ is true, S₂ is also true and S₁ is the cause of S₂.

Reason — The charge given to a conducting sphere resides entirely on its outer surface, so the electric intensity at every interior point is zero, which makes S₁ true. Since

$$\begin{aligned}\text E = -\frac{\text{dV}}{\text{dr}} \\[1em] \text E = 0\end{aligned}$$

the potential gradient inside vanishes, so V is constant throughout the interior and equal to its surface value of 100 V. Hence S₂ is true and follows directly from S₁; the vanishing of the intensity is the cause of the constancy of the potential.

Equipotential surfaces are always spherical.

Reason — Any surface over which the electric potential is the same everywhere is an equipotential surface, so the potential difference between any two of its points is zero and no work is done in moving a charge along it. Since there is no potential gradient along the surface, the electric field can have no component parallel to it, and hence the lines of force are always at right angles to the surface. The shape of the surface, however, depends on the charge distribution — the equipotential surfaces are concentric spheres for a point charge but a family of parallel planes for a uniform electric field. Hence the fourth statement is not true.

is zero

Reason — The electric field is everywhere normal to an equipotential surface, so for every displacement $\text{d}\vec{\text l}$taken along the surface $\vec{\text E}\cdot\text{d}\vec{\text l} = 0$. Hence

$$\begin{aligned}\text W = -\text q\displaystyle\int_{\text A}^{\text B}\vec{\text E}\cdot\text{d}\vec{\text l} \\[1em] \text W = 0\end{aligned}$$

The same conclusion follows from W = q(V_B − V_A), because V_A = V_B on an equipotential surface.

2 pE

Reason — The work done in rotating a dipole in a uniform electric field from an angle θ₁ to an angle θ₂ is

$\text W = \text p\text E(\cos\theta_1 - \cos\theta_2)$

The dipole is in stable equilibrium at θ₁ = 0 and in unstable equilibrium at θ₂ = 180°. Therefore

$$\begin{aligned}\text W = \text p\text E(\cos 0° - \cos 180°) \\[1em] \text W = \text p\text E[1 - (-1)] \\[1em] \text W = 2\text p\text E\end{aligned}$$

1.33 × 10⁷ m/s

Reason —

Given,

• Potential difference, V = 500 V
• Charge on electron, e = 1.6 × 10^-19 C
• Mass of electron, m = 9.0 × 10^-31 kg

The kinetic energy acquired by the electron equals the work done on it,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text e\text V \\[1em] \text v = \sqrt{\frac{2\text e\text V}{\text m}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{\frac{2 \times (1.6\times10^{-19}) \times 500}{9.0\times10^{-31}}} \\[1em] \text v = \sqrt{\frac{1.6\times10^{-16}}{9.0\times10^{-31}}} \\[1em] \text v = \sqrt{1.78\times10^{14}} \\[1em] \text v = 1.33\times10^7\ \text m/\text s\end{aligned}$$

1 : 2

Reason — For a charge falling from rest through a potential difference V,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text q\text V \\[1em] \text v = \sqrt{\frac{2\text q\text V}{\text m}}\end{aligned}$$

Since the mass and the potential difference are the same for both particles, $\text v \propto \sqrt{\text q}$. Hence

$$\begin{aligned}\frac{\text v_1}{\text v_2} = \sqrt{\frac{\text q}{4\text q}} \\[1em] \frac{\text v_1}{\text v_2} = \frac{1}{2}\end{aligned}$$

that is, the ratio of their speeds is 1 : 2.

5.76 J

Reason —

Given,

• q₁ = 8 μC = 8 × 10^-6 C, q₂ = 12 μC = 12 × 10^-6 C
• Initial separation, r₁ = 10 cm = 0.10 m
• Final separation, r₂ = 10 − 4 = 6 cm = 0.06 m

The electric potential energy of a system of two point charges is

$\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}$

The work done equals the increase in this potential energy,

$$\begin{aligned}\text W = \frac{1}{4\pi\epsilon_0}\text q_1\text q_2\left(\frac{1}{\text r_2} - \frac{1}{\text r_1}\right) \\[1em] \text W = (9.0\times10^9)(8\times10^{-6})(12\times10^{-6})\left(\frac{1}{0.06} - \frac{1}{0.10}\right) \\[1em] \text W = 0.864 \times (16.67 - 10) \\[1em] \text W = 0.864 \times 6.67 \\[1em] \text W = 5.76\ \text J\end{aligned}$$

increases

Reason — The potential energy of a charge q at a point of potential V is U = qV. For a positive charge taken from a point of lower potential to a point of higher potential, V increases and so U also increases. This is expected, because a free positive charge moves by itself from higher to lower potential; to carry it the other way an external agent must do work on it, and this work is stored as potential energy of the system.

increase

Reason — For two electrons both charges are negative, so their product is positive and the potential energy

$\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}$

is positive, indicating net repulsion. As r decreases, U increases. Work has to be done by an external agent against the force of repulsion in bringing the electrons closer, and this work is stored as the potential energy of the system.

Both Assertion and Reason are false.

Reason —

Explanation

Assertion (A) is incorrect : For equal steps of potential, the spacing between successive equipotential surfaces varies inversely as the field, since $\text{dr} \propto \frac{1}{\text E}$; a strong field crowds the surfaces together and a weak field spreads them apart. In the given diagram the successive surfaces come closer and closer as we go from A to D. Hence the electric field strength is least at A and greatest at D; it increases from A to D and does not reduce.

Reason (R) is also incorrect : The correct relation is the reverse of the one stated — the electric field intensity is equal to the negative potential gradient,

$\text E = -\frac{\text{dV}}{\text{dr}}$

It is the field that equals the negative gradient of the potential, not the potential difference that equals the negative gradient of the field.

Therefore, both Assertion and Reason are false.

$\frac{\text r_1}{\text r_2}$ > 1

Reason — For a point charge the equipotential surfaces are concentric spheres of potential

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R} \\[1em] \text R = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text V}\end{aligned}$$

Writing $\text k = \frac{1}{4\pi\epsilon_0}$, the radii of the 30 V, 20 V and 10 V surfaces are

$\text R_{30} = \frac{\text{kq}}{30},\qquad \text R_{20} = \frac{\text{kq}}{20},\qquad \text R_{10} = \frac{\text{kq}}{10}$

In the figure r₁ is the separation between the 10 V and 20 V surfaces, and r₂ is the separation between the 20 V and 30 V surfaces. Therefore

$$\begin{aligned}\text r_1 = \text R_{10} - \text R_{20} \\[1em] \text r_1 = \frac{\text{kq}}{10} - \frac{\text{kq}}{20} \\[1em] \text r_1 = \frac{\text{kq}}{20}\end{aligned}$$

and

$$\begin{aligned}\text r_2 = \text R_{20} - \text R_{30} \\[1em] \text r_2 = \frac{\text{kq}}{20} - \frac{\text{kq}}{30} \\[1em] \text r_2 = \frac{\text{kq}}{60}\end{aligned}$$

so that

$$\begin{aligned}\frac{\text r_1}{\text r_2} = \frac{\text{kq}/20}{\text{kq}/60} \\[1em] \frac{\text r_1}{\text r_2} = 3\end{aligned}$$

which is > 1.

This agrees with the general property that equipotential surfaces of equal potential interval lie closer together where the field is strong and farther apart where the field is weak.

Note: The textbook gives option (b), $\frac{\text r_1}{\text r_2}$ < 1. Its figure places 10 V on the innermost and 30 V on the outermost surface, which is impossible for $\text V = \frac{\text{kq}}{\text R}$; option (b) would follow only if the 10 V and 30 V labels were interchanged.

$15\sqrt{2}$ Vm^-1 along PA

Reason —

Given, the side of the square is 0.2 m, V_A = 2 V, V_B = V_D = 5 V and V_C = 8 V.

Along the side AB the potential rises from 2 V to 5 V over 0.2 m, so the component of the field along BA is

$$\begin{aligned}\text E_2 = \frac{\Delta \text V}{\text d} \\[1em] \text E_2 = \frac{5 - 2}{0.2} \\[1em] \text E_2 = 15\ \text V\text m^{-1}\end{aligned}$$

directed from B towards A, since the field points from higher to lower potential.

Similarly along the side AD the potential rises from 2 V to 5 V over 0.2 m, so

$$\begin{aligned}\text E_1 = \frac{5 - 2}{0.2} \\[1em] \text E_1 = 15\ \text V\text m^{-1}\end{aligned}$$

directed from D towards A.

These two components are mutually perpendicular, so the resultant field is

$$\begin{aligned}\text E = \sqrt{\text E_1^2 + \text E_2^2} \\[1em] \text E = \sqrt{15^2 + 15^2} \\[1em] \text E = 15\sqrt{2}\ \text V\text m^{-1}\end{aligned}$$

Its direction bisects the angle between BA and DA, that is, it points along the diagonal from C towards A. As P is the centre of the square, this direction is along PA.

5 V

Reason —

Given, $\text E = \frac{100}{\text x^2}$ N C^-1, with x₁ = 10 m and x₂ = 20 m.

The potential difference between two points is obtained from

$\text V_1 - \text V_2 = \displaystyle\int_{\text x_1}^{\text x_2} \text E\ \text{dx}$

Substituting the given field,

$$\begin{aligned}\text V_1 - \text V_2 = \displaystyle\int_{10}^{20} \frac{100}{\text x^2}\ \text{dx} \\[1em] \text V_1 - \text V_2 = 100\left[-\frac{1}{\text x}\right]_{10}^{20} \\[1em] \text V_1 - \text V_2 = 100\left(-\frac{1}{20} + \frac{1}{10}\right) \\[1em] \text V_1 - \text V_2 = 100 \times \frac{1}{20} \\[1em] \text V_1 - \text V_2 = 5\ \text V\end{aligned}$$

zero

Reason —

Given, V = 5x² − 10x − 9 volt.

The electric field is the negative potential gradient,

$\text E = -\frac{\text{dV}}{\text{dx}}$

Differentiating the given expression term by term,

$$\begin{aligned}\frac{\text{dV}}{\text{dx}} = \frac{\text d}{\text{dx}}(5\text x^2) - \frac{\text d}{\text{dx}}(10\text x) - \frac{\text d}{\text{dx}}(9) \\[1em] \frac{\text{dV}}{\text{dx}} = 10\text x - 10\end{aligned}$$

Therefore

$$\begin{aligned}\text E = -(10\text x - 10) \\[1em] \text E = 10 - 10\text x\end{aligned}$$

At x = 1 m,

$$\begin{aligned}\text E = 10 - 10(1) \\[1em] \text E = 0\end{aligned}$$

$\frac{\sigma}{\epsilon_0}\left(\text R + \text r\right)$

Reason —

Given, two concentric spheres of radii R and r carrying the same surface charge density σ.

The charges on the two spheres are

$\text Q_\text R = \sigma(4\pi\text R^2), \qquad \text Q_\text r = \sigma(4\pi\text r^2)$

The potential inside a charged sphere is the same as that at its surface, so the potential at the common centre due to each sphere is $\frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text{radius}}$. Since potential is a scalar, the two contributions add algebraically,

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\left(\frac{\text Q_\text R}{\text R} + \frac{\text Q_\text r}{\text r}\right) \\[1em] \text V = \frac{1}{4\pi\epsilon_0}\left(\frac{\sigma 4\pi\text R^2}{\text R} + \frac{\sigma 4\pi\text r^2}{\text r}\right) \\[1em] \text V = \frac{\sigma}{\epsilon_0}(\text R + \text r)\end{aligned}$$

$\frac{1}{4\pi\epsilon_0}\frac{\text{qQ}}{2\text R}$

Reason —

Given, a uniformly charged ring of radius R carrying charge Q, and a point charge (− q) on its axis at a distance $\sqrt3 \text R$ from the centre.

Every element of the ring is at the same distance from P,

$$\begin{aligned}\text{distance} = \sqrt{\text R^2 + (\sqrt3 \text R)^2} \\[1em] \text{distance} = \sqrt{\text R^2 + 3\text R^2} \\[1em] \text{distance} = 2\text R\end{aligned}$$

Hence the potential at P due to the ring is

$\text V_\text P = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{2\text R}$

and the potential at infinity is V_∞ = 0. The work done by the external agent is

$$\begin{aligned}\text W = (-\text q)(\text V_\infty - \text V_\text P) \\[1em] \text W = (-\text q)\left(0 - \frac{1}{4\pi\epsilon_0}\frac{\text Q}{2\text R}\right) \\[1em] \text W = \frac{1}{4\pi\epsilon_0}\frac{\text{qQ}}{2\text R}\end{aligned}$$

The work is positive because the charge − q is attracted by the positively charged ring, so the external agent must do work in taking it to infinity.

$\frac{\text V}{6\text r}$

Reason —

Given, a hollow metal sphere of radius r carrying a charge Q, with V_surface − V_3r = V.

Writing $\text k = \frac{1}{4\pi\epsilon_0}$, the potential at the surface and at the external point distant 3r are

$\text V_{\text{surface}} = \frac{\text{kQ}}{\text r}, \qquad \text V_{3\text r} = \frac{\text{kQ}}{3\text r}$

Therefore

$$\begin{aligned}\text V = \frac{\text{kQ}}{\text r} - \frac{\text{kQ}}{3\text r} \\[1em] \text V = \frac{\text{kQ}}{\text r}\left(1 - \frac{1}{3}\right) \\[1em] \text V = \frac{2\text{kQ}}{3\text r} \\[1em] \text{kQ} = \frac{3\text r\text V}{2}\end{aligned}$$

The electric intensity at the external point distant 3r is

$$\begin{aligned}\text E = \frac{\text{kQ}}{(3\text r)^2} \\[1em] \text E = \frac{1}{9\text r^2} \times \frac{3\text r\text V}{2} \\[1em] \text E = \frac{\text V}{6\text r}\end{aligned}$$

None of these

Reason — Statement (i) is false. The restoring force on the charge on the axis of the ring is proportional to the displacement, and hence the motion is simple harmonic, only when the displacement is very small compared with the radius (x << R). At a distance 2R the force is not proportional to the displacement, so the motion is oscillatory but not simple harmonic.

Statement (ii) is also false. A free positive charge moves from a region of higher potential to a region of lower potential, while a free negative charge moves in the opposite sense. Since electrons are negatively charged, they move from a region of lower potential to a region of higher potential.

Hence neither statement is true.

No work is done in this process.

Explanation:
The electrostatic force of attraction which furnishes the centripetal force always remains perpendicular to the direction of velocity of the electron. Alternatively, the circular orbit of the electron around the nucleus is an equipotential surface, so the potential difference between any two of its points is zero. Hence

$$\begin{aligned}\text W = \text q \times \Delta \text V \\[1em] \text W = 0 \qquad (\Delta \text V = 0)\end{aligned}$$

Yes, it can be assumed.

Explanation:
The potential at a point in an electric field is arbitrary and depends on the choice of the zero of reference. If the potential of the earth is taken as +100 volt, the measured value of the potential at every point will be increased by 100 volt. There will be no effect on the potential difference between two points, because the same constant is added at both points and cancels out in the difference. It is the potential difference between two points that matters and not the absolute potential.

The electric potential decreases along an electric line of force.

Explanation:
This follows from the relation $\text E = -\frac{\text{dV}}{\text{dr}}$, in which the negative sign signifies that the potential decreases in the direction of the electric field, that is, along a line of force.

The electron will move towards higher potential and the proton towards lower potential.

Explanation:
A free positive charge such as a proton moves from a region of higher potential to a region of lower potential, since the force qE on it acts along the field. A free negative charge such as an electron experiences a force opposite to the field and therefore moves from lower potential to higher potential.

It increases.

Explanation:
Both the charges are negative, so their product is positive and the potential energy $\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}$ is positive, indicating repulsion. As the separation r decreases, U increases; the work done by the external agent against the repulsion is stored as potential energy of the system.

(i) pE, (ii) 2pE, (iii) 0

Explanation:
The work done in rotating a dipole from the direction of the field through an angle θ is

$\text W = \text p\text E(1 - \cos\theta)$

(i) For θ = 90°, W = pE(1 − cos 90°) = pE(1 − 0) = pE

(ii) For θ = 180°, W = pE(1 − cos 180°) = pE[1 − (−1)] = 2pE

(iii) For θ = 360°, W = pE(1 − cos 360°) = pE(1 − 1) = zero

Zero.

Explanation:
The potential energy of a dipole placed at an angle θ to a uniform electric field is U_θ = − pE cos θ. For θ = 90°,

$$\begin{aligned}\text U_{90°} = -\text p\text E\cos 90° \\[1em] \text U_{90°} = 0\end{aligned}$$

(i) For a point charge, the electric field varies inversely as the square of the distance and the potential inversely as the distance,

$\text E \propto \frac{1}{\text r^2}, \qquad \text V \propto \frac{1}{\text r}$

(ii) The potential will not depend upon r. Inside a charged metallic sphere the electric field is zero, so there is no potential gradient and the potential remains constant, equal to its value at the surface.

The work done is zero in both cases.

Explanation:
The work done in taking a charge from one point to another is W = Q(V_A − V_B), where V_A and V_B are the potentials at the two points. As the charge Q returns to the same point after completing a revolution, V_A = V_B and hence W = 0.

When q is at the centre of the circle, the circle is itself an equipotential surface, so the potential is the same at all points on it and the work done is zero even for an incomplete revolution.

The work done is zero, irrespective of whether the orbit of the electron is circular or elliptical.

Explanation:
The electrostatic field is a conservative field. In a complete orbit the electron returns to its starting point, so the potential difference between the initial and the final position is zero, and hence W = q ΔV = 0.

The distance of the centroid from each corner of an equilateral triangle of side a is $\frac{\text a}{\sqrt3}$.

Since potential is a scalar quantity, the potentials due to the three charges add algebraically,

$$\begin{aligned}\text V = 3 \times \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text a/\sqrt3} \\[1em] \text V = \frac{1}{4\pi\epsilon_0}\frac{3\sqrt3\ \text q}{\text a}\end{aligned}$$

Hence, the electrostatic potential at the centroid is $\frac{1}{4\pi\epsilon_0}\frac{3\sqrt3\ \text q}{\text a}$ volt.

Given,

• q₁ = 5 μC, q₂ = 6 μC, q₃ = − 7 μC
• Radius of the circle, r = 1.0 m

All four charges lie on the circumference, so each is at the same distance r from the centre. As potential is a scalar,

$$\begin{aligned}\text V_\text o = \frac{1}{4\pi\epsilon_0}\frac{\text q_1 + \text q_2 + \text q_3 + \text q_4}{\text r} \\[1em] \text V_\text o = 0\end{aligned}$$

This requires

$$\begin{aligned}\text q_1 + \text q_2 + \text q_3 + \text q_4 = 0 \\[1em] 5 + 6 - 7 + \text q_4 = 0 \\[1em] \text q_4 = -4\ \mu\text C\end{aligned}$$

Hence, the charge q₄ is − 4 μC.

(i) Resultant field = zero, resultant potential = 2V; (ii) resultant field = 2E, resultant potential = zero.

Explanation:
(i) The two charges + q are on opposite sides of P and at equal distances from it. The two fields at P are equal in magnitude but opposite in direction, so the resultant field is zero. The potential is a scalar and the two contributions are equal and of the same sign, so the resultant potential is 2V.

(ii) If the new charge is − q, the two fields at P point the same way and add up to give a resultant field of 2E. The two potentials are equal in magnitude but opposite in sign, so the resultant potential is zero.

The electric field is zero in that region.

Explanation:
If V is constant everywhere, the potential gradient $\frac{\text{dV}}{\text{dr}}$ vanishes in every direction, and since

$\text E = -\frac{\text{dV}}{\text{dr}}$

the electric field intensity must be zero throughout that region.

No, the electric potential is continuous there.

Explanation:
The potential inside a charged conductor is constant and equal to its value at the surface, so there is no sudden jump in V as the surface is crossed. Only the field, which is the potential gradient, changes abruptly from zero inside to a finite value just outside.

No, no charge will flow to the inner sphere.

Explanation:
The potential at every point inside a charged spherical shell is the same as that at its surface. Hence the inner sphere and the shell are already at the same potential, and charge flows between two conductors only when there is a potential difference between them. The entire charge continues to reside on the outer surface of the shell.

A force field is said to be conservative if the work done by an external agent in taking a body between two points in the field depends only on the location of these two points and is independent of the path adopted in joining them.

An equivalent statement is that the work done in taking a body around any closed path in such a field is zero.

Examples : The electrostatic field and the gravitational field are conservative fields.

Electric potential : The minimum amount of work done by an external agent in carrying a unit positive test charge from infinity to a point in the electric field is called the electric potential at that point.

$\text V = \frac{\text W}{\text q_0}$

It is a scalar quantity. Its SI unit is the volt (V), where 1 volt = 1 joule/coulomb.

Equipotential surface : Any surface over which the electric potential is the same everywhere is called an equipotential surface.

For a uniform electric field the equipotential surfaces are a family of planes perpendicular to the lines of force.

In the figure the dotted lines are the cross-sections of these planar equipotential surfaces, drawn at right angles to the uniform field $\vec{\text E} \space$.

(i) Equidistant planes parallel to the x-y plane, since the field is uniform and directed along the z-axis.

(ii) Again planes parallel to the x-y plane, but they become closer together as the field increases in magnitude, because for the same potential interval $\text{dr} \propto \frac{1}{\text E}$.

(iii) Concentric spheres with their centre at the origin, since for a point charge $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r}$ has the same value at all points of a given radius.

(i) On the surface : E is the same everywhere and V is also the same everywhere, since the charge is uniformly distributed over a spherical surface.

(ii) Inside : E is zero at every interior point, and V is the same everywhere and equal to its value at the surface.

(iii) Outside : Both E and V decrease as one moves away from the balloon, E more rapidly than V, since $\text E \propto \frac{1}{\text r^2}$ while $\text V \propto \frac{1}{\text r}$.

Work will be done by some external agent in moving the charge against the electrostatic force.

This work will have no effect upon the field, since the source charges producing the field are unaltered.

However, the work done will be stored as potential energy in the field. Hence the energy of the system consisting of the source charge and the moving charge will increase.

The net work done along the closed rectangular path will be zero.

Suppose the breadth of the rectangular path is d. The force on the charge q is qE. The work done in taking the charge from 1 → 2 is qE × d, and that from 3 → 4 is − qE × d, so these cancel. No work is done in taking the charge from 2 → 3 and from 4 → 1, because along these paths the direction of the force acting on the charge is perpendicular to the displacement.

This result is consistent with the conservative nature of an electrostatic field.

(a) Zero, because the points A and B lie on the circumference of the circle whose centre carries the charge 5 C. The circumference is therefore an equipotential surface, and the potential at A due to the charge 5 C is the same as that at B. Hence

$$\begin{aligned}\text W = \text q(\text V_\text B - \text V_\text A) \\[1em] \text W = 0\end{aligned}$$

(b) Equal work will be done along both the paths. Since A, P and B all lie on the same circle, V_A = V_P = V_B. Therefore the potential difference between A and P is the same as that between P and B, both being zero, and the work done along each part of the path is zero.

Given,

• Two charges, each + q, at (a, 0) and (− a, 0)

At the origin : The distance of the origin from each charge is a. Since potential is a scalar,

$$\begin{aligned}\text V_\text O = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text a} + \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text a} \\[1em] \text V_\text O = \frac{1}{4\pi\epsilon_0}\frac{2\text q}{\text a}\ \text{volt}\end{aligned}$$

At the point (0, b) : The distance of this point from each charge is

$\text r = \sqrt{\text a^2 + \text b^2}$

Therefore

$$\begin{aligned}\text V = 2 \times \frac{1}{4\pi\epsilon_0}\frac{\text q}{\sqrt{\text a^2 + \text b^2}} \\[1em] \text V = \frac{1}{4\pi\epsilon_0}\frac{2\text q}{\sqrt{\text a^2 + \text b^2}}\ \text{volt}\end{aligned}$$

No, identical charges can produce zero field but not zero potential.

The fields due to identical charges can cancel vectorially at a suitable point, but the potential is a scalar and the contribution of every charge of the same sign has the same sign. Hence the potentials add up and cannot cancel.

However, positive and negative charges can be so arranged as to produce both zero field and zero potential at a point, since then the scalar contributions to the potential are of opposite signs and can also cancel.

No, zero potential at a point does not necessarily mean that the electric field is zero at that point.

For example, at a point midway between two equal and opposite charges, and at every point on the equatorial line of a dipole, the potential is zero but the electric field is certainly not zero.

No.

For example, the electric field is zero at a point exactly midway between two equal and similar charges, but the potential at that point is twice that due to a single charge. Similarly the field is zero everywhere inside a charged hollow conductor, while the potential there has the same non-zero value as at the surface.

Given,

• V_A = + V volt, V_B = − V volt
• Distance between the points, d = r metre

(i) Average electric field :

$$\begin{aligned}\text E = \frac{\text V_\text A - \text V_\text B}{\text r} \\[1em] \text E = \frac{\text V - (-\text V)}{\text r} \\[1em] \text E = \frac{2\text V}{\text r}\ \text{volt/metre}\end{aligned}$$

(ii) Change in potential energy :

$$\begin{aligned}\Delta \text U = \text q(\text V_\text B - \text V_\text A) \\[1em] \Delta \text U = \text q(-\text V - \text V) \\[1em] \Delta \text U = -2\text q\text V\end{aligned}$$

The potential energy therefore decreases by 2qV joule.

Note: The textbook answer states that the potential energy will increase by 2qV joule. For a positive charge q moving from the higher potential point A to the lower potential point B, the potential energy actually decreases by 2qV joule; the magnitude of the change is 2qV in either case.

No work is done in moving a charge from one point to another on an equipotential surface, because the potential difference between any two of its points is zero. This means that there is no potential gradient along any direction parallel to the surface. Taking ds as a small displacement along the surface, the component of the field parallel to the surface is

$$\begin{aligned}\text E_\text{parallel} = -\frac{\text{dV}}{\text{ds}} \\[1em] \text E_\text{parallel} = 0\end{aligned}$$

Hence $\vec{\text E} \space$, and with it the lines of force, must be normal to the equipotential surface.

Let m be the mass and e the magnitude of the charge of the electron.

The work done on the electron by the field in taking it through a potential difference V is eV, and since the electron starts from rest, this appears entirely as its kinetic energy,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text e\text V \\[1em] \text v^2 = \frac{2\text e\text V}{\text m} \\[1em] \text v = \sqrt{\frac{2\text e}{\text m}}\ \sqrt{\text V}\end{aligned}$$

For a given particle e and m are constants, so $\sqrt{\frac{2\text e}{\text m}}$ is a constant.

Hence $\text v \propto \sqrt{\text V}$.

Given,

• Both particles are accelerated through the same potential difference V
• m_α = 4m_p and q_α = 2q_p

For a charge q accelerated from rest through a potential difference V,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text q\text V \\[1em] \text v = \sqrt{\frac{2\text q\text V}{\text m}}\end{aligned}$$

Ratio of velocities :

$$\begin{aligned}\frac{\text v_\alpha}{\text v_\text p} = \sqrt{\frac{\text q_\alpha}{\text m_\alpha} \times \frac{\text m_\text p}{\text q_\text p}} \\[1em] \frac{\text v_\alpha}{\text v_\text p} = \sqrt{\frac{2\text q_\text p}{4\text m_\text p} \times \frac{\text m_\text p}{\text q_\text p}} \\[1em] \frac{\text v_\alpha}{\text v_\text p} = \sqrt{\frac{1}{2}} \\[1em] \frac{\text v_\alpha}{\text v_\text p} = \frac{1}{\sqrt2}\end{aligned}$$

Ratio of linear momenta : Since $\text p = \text m\text v = \sqrt{2\text{mqV}}$,

$$\begin{aligned}\frac{\text p_\alpha}{\text p_\text p} = \sqrt{\frac{\text m_\alpha \text q_\alpha}{\text m_\text p \text q_\text p}} \\[1em] \frac{\text p_\alpha}{\text p_\text p} = \sqrt{\frac{4\text m_\text p \times 2\text q_\text p}{\text m_\text p \text q_\text p}} \\[1em] \frac{\text p_\alpha}{\text p_\text p} = \sqrt{8} \\[1em] \frac{\text p_\alpha}{\text p_\text p} = 2\sqrt2\end{aligned}$$

Hence $\frac{\text v_\alpha}{\text v_\text p} = \frac{1}{\sqrt2}$ and $\frac{\text p_\alpha}{\text p_\text p} = 2\sqrt2$.

The dipole is held along the Z-axis with its centre at the origin. Therefore the X-axis is the equatorial line of the dipole, and the potential at all points of this line is zero.

Both the initial point (5, 0, 0) and the final point (− 7, 0, 0) lie on the X-axis, so

$$\begin{aligned}\text V_1 = \text V_2 = 0 \\[1em] \text W = \text q_0(\text V_2 - \text V_1) \\[1em] \text W = 0\end{aligned}$$

Hence, no work is done.

The answer will remain zero even if the path of the test charge is changed, because the work done in an electrostatic field between two given points is independent of the path connecting them.

All points in the space inside a charged metallic sphere are at the same electric potential, because the electric field there is zero everywhere. As there is no potential difference between any two interior points, no work is done in moving a charge between them.

Outside the sphere, however, the potential varies as $\text V \propto \frac{1}{\text r}$, so there is a potential gradient and hence an electric field at every point. Moving a positive charge towards the positively charged sphere means moving it towards a region of higher potential, against the force of repulsion. Work must therefore be done by an external agent.

No.

Given, q = 1 C and R = 1 cm = 0.01 m. The electric field on the surface of the sphere would be

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2} \\[1em] \text E = \frac{(9\times10^9)(1)}{(0.01)^2} \\[1em] \text E = 9\times10^{13}\ \text N\text C^{-1}\ (\text{or}\ \text V\text m^{-1})\end{aligned}$$

The dielectric strength of air is about 3 × 10⁶ V m^-1. If the intensity of the electric field in air is greater than this value, the air will be ionised and there will be leakage of charge from the sphere into the air.

Hence the sphere will not be able to hold a charge of 1 coulomb.

Let a point charge + Q coulomb be held at a point O in vacuum, and let P be the point, at a distance r from O, where the electric potential is required. The potential at P is the work done per unit charge in bringing a test charge + q₀ from infinity to P, so this work has to be found first.

Consider the test charge + q₀ at an intermediate point A, at a distance x from O on the far side of P. According to Coulomb's law, the electric force on q₀ has the magnitude

$\text F = \frac{1}{4\pi\epsilon_0}\frac{\text Q\text q_0}{\text x^2}\ \text{newton} \qquad \dots(\text i)$

This force $\vec{\text F} \space$points away from O. When the test charge is moved through a small distance dx to the neighbouring point B, its displacement is opposite to the force, which is why a negative sign appears in the work done,

$\text{dW} = \text F(-\text{dx})$

Adding up all such elements from x = ∞ to x = r gives the total work done in bringing + q₀ up to the point P,

$$\begin{aligned}\text W = -\displaystyle\int_\infty^\text r \text F\ \text{dx} \\[1em] \text W = -\frac{\text Q\text q_0}{4\pi\epsilon_0}\displaystyle\int_\infty^\text r \frac{1}{\text x^2}\ \text{dx} \\[1em] \text W = -\frac{\text Q\text q_0}{4\pi\epsilon_0}\left[-\frac{1}{\text x}\right]_\infty^\text r \\[1em] \text W = -\frac{\text Q\text q_0}{4\pi\epsilon_0}\left(-\frac{1}{\text r} + \frac{1}{\infty}\right) \\[1em] \text W = \frac{1}{4\pi\epsilon_0}\frac{\text Q\text q_0}{\text r}\end{aligned}$$

Since the electric potential is the work done per unit test charge,

$$\begin{aligned}\text V = \frac{\text W}{\text q_0} \\[1em] \text V = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r}\ \text{volt}\end{aligned}$$

Hence, the electric potential at a point distant r from a point charge Q is $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r}$.

Potential difference : The work done by an external agent in carrying a unit positive test charge from one point to the other point in an electric field is called the potential difference between those points.

If W is the work done in carrying a test charge q₀ from B to A, then

$\text V_\text A - \text V_\text B = \frac{\text W}{\text q_0}$

Its SI unit is the volt and it is a scalar quantity.

Relation between electric field and potential difference :

Take a point charge + q at a point O, producing an electric field E along the X-axis. Let A and B be two neighbouring points on this axis at distances r and r + dr from O, the separation dr being vanishingly small. Let V and V + dV be the electric potentials at A and B respectively, so that dV = V_B − V_A is the change in potential over the distance dr.

A small positive test charge q₀ is now carried from B to A. The field exerts on it a force F along the direction of the field, where

$\text F = \text q_0\text E \qquad \dots(\text i)$

In moving q₀ from B to A, an external agent has to work against the force F, so it applies an equal force directed towards A. The displacement from B to A, of magnitude dr, is in the direction of this applied force, so the work done by the agent is positive. If this work is dW, then, using Eq. (i),

$$\begin{aligned}\text{dW} = \text F\ \text{dr} \\[1em] \text{dW} = \text q_0\text E\ \text{dr} \\[1em] \frac{\text{dW}}{\text q_0} = \text E\ \text{dr} \qquad \dots(\text{ii})\end{aligned}$$

From the definition of potential difference, the work done per unit charge in going from B to A is V_A − V_B, so

$$\begin{aligned}\frac{\text{dW}}{\text q_0} = \text V_\text A - \text V_\text B \\[1em] \frac{\text{dW}}{\text q_0} = \text V - (\text V + \text{dV}) \\[1em] \frac{\text{dW}}{\text q_0} = -\text{dV} \qquad \dots(\text{iii})\end{aligned}$$

Equating the right-hand sides of Eqs. (ii) and (iii),

$$\begin{aligned}-\text E\ \text{dr} = \text{dV} \\[1em] \text E = -\frac{\text{dV}}{\text{dr}}\end{aligned}$$

Here $\frac{\text{dV}}{\text{dr}}$, the change of potential per unit distance, is called the potential gradient. The relation therefore shows that the electric field intensity in any given direction equals the negative of the potential gradient in that direction, the negative sign indicating that the potential falls as one moves along the electric field.

The relation also shows that the unit of electric field may be written as volt/metre, so that 1 N-C^-1 = 1 V-m^-1.

Equipotential surface : Any surface over which the electric potential is the same everywhere is called an equipotential surface.

Equipotential surfaces for a point charge : An isolated point charge + q produces at a distance r the potential

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r}$

Since V depends only on r, every point of a sphere of radius r centred on + q is at the same potential $\frac{\text q}{4\pi\epsilon_0\text r}$, so such a sphere is an equipotential surface. The equipotential surfaces of a point charge are therefore concentric spheres centred on the charge, the potential of each being inversely proportional to its radius r.

Electric field is perpendicular to an equipotential surface : All points of an equipotential surface are at the same potential, so carrying a charge between any two of them requires no work. Consequently the potential does not change along any direction lying in the surface, and the component of the field in such a direction, being the negative potential gradient, is zero,

$$\begin{aligned}\text E_{\text{parallel}} = -\frac{\text{dV}}{\text{dr}} \\[1em] \text E_{\text{parallel}} = 0\end{aligned}$$

With no component along the surface, the electric field $\vec{\text E} \space$, and with it every line of force, must cross the equipotential surface at right angles. For a point charge this is seen directly: the radial lines of force cut each of the concentric spherical equipotential surfaces normally.

Electric potential energy : The electric potential energy of a system of charges is the work done in bringing those charges from infinity to their positions near each other in the system.

Expression for two point charges :

Consider two charges + q₁ and + q₂ coulomb in vacuum, at the points A and B, a distance r metre apart, forming the system AB.

To find the energy of this system, imagine that + q₁ is already at A while + q₂ is at infinity. The charge + q₁ sets up at B the electric potential

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q_1}{\text r}\ \text{volt}$

The work needed to bring q₂ from infinity to B is the product of the charge and this potential (= charge × potential),

$$\begin{aligned}\text W = \text q_2\text V \\[1em] \text W = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}\ \text{joule}\end{aligned}$$

This work is stored as the electric potential energy U of the system,

$\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}\ \text{joule}$

The charges q₁ and q₂ are substituted in this formula along with their signs. U therefore comes out positive for like charges, showing net repulsion, and negative for unlike charges, showing net attraction.

Generalisation for n point charges : For a system of n charges the potential energy is found pair by pair. There are $\frac{\text n(\text n-1)}{2}$ distinct pairs; the energy of each pair is obtained from the two-charge formula and all the terms are added algebraically,

$\text U = \frac{1}{4\pi\epsilon_0}\left[\frac{\text q_1\text q_2}{\text r_{12}} + \frac{\text q_1\text q_3}{\text r_{13}} + \dots + \frac{\text q_{\text n-1}\text q_\text n}{\text r_{(\text n-1)\text n}}\right]$

the sum running over all the $\frac{\text n(\text n-1)}{2}$ pairs.

For instance, three charges q₁, q₂ and q₃ at the corners of a triangle have the potential energy

$\text U = \frac{1}{4\pi\epsilon_0}\left[\frac{\text q_1\text q_2}{\text r_{12}} + \frac{\text q_2\text q_3}{\text r_{23}} + \frac{\text q_1\text q_3}{\text r_{13}}\right]$

A dipole in a uniform electric field rests in equilibrium with its moment along the field, and turning it away from this position requires work. Let a dipole of moment p (= q × 2l) be turned through an angle θ from its equilibrium position in a uniform electric field E. The couple acting on the dipole does not stay constant but changes as the dipole turns.

When the dipole is at some angle θ to the field E, the torque on it is

$\tau = \text p\text E\sin\theta$

To turn it from this position through a further, infinitesimally small angle dθ, the work required is

$$\begin{aligned}\text{dW} = \text{torque} \times \text{angular displacement} \\[1em] \text{dW} = \tau\ \text d\theta \\[1em] \text{dW} = \text p\text E\sin\theta\ \text d\theta\end{aligned}$$

Integrating between the limits θ₁ and θ₂ gives the work done in turning the dipole from θ₁ to θ₂,

$$\begin{aligned}\text W = \displaystyle\int_{\theta_1}^{\theta_2}\text p\text E\sin\theta\ \text d\theta \\[1em] \text W = \text p\text E\left[-\cos\theta\right]_{\theta_1}^{\theta_2} \\[1em] \text W = \text p\text E\left[-\cos\theta_2 + \cos\theta_1\right] \\[1em] \text W = \text p\text E(\cos\theta_1 - \cos\theta_2)\end{aligned}$$

For a rotation that starts from the equilibrium position (θ₁ = 0) and ends at an angle θ with the field,

$\text W = \text p\text E(1 - \cos\theta)$

Potential energy of a point charge in an external field : Let V be the potential at a point P arising from an external electric field. If a charge q is brought slowly from infinity to P, the work done by the external agent is

$$\begin{aligned}\text W = \text q\ \Delta \text V \\[1em] \text W = \text q(\text V_\text P - \text V_\infty) \\[1em] \text W = \text q(\text V - 0) \\[1em] \text W = \text q\text V\end{aligned}$$

This work is stored as the electrostatic potential energy of the charge q at the point P in the external field, U = qV.

Potential energy of a dipole in a uniform electric field :

The potential energy of a dipole in an electric field equals the work done in bringing it from infinity to its position inside the field.

Imagine the dipole being carried in from infinity into a uniform electric field E, with its dipole moment p kept along the field throughout. The field pushes the charge + q along E with a force F (= qE) and pulls the charge − q the opposite way with an equal force F (= qE). So, while the dipole is brought in, an external agent does work on + q, whereas on − q the work is done by the field itself.

In the course of bringing the dipole in, the charge − q covers a distance 2l more than the charge + q, so the work associated with − q is the larger. The 'net' work done in bringing the dipole from infinity into the field is therefore

$$\begin{aligned}\text W = \text{force on charge}\ (-\text q) \times \text{additional distance moved} \\[1em] \text W = -\text q\text E \times 2\text l \\[1em] \text W = -\text p\text E\end{aligned}$$

where p (= q × 2l) is the moment of the electric dipole. This net work is the potential energy U₀ of the dipole lying parallel to the field,

$\text U_0 = -\text p\text E$

and in this orientation the dipole is in stable equilibrium in the field.

If the dipole is next turned through an angle θ in the field, the work that must be done on it is

$\text W = \text p\text E(1 - \cos\theta)$

and this work raises its potential energy. In the orientation θ, therefore,

$$\begin{aligned}\text U_\theta = \text U_0 + \text W \\[1em] \text U_\theta = -\text p\text E + \text p\text E(1 - \cos\theta) \\[1em] \text U_\theta = -\text p\text E\cos\theta\end{aligned}$$

which in vector notation reads

$\text U = -\vec{\text p}\cdot\vec{\text E}$

Consider an electric dipole AB made up of the charges − q and + q coulomb, a small distance 2l metre apart in vacuum. Let P lie on the axis of the dipole at a distance r metre from its mid-point O.

The point P is at a distance (r − l) from the charge + q and (r + l) from the charge − q. The charge + q alone gives at P the potential

$\text V_1 = \frac{1}{4\pi\epsilon_0}\frac{\text q}{(\text r - \text l)}$

while the charge − q gives

$\text V_2 = -\frac{1}{4\pi\epsilon_0}\frac{\text q}{(\text r + \text l)}$

As electric potential is a scalar quantity, the net potential V at P is simply the algebraic sum of V₁ and V₂,

$$\begin{aligned}\text V = \text V_1 + \text V_2 \\[1em] \text V = \frac{\text q}{4\pi\epsilon_0}\left[\frac{1}{(\text r - \text l)} - \frac{1}{(\text r + \text l)}\right] \\[1em] \text V = \frac{\text q}{4\pi\epsilon_0}\left[\frac{(\text r + \text l) - (\text r - \text l)}{\text r^2 - \text l^2}\right] \\[1em] \text V = \frac{\text q}{4\pi\epsilon_0}\frac{2\text l}{\text r^2 - \text l^2} \\[1em] \text V = \frac{1}{4\pi\epsilon_0}\frac{2\text q\text l}{\text r^2 - \text l^2}\end{aligned}$$

Writing 2ql = p, the electric dipole moment,

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2 - \text l^2}$

For a distant point, r >> 2l, the term l² is negligible compared with r², and the potential at P on the axis of the dipole becomes

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2}\ \text{volt}$

(i) The response of the student is correct.

The potential of a charged sphere of radius R is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R}$

For a given sphere V is proportional to q, so adding charge to the sphere raises its potential.

(ii) The response is correct.

The earth is taken to be at zero potential. A conductor connected to the earth comes to the potential of the earth, so earthing brings the potential of the sphere down to zero.

(iii) The response is incorrect.

At any point inside a charged spherical shell the potential has the same value as on its surface. The inner sphere A therefore receives the full surface potential of B in addition to its own, while at B the charge on A acts as if placed at the centre. Taking $\text k = \frac{1}{4\pi\epsilon_0}$,

$\text V_\text A = \text k\left(\frac{\text q_1}{\text R_1} + \frac{\text q_2}{\text R_2}\right), \qquad \text V_\text B = \text k\left(\frac{\text q_2}{\text R_2} + \frac{\text q_1}{\text R_2}\right)$

Since R₂ > R₁,

$\frac{\text q_1}{\text R_1} \gt \frac{\text q_1}{\text R_2} \quad \Rightarrow \quad \text V_\text A \gt \text V_\text B$

Thus it is the inner sphere A, not the sphere B, that is at the higher electric potential.

Given,

• Radius of the inner shell = r, radius of the outer shell = R, with R > r
• Charge on the outer shell = q, inner shell initially uncharged and then grounded

(a) Charge on the inner shell :

Let the charge induced on the grounded inner shell be q'. Since the inner shell is earthed, its potential must be zero. The potential of the inner shell is the sum of the potential due to its own charge q' and that due to the charge q on the outer shell (which is the same at all interior points),

$$\begin{aligned}\text V_{\text{inner}} = \frac{1}{4\pi\epsilon_0}\left(\frac{\text q'}{\text r} + \frac{\text q}{\text R}\right) \\[1em] \frac{1}{4\pi\epsilon_0}\left(\frac{\text q'}{\text r} + \frac{\text q}{\text R}\right) = 0 \\[1em] \frac{\text q'}{\text r} = -\frac{\text q}{\text R} \\[1em] \text q' = -\text q\left(\frac{\text r}{\text R}\right)\end{aligned}$$

(b) Potential of the outer sphere :

The potential of the outer shell is due to its own charge q and the charge q' on the inner shell, both treated as being at the common centre,

$$\begin{aligned}\text V_{\text{outer}} = \frac{1}{4\pi\epsilon_0}\left(\frac{\text q}{\text R} + \frac{\text q'}{\text R}\right) \\[1em] \text V_{\text{outer}} = \frac{1}{4\pi\epsilon_0\text R}\left(\text q - \frac{\text{qr}}{\text R}\right) \\[1em] \text V_{\text{outer}} = \frac{\text q(\text R - \text r)}{4\pi\epsilon_0\text R^2}\end{aligned}$$

Hence, the charge on the inner shell is $-\text q\left(\frac{\text r}{\text R}\right)$ and the potential of the outer sphere is $\frac{\text q(\text R - \text r)}{4\pi\epsilon_0\text R^2}$.

Electric potential : The work done by an external agent in carrying a unit positive test charge from infinity to a point in the electric field is called the electric potential at that point,

$\text V = \frac{\text W}{\text q_0}$

SI unit : volt (V), where 1 volt = 1 joule/coulomb. Thus, if 1 joule of work is done in carrying a test charge of 1 coulomb from infinity to a point in an electric field, then the potential at that point will be 1 volt.

Dimensional formula :

$$\begin{aligned}\text{Dimensions of electric potential} = \frac{\text{dimensions of work}}{\text{dimensions of charge}} \\[1em] \text{Dimensions of electric potential} = \frac{[\text M\text L^2\text T^{-2}]}{[\text A\text T]} \\[1em] \text{Dimensions of electric potential} = [\text M\text L^2\text T^{-3}\text A^{-1}]\end{aligned}$$

Nature : Electric potential is a scalar quantity.

Example of V = 0 but E ≠ 0 : At every point on the equatorial line of an electric dipole, the potentials due to + q and − q are equal and opposite, so V = 0. The fields due to the two charges, however, add up and the resultant electric field is not zero.

Example of V ≠ 0 but E = 0 : At every point inside a charged hollow metallic sphere the electric field is zero, but the potential is constant and equal to the non-zero value $\frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R}$ at the surface.

Method of transferring the whole charge : The charged hollow sphere is placed inside the larger hollow sphere and the two are connected by a conducting wire. The charge given to a conductor always resides on its outer surface. Hence the entire charge of the inner sphere flows through the wire to the outer surface of the larger sphere, and the inner sphere is left completely uncharged.

The transfer is complete no matter how high the potential of the outer sphere already is, because the potential of the inner sphere always exceeds that of the outer one so long as it carries any charge. This is the principle used in the Van de Graaff generator to build up very high potentials.

Potential difference : The work done by an external agent in carrying a unit positive test charge from one point to the other point in an electric field is called the potential difference between those points,

$\text V_\text A - \text V_\text B = \frac{\text W}{\text q_0}$

Unit : volt (V), the same as that of potential.

Value on an equipotential surface : By definition the potential is the same at every point of an equipotential surface. Hence the potential difference between any two points on it, however far apart they are, is zero.

Electric potential : The work done by an external agent in carrying a unit positive test charge from infinity to a point in the electric field is called the electric potential at that point. It is a scalar quantity and its SI unit is the volt.

Relation between E and V for a point charge :

Take a point charge + q at O whose electric field is directed along the X-axis. Let A and B be neighbouring points on the axis at distances r and r + dr from O, with dr vanishingly small, and let V and V + dV be the electric potentials at A and B respectively, so that dV = V_B − V_A is the change in potential over the distance dr.

A small positive test charge q₀ is taken from B to A. The field acts on it with the force

$\text F = \text q_0\text E \qquad \dots(\text i)$

In moving q₀ from B to A the external agent works against this force, applying an equal force directed towards A. The displacement from B to A, of magnitude dr, is along this applied force, so the work done by the agent is positive. If the work done is dW, then

$$\begin{aligned}\text{dW} = \text F\ \text{dr} \\[1em] \text{dW} = \text q_0\text E\ \text{dr} \\[1em] \frac{\text{dW}}{\text q_0} = \text E\ \text{dr} \qquad \dots(\text{ii})\end{aligned}$$

The work done per unit charge from B to A is, by definition, the potential difference V_A − V_B, so

$$\begin{aligned}\frac{\text{dW}}{\text q_0} = \text V_\text A - \text V_\text B \\[1em] \frac{\text{dW}}{\text q_0} = \text V - (\text V + \text{dV}) \\[1em] \frac{\text{dW}}{\text q_0} = -\text{dV} \qquad \dots(\text{iii})\end{aligned}$$

Equating the two expressions for $\frac{\text{dW}}{\text q_0}$ in Eqs. (ii) and (iii),

$\text E = -\frac{\text{dV}}{\text{dr}}$

Electric field is the gradient of electric potential : The quantity $\frac{\text{dV}}{\text{dr}}$ measures how fast the potential changes with distance and is called the potential gradient. The relation $\text E = -\frac{\text{dV}}{\text{dr}}$ thus expresses the electric field intensity in a given direction as the negative of the potential gradient in that direction; the negative sign shows that the potential falls along the direction of the electric field.

As a check, apply this to a point charge q, whose potential at a distance r is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r}$

The field of a point charge is radial by symmetry, so its magnitude is the negative potential gradient taken along r,

$$\begin{aligned}\text E = -\frac{\text{dV}}{\text{dr}} \\[1em] \text E = -\frac{\text d}{\text{dr}}\left(\frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r}\right) \\[1em] \text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2}\end{aligned}$$

This is the known expression for the intensity due to a point charge. If the potential changes along all three axes, V = f(x, y, z), its gradient is written as

$\text{gradient}\ \text V = \hat{\text i}\frac{\partial \text V}{\partial \text x} + \hat{\text j}\frac{\partial \text V}{\partial \text y} + \hat{\text k}\frac{\partial \text V}{\partial \text z}$

Electric potential energy : The electric potential energy of a system of two point charges is the work done in bringing the two charges from infinity to their given separation.

Expression :

Let the charges + q₁ and + q₂ coulomb be at the points A and B in vacuum, r metre apart, forming the system AB. Assume that + q₁ is fixed at A and that + q₂ starts at infinity.

At B the charge + q₁ produces the electric potential

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q_1}{\text r}\ \text{volt}$

Using the definition of electric potential, the work done (= charge × potential) in bringing q₂ from infinity to B is

$$\begin{aligned}\text W = \text q_2\text V \\[1em] \text W = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}\ \text{joule}\end{aligned}$$

This work is stored as the electric potential energy U of the system,

$\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}\ \text{joule}$

In using this formula, q₁ and q₂ are substituted with their signs.

Effect of reducing the distance :

For two similar charges the product q₁q₂ is positive, so U is positive and indicates net repulsion. As r is reduced, U increases; work has to be done by an external agent against the repulsion, and this is stored as potential energy.

For two opposite charges the product q₁q₂ is negative, so U is negative and indicates net attraction. As r is reduced, U becomes more and more negative, that is, it decreases; the charges attract each other, so the field itself does the work and the potential energy of the system falls.

Work done in rotating an electric dipole : Consider a dipole of moment p (= q × 2l) in a uniform electric field E, turned through an angle θ away from its equilibrium position. As it turns, the couple acting on it keeps changing.

At an instant when the dipole makes an angle θ with the field E, the torque on it is

$\tau = \text p\text E\sin\theta$

Turning it through a further, infinitesimally small angle dθ needs the work

$$\begin{aligned}\text{dW} = \tau\ \text d\theta \\[1em] \text{dW} = \text p\text E\sin\theta\ \text d\theta\end{aligned}$$

The work done in turning the dipole from the angle θ₁ to θ₂ is found by integrating,

$$\begin{aligned}\text W = \displaystyle\int_{\theta_1}^{\theta_2}\text p\text E\sin\theta\ \text d\theta \\[1em] \text W = \text p\text E\left[-\cos\theta\right]_{\theta_1}^{\theta_2} \\[1em] \text W = \text p\text E(\cos\theta_1 - \cos\theta_2)\end{aligned}$$

When the rotation begins at the equilibrium position (θ₁ = 0) and ends at an angle θ with the field,

$\text W = \text p\text E(1 - \cos\theta)$

Total electric potential energy of the dipole : A dipole lying parallel to the field has the potential energy U₀ = − pE. Turning it through an angle θ requires the work W found above, which adds to this energy, so

$$\begin{aligned}\text U_\theta = \text U_0 + \text W \\[1em] \text U_\theta = -\text p\text E + \text p\text E(1 - \cos\theta) \\[1em] \text U_\theta = -\text p\text E\cos\theta \qquad \text{or} \qquad \text U = -\vec{\text p}\cdot\vec{\text E}\end{aligned}$$

Stable equilibrium : This occurs when the dipole moment $\vec{\text p}$ points along the electric intensity $\vec{\text E} \space$, that is θ = 0. Then

$$\begin{aligned}\text U_{0°} = -\text p\text E\cos 0° \\[1em] \text U_{0°} = -\text p\text E \qquad (\text{least})\end{aligned}$$

Since the potential energy has its minimum value here, the dipole is in stable equilibrium.

Most unstable equilibrium : This occurs when the dipole has been turned through 180° from the stable position, that is θ = 180°. Then

$$\begin{aligned}\text U_{180°} = -\text p\text E\cos 180° \\[1em] \text U_{180°} = +\text p\text E \qquad (\text{greatest})\end{aligned}$$

Since the potential energy has its maximum value here, the dipole is in unstable equilibrium.

(i) The potential at a general point due to a short dipole is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p\cos\theta}{\text r^2}$

Comparing this with the given expression, n = 2.

(ii) The potential is zero wherever cos θ = 0, that is θ = 90°. Hence the locus is a straight line passing through the mid-point of the dipole and perpendicular to the dipole axis, that is, the equatorial line of the dipole.

(iii) On the equatorial line the potential is zero at every point, so the potential difference between the point at r = 5 cm and that at r = 3 cm is zero.

$$\begin{aligned}\text W = \text q\ \Delta \text V \\[1em] \text W = (4\times10^{-6}) \times 0 \\[1em] \text W = 0\end{aligned}$$

Hence, the work done is zero.

(iv) After rotation through the angle θ, the point P lies on the axis of the dipole. Hence θ = 0 and cos θ = 1, giving

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2} \\[1em] \text V = \frac{1}{4\pi\epsilon_0}\frac{\text q(2\text l)}{\text r^2}\end{aligned}$$

(v) In stable equilibrium the dipole moment is along the field, so θ = 0 and

$$\begin{aligned}\text U = -\text p\text E\cos\theta \\[1em] \text U = -\text p\text E \\[1em] \text U = -\text q(2\text l)\text E\end{aligned}$$

(i) Ionisation, electrostatic attraction and electric discharge.

(ii) The discharge electrodes are fed from a high DC voltage source, so the potential difference V between the mesh and the plates stays fixed when the dielectric is introduced. Since E = V/d and the separation d is unchanged, the electric field between the plates does not change. The force on the charged particles therefore remains the same, and on this basis the efficiency of the device does not decrease.

Note: The textbook states that the field between the plates reduces and the efficiency decreases. That holds only for isolated plates carrying a fixed charge, where E falls to E/K; it does not apply here, because the passage states that the electrodes are kept at a fixed DC voltage.

(iii)

Given,

• Potential difference, V = 1.50 × 10⁴ V
• Separation between the plates, d = 2 cm = 2 × 10^-2 m

For two parallel plates the field between them is uniform, so

$$\begin{aligned}\text E = \frac{\text V}{\text d} \\[1em] \text E = \frac{1.50\times10^4}{2\times10^{-2}} \\[1em] \text E = 7.5\times10^5\ \text N/\text C\end{aligned}$$

(iv) Air purifier or electrostatic air cleaner.

Given,

• The line for Q₁ makes an angle of 60° with the $\frac{1}{\text r}$ axis
• The line for Q₂ makes an angle of 30° with the $\frac{1}{\text r}$ axis

The potential at a distance r due to a point charge Q is

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r} \\[1em] \text V = \left(\frac{\text Q}{4\pi\epsilon_0}\right)\frac{1}{\text r}\end{aligned}$$

This is a straight line through the origin when V is plotted against $\frac{1}{\text r}$, whose slope is $\frac{\text Q}{4\pi\epsilon_0}$. Hence the slope is directly proportional to the charge,

$$\begin{aligned}\frac{\text Q_1}{\text Q_2} = \frac{\tan 60°}{\tan 30°} \\[1em] = \frac{\sqrt3}{\frac{1}{\sqrt3}} \\[1em] = \sqrt3 \times \sqrt3 \\[1em] \frac{\text Q_1}{\text Q_2} = 3\end{aligned}$$

Hence, $\frac{\text Q_1}{\text Q_2}$ = 3 : 1.

Given,

• Charge, q₀ = 1.3 × 10^-7 C
• Work done, W = 6.5 × 10^-5 J

By the definition of electric potential,

$$\begin{aligned}\text V = \frac{\text W}{\text q_0} \\[1em] = \frac{6.5\times10^{-5}}{1.3\times10^{-7}} \\[1em] = 5 \times 10^2 \\[1em] \text V = 500\ \text V\end{aligned}$$

Hence, the electric potential at the point is 500 V.

Given,

• Work done, W = 5 J
• Charge, q₀ = 0.5 C

By the definition of potential difference,

$$\begin{aligned}\text V_\text A - \text V_\text B = \frac{\text W}{\text q_0} \\[1em] = \frac{5\ \text J}{0.5\ \text C} \\[1em] \text V_\text A - \text V_\text B = 10\ \text V\end{aligned}$$

Hence, the potential difference between the points is 10 V.

Given,

• Radius of the circle, r = 10 cm = 0.10 m
• Charge on the wire, q = 250 μC = 250 × 10^-6 C

Electric potential at the centre : Every element of the charged wire is at the same distance r from the centre, and potential is a scalar quantity. Hence

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] = \frac{(9.0\times10^9)(250\times10^{-6})}{0.10} \\[1em] = \frac{2.25\times10^6}{0.10} \\[1em] \text V = 2.25\times10^7\ \text V\end{aligned}$$

Intensity of electric field at the centre : The field is a vector quantity. For every small element of the ring there is a diametrically opposite element which produces an equal and opposite field at the centre. Hence all the contributions cancel in pairs and

$\text E = 0$

Hence, the electric potential at the centre is 2.25 × 10⁷ V and the intensity of the electric field is zero.

Given,

• Charge at O, q = + 10 μμC = 10 × 10^-12 C
• OA = 0.10 m, OB = 0.20 m

(i) Electric potentials at A and B :

$$\begin{aligned}\text V_\text A = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text{OA}} \\[1em] = \frac{(9.0\times10^9)(10\times10^{-12})}{0.10} \\[1em] = \frac{9\times10^{-2}}{0.10} \\[1em] \text V_\text A = 0.90\ \text V \\[1em] \text V_\text B = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text{OB}} \\[1em] = \frac{(9.0\times10^9)(10\times10^{-12})}{0.20} \\[1em] = \frac{9\times10^{-2}}{0.20} \\[1em] \text V_\text B = 0.45\ \text V\end{aligned}$$

(ii) Work done in taking a charge of − 10 μμC from A to B :

$$\begin{aligned}\text W = \text q_0(\text V_\text B - \text V_\text A) \\[1em] = (-10\times10^{-12})(0.45 - 0.90) \\[1em] = (-10\times10^{-12})(-0.45) \\[1em] \text W = 4.5\times10^{-12}\ \text J\end{aligned}$$

Hence, V_A = 0.90 V, V_B = 0.45 V and the work done is 4.5 × 10^-12 J.

Given,

• q_A = 8.0 μC = 8.0 × 10^-6 C, q_B = − 2.0 μC = − 2.0 × 10^-6 C
• AB = 1.00 m, so AC = CB = 0.50 m
• AD = 0.80 m, DB = 0.20 m
• Charge to be moved, q₀ = 0.02 μC = 0.02 × 10^-6 C

Potential at the mid-point C : Potential is a scalar, so the contributions add algebraically,

$$\begin{aligned}\text V_\text C = \frac{1}{4\pi\epsilon_0}\left[\frac{\text q_\text A}{\text{AC}} + \frac{\text q_\text B}{\text{CB}}\right] \\[1em] = (9.0\times10^9)\left[\frac{8.0\times10^{-6}}{0.50} + \frac{-2.0\times10^{-6}}{0.50}\right] \\[1em] = (9.0\times10^9)\left[\frac{6.0\times10^{-6}}{0.50}\right] \\[1em] = (9.0\times10^9)(1.2\times10^{-5}) \\[1em] \text V_\text C = 1.08\times10^5\ \text V\end{aligned}$$

Potential at D :

$$\begin{aligned}\text V_\text D = (9.0\times10^9)\left[\frac{8.0\times10^{-6}}{0.80} + \frac{-2.0\times10^{-6}}{0.20}\right] \\[1em] = (9.0\times10^9)\left[1.0\times10^{-5} - 1.0\times10^{-5}\right] \\[1em] \text V_\text D = 0\end{aligned}$$

Work done in taking 0.02 μC from D to C :

$$\begin{aligned}\text W = \text q_0(\text V_\text C - \text V_\text D) \\[1em] = (0.02\times10^{-6})(1.08\times10^5 - 0) \\[1em] \text W = 2.16\times10^{-3}\ \text J\end{aligned}$$

Hence, V_C = 1.08 × 10⁵ V, V_D = zero and the work done is 2.16 × 10^-3 J.

Given,

• q₁ = + 20 μC = 20 × 10^-6 C, q₂ = − 10 μC = − 10 × 10^-6 C
• Separation between the charges = 6 cm, so half the separation = 3 cm = 0.03 m
• Distance of the point from the mid-point along the right bisector = 4 cm = 0.04 m

The point on the right bisector is equidistant from both the charges,

$$\begin{aligned}\text r = \sqrt{(0.03)^2 + (0.04)^2} \\[1em] = \sqrt{0.0009 + 0.0016} \\[1em] = \sqrt{0.0025} \\[1em] \text r = 0.05\ \text m\end{aligned}$$

Since potential is a scalar,

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q_1 + \text q_2}{\text r} \\[1em] = \frac{(9.0\times10^9)(20\times10^{-6} - 10\times10^{-6})}{0.05} \\[1em] = \frac{(9.0\times10^9)(10\times10^{-6})}{0.05} \\[1em] = \frac{9\times10^4}{0.05} \\[1em] \text V = 1.8\times10^6\ \text V\end{aligned}$$

Hence, the electric potential at the point is 1.8 × 10⁶ V.

Given,

• V = 10³ V at r₁ = 0.3 m
• Distance at which E is required, r₂ = 0.4 m

For a point charge,

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r_1} \\[1em] \Rightarrow \quad \frac{\text q}{4\pi\epsilon_0} = \text V\text r_1 \\[1em] = 10^3 \times 0.3 \\[1em] \frac{\text q}{4\pi\epsilon_0} = 300\ \text N\text{-m}^2\text C^{-1}\end{aligned}$$

The electric field intensity at a distance r₂ is

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r_2^2} \\[1em] = \frac{300}{(0.4)^2} \\[1em] = \frac{300}{0.16} \\[1em] \text E = 1875\ \text V\text{-m}^{-1}\end{aligned}$$

Hence, the electric field intensity at 0.4 m is 1875 V-m^-1.

Given,

• E = 20 N-C^-1
• V = 10 J-C^-1

For a point charge,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2}, \qquad \text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r}$

Dividing the second by the first,

$$\begin{aligned}\frac{\text V}{\text E} = \text r \\[1em] \Rightarrow \quad \text r = \frac{10}{20} \\[1em] \text r = 0.5\ \text m\end{aligned}$$

The magnitude of the charge is

$$\begin{aligned}\text q = \frac{\text V\text r}{\left(\frac{1}{4\pi\epsilon_0}\right)} \\[1em] = \frac{10 \times 0.5}{9.0\times10^9} \\[1em] \text q = 0.55\times10^{-9}\ \text C\end{aligned}$$

Hence, the point is 0.5 m from the charge and the magnitude of the charge is 0.55 × 10^-9 C.

Given,

• Source charge, q = 6.0 × 10^-8 C at the origin
• Initial position of the electron, r₁ = 3 m; final position, r₂ = 6 m
• Charge on the electron, q₀ = − 1.6 × 10^-19 C

The potentials at the two points are

$$\begin{aligned}\text V_1 = \frac{(9.0\times10^9)(6.0\times10^{-8})}{3} \\[1em] = \frac{540}{3} \\[1em] \text V_1 = 180\ \text V \\[1em] \text V_2 = \frac{(9.0\times10^9)(6.0\times10^{-8})}{6} \\[1em] = \frac{540}{6} \\[1em] \text V_2 = 90\ \text V\end{aligned}$$

The work done is

$$\begin{aligned}\text W = \text q_0(\text V_2 - \text V_1) \\[1em] = (-1.6\times10^{-19})(90 - 180) \\[1em] = (-1.6\times10^{-19})(-90) \\[1em] \text W = 1.44\times10^{-17}\ \text J\end{aligned}$$

Hence, the work done is 1.44 × 10^-17 J.

Given,

• Charge at each corner, q = 250 μC = 250 × 10^-6 C
• Distance of the centroid from each corner, r = 18 cm = 0.18 m

Potential is a scalar quantity, so the potentials due to the three charges add algebraically,

$$\begin{aligned}\text V = 3 \times \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] = 3 \times \frac{(9.0\times10^9)(250\times10^{-6})}{0.18} \\[1em] = 3 \times \frac{2.25\times10^6}{0.18} \\[1em] = 3 \times 1.25\times10^7 \\[1em] \text V = 3.75\times10^7\ \text V\end{aligned}$$

Hence, the resultant electric potential at the centroid is 3.75 × 10⁷ V.

Given,

• Charge on the electron, e = 1.6 × 10^-19 C
• Potential difference, V = 1000 V

The kinetic energy gained by a charge accelerated from rest through a potential difference V is

$$\begin{aligned}\text K = \text q\text V \\[1em] = (1.6\times10^{-19})(1000) \\[1em] \text K = 1.6\times10^{-16}\ \text J\end{aligned}$$

Hence, the kinetic energy gained by the electron is 1.6 × 10^-16 J.

Given,

• Potential difference, V = 10⁶ V
• Charge on α-particle = 3.2 × 10^-19 C, mass = 6.6 × 10^-27 kg
• Charge on proton = 1.6 × 10^-19 C, mass = 1.7 × 10^-27 kg

(i) Kinetic energy of the α-particle :

$$\begin{aligned}\text K_\alpha = \text q_\alpha\text V \\[1em] = (3.2\times10^{-19})(10^6) \\[1em] \text K_\alpha = 3.2\times10^{-13}\ \text J\end{aligned}$$

(ii) Kinetic energy of the proton :

$$\begin{aligned}\text K_\text p = \text q_\text p\text V \\[1em] = (1.6\times10^{-19})(10^6) \\[1em] \text K_\text p = 1.6\times10^{-13}\ \text J\end{aligned}$$

(iii) Comparison of velocities : Since $\text K = \frac{1}{2}\text m\text v^2$,

$$\begin{aligned}\text v = \sqrt{\frac{2\text K}{\text m}} \\[1em] \frac{\text v_\alpha}{\text v_\text p} = \sqrt{\frac{\text K_\alpha}{\text K_\text p} \times \frac{\text m_\text p}{\text m_\alpha}} \\[1em] = \sqrt{\frac{3.2\times10^{-13}}{1.6\times10^{-13}} \times \frac{1.7\times10^{-27}}{6.6\times10^{-27}}} \\[1em] = \sqrt{2 \times 0.258} \\[1em] = \sqrt{0.515} \\[1em] \frac{\text v_\alpha}{\text v_\text p} = 0.72\end{aligned}$$

Since this ratio is less than 1, v_p > v_α.

Hence, K_α = 3.2 × 10^-13 J, K_p = 1.6 × 10^-13 J and the proton will gain more velocity.

Given,

• The uniform field E is directed along the positive X-axis
• A is the point (6, 0) and C is the point (2, 0)

(i) Potential difference between A and C :

The electrostatic field is conservative, so the work done depends only on the initial and final positions and not on the path A → B → C. Since C and A both lie on the X-axis, only the displacement along the field matters,

$$\begin{aligned}\text{AC} = 6 - 2 \\[1em] \text{AC} = 4\ \text{units}\end{aligned}$$

In a uniform electric field the potential difference between two points separated by a distance d along the field is

$$\begin{aligned}\text V_\text C - \text V_\text A = \text E \times \text{AC} \\[1em] \text V_\text C - \text V_\text A = 4\text E\end{aligned}$$

(ii) The electric potential is more at C, because the potential always decreases in the direction of the electric field. Since the field points from C towards A, C is at the higher potential.

The potential at a distance r due to a point charge q is

$\text V = \left(\frac{\text q}{4\pi\epsilon_0}\right)\frac{1}{\text r}$

which is a straight line through the origin of slope $\frac{\text q}{4\pi\epsilon_0}$ when V is plotted against $\frac{1}{\text r}$.

(i) The line for q₁ has a positive slope and the line for q₂ has a negative slope. Hence q₁ is positive and q₂ is negative.

(ii) The magnitude of the slope of a line is proportional to the magnitude of the charge. From the graph the angle β made by the line of q₂ is greater than the angle α made by the line of q₁, so the magnitude of the slope for q₂ is greater.

Hence |q₂| > |q₁|, that is, q₂ has the larger magnitude.

Given,

• E = 10r + 5 N C^-1
• r₁ = 1 m, r₂ = 10 m

The change in potential is obtained from

$$\begin{aligned}\text{dV} = -\text E\ \text{dr} \\[1em] \text V_2 - \text V_1 = -\displaystyle\int_{1}^{10}(10\text r + 5)\ \text{dr}\end{aligned}$$

Integrating term by term,

$$\begin{aligned}= -\left[\frac{10\text r^2}{2} + 5\text r\right]_{1}^{10} \\[1em] = -\left[5\text r^2 + 5\text r\right]_{1}^{10}\end{aligned}$$

Evaluating the limits,

$$\begin{aligned}= -\left[\left(5(10)^2 + 5(10)\right) - \left(5(1)^2 + 5(1)\right)\right] \\[1em] = -\left[(500 + 50) - (5 + 5)\right] \\[1em] = -\left[550 - 10\right] \\[1em] \text V_2 - \text V_1 = -540\ \text V\end{aligned}$$

Hence, the change in electric potential is − 540 V; the potential decreases by 540 V.

The electric field is the negative slope of the V-x graph,

$\text E = -\frac{\text{dV}}{\text{dx}}$

From x = 0 to x = 1 m : V rises uniformly from 0 to 100 V, so the slope is $\frac{100 - 0}{1 - 0} = 100$ V m^-1 and

$\text E = -100\ \text V\text{-m}^{-1}$

From x = 1 m to x = 2 m : V is constant at 100 V, so the slope is zero and

$\text E = 0$

From x = 2 m to x = 3 m : V falls uniformly from 100 V to 0, so the slope is $\frac{0 - 100}{3 - 2} = -100$ V m^-1 and

$\text E = +100\ \text V\text{-m}^{-1}$

The graph of E against x is therefore a step graph, with E = − 100 V-m^-1 for 0 < x < 1, E = 0 for 1 < x < 2 and E = + 100 V-m^-1 for 2 < x < 3.

The point x = 2.5 m lies in the third region.

Hence, the electric field strength at x = 2.5 m is 100 V-m^-1.

Given, V = 3x² volt.

Since V depends only on x, the electric field has only an x-component,

$\vec{\text E} = -\frac{\text{dV}}{\text{dx}}\hat{\text i}$

Differentiating,

$$\begin{aligned}\frac{\text{dV}}{\text{dx}} = \frac{\text d}{\text{dx}}(3\text x^2) \\[1em] \frac{\text{dV}}{\text{dx}} = 6\text x\end{aligned}$$

Therefore

$\vec{\text E} = -6\text x\ \hat{\text i}$

At the point (1 m, 0, 2 m), x = 1 m,

$\vec{\text E} = -6\ \hat{\text i}\ \text V\text{-m}^{-1}$

Hence, the electric field at the given point is 6 V-m^-1, directed along the negative x-axis.

Note: The printed answer gives 6 V-m^-1 along the x-axis. Since $\vec{\text E} = -\frac{\text{dV}}{\text{dx}}\hat{\text i}$, the field is 6 V-m^-1 along the negative x-axis.

Given,

$$\begin{aligned}\text V = \text x(\text y^2 - 5\text x) \\[1em] \text V = \text x\text y^2 - 5\text x^2\end{aligned}$$

The Cartesian components of the field are the negative partial derivatives of V,

$\text E_\text x = -\frac{\partial \text V}{\partial \text x}, \qquad \text E_\text y = -\frac{\partial \text V}{\partial \text y}, \qquad \text E_\text z = -\frac{\partial \text V}{\partial \text z}$

x-component : Differentiating partially with respect to x, treating y as constant,

$$\begin{aligned}\frac{\partial \text V}{\partial \text x} = \text y^2 - 10\text x \\[1em] \Rightarrow \quad \text E_\text x = -(\text y^2 - 10\text x) \\[1em] \text E_\text x = 10\text x - \text y^2\end{aligned}$$

y-component : Differentiating partially with respect to y, treating x as constant,

$$\begin{aligned}\frac{\partial \text V}{\partial \text y} = 2\text x\text y \\[1em] \Rightarrow \quad \text E_\text y = -2\text x\text y\end{aligned}$$

z-component : V does not contain z, so

$$\begin{aligned}\frac{\partial \text V}{\partial \text z} = 0 \\[1em] \Rightarrow \quad \text E_\text z = 0\end{aligned}$$

Hence, E_x = 10x − y², E_y = − 2xy and E_z = 0.

Given,

• The equipotential surfaces are of 30 V, 40 V and 50 V
• They cut the x-axis at x = 20 cm, 30 cm and 40 cm, each making an angle of 30° with the x-axis

The distance between two successive surfaces measured along the x-axis is 10 cm. The perpendicular distance between them is therefore

$$\begin{aligned}\text d = 10\sin 30° \\[1em] = 10 \times \frac{1}{2} \\[1em] = 5\ \text{cm} \\[1em] \text d = 0.05\ \text m\end{aligned}$$

The potential difference between two successive surfaces is

$$\begin{aligned}\Delta \text V = 40 - 30 \\[1em] \Delta \text V = 10\ \text V\end{aligned}$$

Hence the magnitude of the electric field is

$$\begin{aligned}\text E = \frac{\Delta \text V}{\text d} \\[1em] = \frac{10}{0.05} \\[1em] \text E = 200\ \text V\text{-m}^{-1}\end{aligned}$$

The field is normal to the equipotential surfaces and directed from the higher potential surface towards the lower potential surface, that is, towards the region of decreasing x.

Hence, the magnitude of the electric field is 200 V-m^-1, making an angle of 120° with the x-axis.

Given,

• $\vec{\text E} = \text C\text x\hat{\text i}$ with C = 10 V/m²
• V = 0 at the point (10 m, 20 m)

The field has only an x-component, so only the x-coordinate matters. Using

$$\begin{aligned}\text{dV} = -\text E_\text x\ \text{dx} \\[1em] \displaystyle\int_{0}^{\text V}\text{dV} = -\displaystyle\int_{\text x = 10}^{\text x = 0}\text C\text x\ \text{dx}\end{aligned}$$

Integrating,

$$\begin{aligned}\text V = -\text C\left[\frac{\text x^2}{2}\right]_{10}^{0} \\[1em] = -10\left(\frac{0^2}{2} - \frac{10^2}{2}\right) \\[1em] = -10(0 - 50) \\[1em] \text V = 500\ \text V\end{aligned}$$

Hence, the potential at the origin is 500 V.

Given,

• Uniform electric field, E = 10 N/C directed horizontally
• The triangle ABC is equilateral with each side 2 m, B and C at the base and A at the apex

In a uniform electric field only the displacement measured along the direction of the field is effective,

$\text V_\text A - \text V_\text B = -\text E(\text x_\text A - \text x_\text B)$

where x_A and x_B are the positions of A and B measured along the direction of the field. Since the triangle is equilateral with side 2 m, the foot of the perpendicular from A to BC is at the mid-point of BC, so the horizontal separation between B and A is

$$\begin{aligned}\text x_\text A - \text x_\text B = \frac{2}{2} \\[1em] \text x_\text A - \text x_\text B = 1\ \text m\end{aligned}$$

Moving from A to B is a displacement of 1 m opposite to the field direction, hence

$$\begin{aligned}\text V_\text A - \text V_\text B = -\text E(\text x_\text A - \text x_\text B) \\[1em] = -10 \times 1 \\[1em] \text V_\text A - \text V_\text B = -10\ \text V\end{aligned}$$

Hence, V_A − V_B = − 10 V.

Given,

• Work done, W = 10 J
• Charge, q = − 200 C
• Separation between the points, d = 4 cm = 0.04 m

(i) and (ii) Potential difference :

$$\begin{aligned}\text V_\text B - \text V_\text A = \frac{\text W}{\text q} \\[1em] = \frac{10}{-200} \\[1em] \text V_\text B - \text V_\text A = -0.05\ \text V\end{aligned}$$

Since V_B − V_A is negative, V_A > V_B. Hence A is at the higher potential and the magnitude of the potential difference is 0.05 V.

(iii) Component of the electric field along AB :

$$\begin{aligned}\text E = \frac{\text V_\text A - \text V_\text B}{\text d} \\[1em] = \frac{0.05}{0.04} \\[1em] \text E = 1.25\ \text V\text{-m}^{-1}\end{aligned}$$

Hence, A is at the higher potential, the PD is 0.05 V and the component of the field parallel to AB is 1.25 V-m^-1.

Given,

• Separation between the plates, d = 2 cm = 2 × 10^-2 m
• Potential difference, V = 1000 V
• Charge on the proton, q = 1.6 × 10^-19 C

(i) Intensity of the electric field : The plates are long and close together, so the field between them is uniform,

$$\begin{aligned}\text E = \frac{\text V}{\text d} \\[1em] = \frac{1000}{2\times10^{-2}} \\[1em] \text E = 50000\ \text V\text{-m}^{-1}\end{aligned}$$

(ii) Force on the proton :

$$\begin{aligned}\text F = \text q\text E \\[1em] = (1.6\times10^{-19})(5\times10^4) \\[1em] \text F = 8.0\times10^{-15}\ \text N\end{aligned}$$

Since the electric field between the plates is uniform at all places, the proton will experience the same force at each place between the plates.

Hence, E = 50000 V-m^-1 and F = 8.0 × 10^-15 N.

Given,

• Separation between the plates, d = 2.0 cm = 2.0 × 10^-2 m
• Potential difference, V = 120 V
• Charge on the electron, e = 1.6 × 10^-19 C

(i) Electric field between the plates :

$$\begin{aligned}\text E = \frac{\text V}{\text d} \\[1em] = \frac{120}{2.0\times10^{-2}} \\[1em] \text E = 6.0\times10^3\ \text V\text{-m}^{-1}\end{aligned}$$

(ii) Force on the electron :

$$\begin{aligned}\text F = \text e\text E \\[1em] = (1.6\times10^{-19})(6.0\times10^3) \\[1em] \text F = 9.6\times10^{-16}\ \text N\end{aligned}$$

(iii) Increase in electrical energy :

$$\begin{aligned}\text W = \text e\text V \\[1em] = (1.6\times10^{-19})(120) \\[1em] \text W = 1.92\times10^{-17}\ \text J\end{aligned}$$

Hence, E = 6.0 × 10³ V-m^-1, F = 9.6 × 10^-16 N and the increase in electrical energy is 1.92 × 10^-17 J.

Given,

• Mutual potential energy, U = 2.4 × 10⁵ eV
• Charge on each proton, e = 1.6 × 10^-19 C

Converting the energy into joule,

$$\begin{aligned}\text U = (2.4\times10^5)(1.6\times10^{-19}) \\[1em] \text U = 3.84\times10^{-14}\ \text J\end{aligned}$$

The potential energy of the two-proton system is

$$\begin{aligned}\text U = \frac{1}{4\pi\epsilon_0}\frac{\text e^2}{\text r} \\[1em] \Rightarrow \quad \text r = \frac{1}{4\pi\epsilon_0}\frac{\text e^2}{\text U}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text r = \frac{(9.0\times10^9)(1.6\times10^{-19})^2}{3.84\times10^{-14}} \\[1em] = \frac{(9.0\times10^9)(2.56\times10^{-38})}{3.84\times10^{-14}} \\[1em] = \frac{2.304\times10^{-28}}{3.84\times10^{-14}} \\[1em] \text r = 6.0\times10^{-15}\ \text m\end{aligned}$$

Hence, the distance between the protons is 6.0 × 10^-15 m.

Given,

• Radius of the orbit, r = 5.3 × 10^-11 m
• Charge on the hydrogen nucleus = + 1.6 × 10^-19 C (Z = 1)
• Charge on the electron = − 1.6 × 10^-19 C

Electric potential at this radius :

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text{Ze}}{\text r} \\[1em] = \frac{(9.0\times10^9)(1.6\times10^{-19})}{5.3\times10^{-11}} \\[1em] = \frac{1.44\times10^{-9}}{5.3\times10^{-11}} \\[1em] \text V = 27.2\ \text V\end{aligned}$$

Electric potential energy of the atom :

$$\begin{aligned}\text U = (-\text e)\text V \\[1em] = (-1.6\times10^{-19}\ \text C)(27.2\ \text V) \\[1em] \text U = -4.35\times10^{-18}\ \text J\end{aligned}$$

Expressed in electron-volt, since U = charge × potential and the charge is one electronic charge,

$\text U = -27.2\ \text{eV}$

Potential due to a helium nucleus at the same radius : For helium Z = 2, so the charge on the nucleus is doubled and

$$\begin{aligned}\text V' = 2 \times 27.2 \\[1em] \text V' = 54.4\ \text V\end{aligned}$$

Hence, V = 27.2 V, U = − 27.2 eV and the potential due to a helium nucleus is 54.4 V.

Given,

• q₁ = − 1 μC = − 1 × 10^-6 C, q₂ = + 2 μC = 2 × 10^-6 C, q₃ = − 1 μC = − 1 × 10^-6 C
• r₁₂ = 1 m, r₂₃ = 1 m, and hence r₁₃ = 2 m

The number of pairs is $\frac{\text n(\text n - 1)}{2} = \frac{3 \times 2}{2} = 3$. The electrostatic energy of the system is the algebraic sum of the potential energies of all the pairs,

$\text U = \frac{1}{4\pi\epsilon_0}\left[\frac{\text q_1\text q_2}{\text r_{12}} + \frac{\text q_2\text q_3}{\text r_{23}} + \frac{\text q_1\text q_3}{\text r_{13}}\right]$

Substituting the values,

$$\begin{aligned}= (9.0\times10^9)\Bigg[\frac{(-1\times10^{-6})(2\times10^{-6})}{1} \\[1em] + \frac{(2\times10^{-6})(-1\times10^{-6})}{1} \\[1em] + \frac{(-1\times10^{-6})(-1\times10^{-6})}{2}\Bigg] \\[1em] = (9.0\times10^9)\left[-2\times10^{-12} - 2\times10^{-12} + 0.5\times10^{-12}\right] \\[1em] = (9.0\times10^9)(-3.5\times10^{-12}) \\[1em] \text U = -3.15\times10^{-2}\ \text J\end{aligned}$$

Hence, the electrostatic energy of the system is − 31.5 mJ.

Given,

• Charge on the up quark, $\text q_\text u = \frac{2}{3}\text e$
• Charge on each down quark, $\text q_\text d = -\frac{1}{3}\text e$
• Side of the triangle, a = 10^-15 m

There are three pairs of quarks, all separated by the same distance a. Hence

$\text U = \frac{1}{4\pi\epsilon_0 \text a}\left[\text q_\text u\text q_{\text d1} + \text q_\text u\text q_{\text d2} + \text q_{\text d1}\text q_{\text d2}\right]$

Substituting the quark charges,

$$\begin{aligned}= \frac{\text e^2}{4\pi\epsilon_0 \text a}\left[\left(\frac{2}{3}\right)\left(-\frac{1}{3}\right) + \left(\frac{2}{3}\right)\left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\right] \\[1em] = \frac{\text e^2}{4\pi\epsilon_0 \text a}\left[-\frac{2}{9} - \frac{2}{9} + \frac{1}{9}\right] \\[1em] \text U = -\frac{1}{3}\frac{\text e^2}{4\pi\epsilon_0 \text a}\end{aligned}$$

Substituting the numerical values,

$$\begin{aligned}\text U = -\frac{1}{3} \times \frac{(9.0\times10^9)(1.6\times10^{-19})^2}{10^{-15}} \\[1em] = -\frac{1}{3} \times \frac{2.304\times10^{-28}}{10^{-15}} \\[1em] = -\frac{2.304\times10^{-13}}{3} \\[1em] \text U = -7.68\times10^{-14}\ \text J\end{aligned}$$

Converting into MeV, using 1 MeV = 1.6 × 10^-13 J,

$$\begin{aligned}\text U = -\frac{7.68\times10^{-14}}{1.6\times10^{-13}} \\[1em] \text U = -0.48\ \text{MeV}\end{aligned}$$

Hence, the electrostatic potential energy of the neutron is − 0.48 MeV.

Given,

• Charges of 4 μC, 9 μC and 36 μC on a line AB of length 3 m

For minimum potential energy, the two charges of largest magnitude must be farthest apart. Hence 9 μC is placed at A and 36 μC at B, and 4 μC is placed in between, at a distance x metre from A and (3 − x) metre from B.

The potential energy of the system is

$\text U = \text k\left[\frac{9 \times 4}{\text x} + \frac{4 \times 36}{(3 - \text x)} + \frac{9 \times 36}{3}\right]\times10^{-12}$

where $\text k = \frac{1}{4\pi\epsilon_0}$.

For U to be minimum, $\frac{\text{dU}}{\text{dx}} = 0$. Differentiating,

$$\begin{aligned}\frac{\text{dU}}{\text{dx}} = \text k\left[-\frac{36}{\text x^2} + \frac{144}{(3 - \text x)^2}\right]\times10^{-12} = 0 \\[1em] \Rightarrow \quad \frac{144}{(3 - \text x)^2} = \frac{36}{\text x^2} \\[1em] \Rightarrow \quad 4\text x^2 = (3 - \text x)^2\end{aligned}$$

Taking the square root,

$$\begin{aligned}2\text x = 3 - \text x \\[1em] \Rightarrow \quad 3\text x = 3 \\[1em] \Rightarrow \quad \text x = 1\ \text m\end{aligned}$$

Hence, 9 μC is placed at A, 36 μC at B, and 4 μC is placed on AB at a distance of 1 m from the 9 μC charge (that is, 2 m from the 36 μC charge).

Given, a square of side a with charges of magnitude q at its corners, the diagonal being $\text a\sqrt2$.

(i) Figure (a) : Here the charges + q and − q alternate around the square, so the four sides join unlike charges and the two diagonals join like charges,

$$\begin{aligned}\text U_\text a = \frac{1}{4\pi\epsilon_0}\left[4\left(\frac{-\text q^2}{\text a}\right) + 2\left(\frac{+\text q^2}{\text a\sqrt2}\right)\right] \\[1em] = \frac{\text q^2}{4\pi\epsilon_0\text a}\left[-4 + \frac{2}{\sqrt2}\right] \\[1em] = \frac{\text q^2}{4\pi\epsilon_0\text a}\left[-4 + \sqrt2\right] \\[1em] \text U_\text a = -\frac{\text q^2}{4\pi\epsilon_0\text a}(4 - \sqrt2)\end{aligned}$$

Figure (b) : Here the two charges + q lie on one side of the square and the two charges − q on the other. Of the four sides, two join like charges and two join unlike charges, so their contributions cancel. Both diagonals join unlike charges,

$$\begin{aligned}\text U_\text b = \frac{1}{4\pi\epsilon_0}\left[\left(\frac{2\text q^2}{\text a} - \frac{2\text q^2}{\text a}\right) + 2\left(\frac{-\text q^2}{\text a\sqrt2}\right)\right] \\[1em] = -\frac{\text q^2}{4\pi\epsilon_0\text a}\left(\frac{2}{\sqrt2}\right) \\[1em] \text U_\text b = -\frac{\text q^2}{4\pi\epsilon_0\text a}(\sqrt2)\end{aligned}$$

(ii) All the four corners are equidistant from the centre of the square, and the algebraic sum of the four charges is zero in each case. Since potential is a scalar,

$$\begin{aligned}\text V_{\text{centre}} = \frac{1}{4\pi\epsilon_0}\frac{(\text q - \text q + \text q - \text q)}{\text a/\sqrt2} \\[1em] \text V_{\text{centre}} = 0\end{aligned}$$

Therefore the extra work needed is

$$\begin{aligned}\text W = \text q_0 \text V_{\text{centre}} \\[1em] \text W = 0\end{aligned}$$

Hence, $\text U_\text a = -\frac{\text q^2}{4\pi\epsilon_0\text a}(4 - \sqrt2)$, $\text U_\text b = -\frac{\text q^2}{4\pi\epsilon_0\text a}(\sqrt2)$ and no extra work is needed in bringing q₀ to the centre.

Given,

• Separation between the protons, r = 1.0 × 10^-10 m
• Charge on each proton, e = 1.6 × 10^-19 C

The potential energy of the two-proton system is

$$\begin{aligned}\text U = \frac{1}{4\pi\epsilon_0}\frac{\text e^2}{\text r} \\[1em] = \frac{(9.0\times10^9)(1.6\times10^{-19})^2}{1.0\times10^{-10}} \\[1em] = \frac{2.304\times10^{-28}}{1.0\times10^{-10}} \\[1em] \text U = 2.304\times10^{-18}\ \text J\end{aligned}$$

When the protons are far apart their potential energy becomes zero, so this entire energy appears as kinetic energy.

Case 1 — both protons free : By symmetry the two protons have equal masses and equal charges, so the potential energy is equally divided between them,

$$\begin{aligned}\text K = \frac{2.304\times10^{-18}}{2} \\[1em] \text K = 1.15\times10^{-18}\ \text J\end{aligned}$$

Case 2 — one proton fixed : The whole of the potential energy appears as the kinetic energy of the freed proton,

$\text K = 2.30\times10^{-18}\ \text J$

Hence, each proton acquires 1.15 × 10^-18 J in the first case, and the freed proton acquires 2.30 × 10^-18 J in the second case.

Given,

• Speed of each electron, v = 10⁶ m s^-1
• Mass of electron, m = 9.0 × 10^-31 kg
• Charge on electron, e = 1.6 × 10^-19 C

At the distance of closest approach both electrons come momentarily to rest, so the whole of the initial kinetic energy is converted into the potential energy of the system.

Total initial kinetic energy,

$$\begin{aligned}\text K = 2 \times \frac{1}{2}\text m\text v^2 \\[1em] = \text m\text v^2 \\[1em] = (9.0\times10^{-31})(10^6)^2 \\[1em] \text K = 9.0\times10^{-19}\ \text J\end{aligned}$$

At the closest approach r,

$$\begin{aligned}\frac{1}{4\pi\epsilon_0}\frac{\text e^2}{\text r} = \text K \\[1em] \text r = \frac{(9.0\times10^9)(1.6\times10^{-19})^2}{9.0\times10^{-19}} \\[1em] = \frac{2.304\times10^{-28}}{9.0\times10^{-19}} \\[1em] \text r = 2.56\times10^{-10}\ \text m\end{aligned}$$

Hence, the closest approach between the electrons is 2.56 × 10^-10 m.

Given,

• Charge, q = 1.0 × 10^-6 C
• Separation, 2l = 2.0 cm = 2.0 × 10^-2 m
• Electric field, E = 1.0 × 10⁵ V-m^-1

The electric dipole moment is

$$\begin{aligned}\text p = \text q \times 2\text l \\[1em] = (1.0\times10^{-6})(2.0\times10^{-2}) \\[1em] \text p = 2.0\times10^{-8}\ \text C\text{-m}\end{aligned}$$

Maximum torque : The torque on a dipole is τ = pE sin θ, which is maximum when θ = 90°,

$$\begin{aligned}\tau_{\text{max}} = \text p\text E \\[1em] = (2.0\times10^{-8})(1.0\times10^5) \\[1em] \tau_{\text{max}} = 2.0\times10^{-3}\ \text N\text{-m}\end{aligned}$$

Work done in rotating it through 180° :

$$\begin{aligned}\text W = \text p\text E(1 - \cos 180°) \\[1em] = 2\text p\text E \\[1em] = 2 \times 2.0\times10^{-3} \\[1em] \text W = 4.0\times10^{-3}\ \text J\end{aligned}$$

Hence, the maximum torque is 2.0 × 10^-3 N-m and the work done is 4.0 × 10^-3 J.

Given,

• Length of the dipole, 2l = 2 cm = 2 × 10^-2 m
• Angle with the field, θ = 60°
• Electric field, E = 10⁵ N-C^-1
• Torque, $\tau = 8\sqrt3$ N-m

(i) Magnitude of the charge : The torque on a dipole is

$$\begin{aligned}\tau = \text p\text E\sin\theta \\[1em] 8\sqrt3 = \text p \times 10^5 \times \sin 60° \\[1em] = \text p \times 10^5 \times \frac{\sqrt3}{2} \\[1em] \Rightarrow \quad \text p = \frac{8\sqrt3 \times 2}{\sqrt3 \times 10^5} \\[1em] = \frac{16}{10^5} \\[1em] \text p = 1.6\times10^{-4}\ \text C\text{-m}\end{aligned}$$

Since p = q × 2l,

$$\begin{aligned}\text q = \frac{\text p}{2\text l} \\[1em] = \frac{1.6\times10^{-4}}{2\times10^{-2}} \\[1em] \text q = 8\times10^{-3}\ \text C\end{aligned}$$

(ii) Potential energy of the dipole :

$$\begin{aligned}\text U = -\text p\text E\cos\theta \\[1em] = -(1.6\times10^{-4})(10^5)\cos 60° \\[1em] = -(1.6\times10^{-4})(10^5)(0.5) \\[1em] \text U = -8\ \text J\end{aligned}$$

Hence, the charge on the dipole is 8 × 10^-3 C and its potential energy is − 8 J.

Given,

• Charge, q = 3 μC = 3 × 10^-6 C
• Separation, 2l = 3 × 10^-3 m
• Distance of the point, r = 0.6 m

The electric dipole moment is

$$\begin{aligned}\text p = \text q \times 2\text l \\[1em] = (3\times10^{-6})(3\times10^{-3}) \\[1em] \text p = 9\times10^{-9}\ \text C\text{-m}\end{aligned}$$

(i) Broad-side-on (equatorial) position : A point on the equatorial line is equidistant from both the charges, so the potentials due to + q and − q are equal and opposite,

$\text V = 0$

(ii) After rotating the dipole through 90° : The same point now lies on the axis of the dipole (end-on position), so

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2} \\[1em] = \frac{(9.0\times10^9)(9\times10^{-9})}{(0.6)^2} \\[1em] = \frac{81}{0.36} \\[1em] \text V = 225\ \text V\end{aligned}$$

Hence, the potential is zero in the broad-side-on position and 225 V after the dipole is rotated through 90°.

Given,

• Dipole moment of one molecule, p = 10^-30 C-m
• Electric field, E = 10⁷ V m^-1
• Change in the direction of the field, θ = 60°
• Number of molecules in one mole, N = 6 × 10²³

Initially every dipole is aligned along the field, that is at θ₁ = 0. When the field direction is turned through 60°, each dipole momentarily makes an angle θ₁ = 60° with the new direction and then realigns itself to θ₂ = 0.

The energy released by one molecule in realigning is

$$\begin{aligned}\text W = \text p\text E(\cos\theta_1 - \cos\theta_2) \\[1em] = \text p\text E(\cos 60° - \cos 0°)\end{aligned}$$

Taking the magnitude of the energy given out,

$$\begin{aligned}|\text W| = \text p\text E(1 - \cos 60°) \\[1em] = (10^{-30})(10^7)(1 - 0.5) \\[1em] = (10^{-23})(0.5) \\[1em] |\text W| = 5\times10^{-24}\ \text J\end{aligned}$$

For one mole the total energy released is

$$\begin{aligned}\text Q = \text N|\text W| \\[1em] = (6\times10^{23})(5\times10^{-24}) \\[1em] \text Q = 3\ \text J\end{aligned}$$

Hence, the heat released by the substance is 3 J.

Note: The printed answer gives −3 J, which is the change in potential energy. The heat released is its magnitude, 3 J.

Given,

• Radius of the nucleus, R = 3.4 × 10^-14 m
• Atomic number, Z = 47
• Charge on proton, e = 1.6 × 10^-19 C

The charge on the nucleus is

$$\begin{aligned}\text q = \text{Ze} \\[1em] = 47 \times 1.6\times10^{-19} \\[1em] \text q = 7.52\times10^{-18}\ \text C\end{aligned}$$

The potential at the surface of the nucleus is

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R} \\[1em] = \frac{(9.0\times10^9)(7.52\times10^{-18})}{3.4\times10^{-14}} \\[1em] = \frac{6.768\times10^{-8}}{3.4\times10^{-14}} \\[1em] \text V = 2.0\times10^6\ \text V\end{aligned}$$

Hence, the electric potential at the surface of the silver nucleus is 2.0 × 10⁶ V.

Given,

• Radius of the conductor, R = 12 cm = 0.12 m
• Charge, q = 2.4 × 10^-9 C

(i) Potential on the surface :

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R} \\[1em] = \frac{(9.0\times10^9)(2.4\times10^{-9})}{0.12} \\[1em] = \frac{21.6}{0.12} \\[1em] \text V = 180\ \text V\end{aligned}$$

(ii) Intensity of the electric field at 4 m : For an external point the charged sphere behaves as if the whole charge were placed at its centre,

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2} \\[1em] = \frac{(9.0\times10^9)(2.4\times10^{-9})}{(4)^2} \\[1em] = \frac{21.6}{16} \\[1em] \text E = 1.35\ \text V\text{-m}^{-1}\end{aligned}$$

Hence, the potential on the surface is 180 V and the intensity of the field at 4 m is 1.35 V-m^-1.

Given,

• Radius of the sphere, R = 12 cm = 0.12 m
• Potential at r = 6.0 cm (a point inside the sphere) is 100 V

Potential at the surface : The point at 6.0 cm lies inside the hollow sphere, where the electric field is zero and therefore the potential is constant and equal to its value at the surface. Hence

$\text V_{\text{surface}} = 100\ \text V$

From this,

$$\begin{aligned}\frac{1}{4\pi\epsilon_0}\text q = \text V_{\text{surface}} \times \text R \\[1em] = 100 \times 0.12 \\[1em] \frac{1}{4\pi\epsilon_0}\text q = 12\ \text N\text{-m}^2\text C^{-1}\end{aligned}$$

Potential at 20 cm from the centre : This point is outside the sphere, so

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] = \frac{12}{0.20} \\[1em] \text V = 60\ \text V\end{aligned}$$

Hence, the potential at the surface is 100 V and that at 20 cm from the centre is 60 V.

Given,

• Radius of the sphere = 2.0 cm
• V = 400 V at r₁ = 10 cm = 0.10 m
• Distance at which E is required, r₂ = 20 cm = 0.20 m

Both points lie outside the sphere, so the sphere behaves as if the whole charge were at its centre,

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r_1} \\[1em] \Rightarrow \quad \frac{\text q}{4\pi\epsilon_0} = \text V\text r_1 \\[1em] = 400 \times 0.10 \\[1em] \frac{\text q}{4\pi\epsilon_0} = 40\ \text N\text{-m}^2\text C^{-1}\end{aligned}$$

The electric field at r₂ is

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r_2^2} \\[1em] = \frac{40}{(0.20)^2} \\[1em] = \frac{40}{0.04} \\[1em] \text E = 1000\ \text V\text{-m}^{-1}\end{aligned}$$

Hence, the electric field at 20 cm from the centre is 1000 V-m^-1.

Given,

• Number of drops, n = 8
• Radius of each small drop, r = 1 mm = 10^-3 m
• Charge on each small drop, q = 10^-10 C

The volume of the bigger drop equals the total volume of the eight small drops,

$$\begin{aligned}\frac{4}{3}\pi\text R^3 = 8 \times \frac{4}{3}\pi\text r^3 \\[1em] \Rightarrow \quad \text R = (8)^{1/3}\text r \\[1em] = 2\text r \\[1em] \text R = 2\times10^{-3}\ \text m\end{aligned}$$

The charge on the bigger drop is

$$\begin{aligned}\text Q = 8\text q \\[1em] \text Q = 8\times10^{-10}\ \text C\end{aligned}$$

The potential at the surface of the bigger drop is

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text R} \\[1em] = \frac{(9.0\times10^9)(8\times10^{-10})}{2\times10^{-3}} \\[1em] = \frac{7.2}{2\times10^{-3}} \\[1em] \text V = 3600\ \text V\end{aligned}$$

Hence, the potential of the bigger drop is 3600 V.

Given,

• Number of drops, n = 27
• Potential of each small drop, V = 10 V

Let q be the charge and r the radius of each small drop, and Q and R those of the big drop. Since the volume remains the same,

$$\begin{aligned}\frac{4}{3}\pi\text R^3 = 27 \times \frac{4}{3}\pi\text r^3 \\[1em] \Rightarrow \quad \text R = 3\text r\end{aligned}$$

The total charge is conserved, so Q = 27q.

For a small drop,

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} = 10\ \text V$

For the big drop,

$$\begin{aligned}\text V' = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text R} \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{27\text q}{3\text r} \\[1em] = 9 \times \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] = 9\text V \\[1em] = 9 \times 10 \\[1em] \text V' = 90\ \text V\end{aligned}$$

Hence, the potential of the large drop is 90 V.

Given,

• Number of drops, n = 64
• Potential of the bigger drop, V_b = 64 V

If r and q are the radius and charge of a small drop, and R and Q those of the big drop, then from the conservation of volume and charge,

$$\begin{aligned}\frac{4}{3}\pi\text R^3 = 64 \times \frac{4}{3}\pi\text r^3 \\[1em] \Rightarrow \quad \text R = 4\text r, \qquad \text Q = 64\text q\end{aligned}$$

The potential of the big drop is

$$\begin{aligned}\text V_\text b = \frac{1}{4\pi\epsilon_0}\frac{64\text q}{4\text r} \\[1em] = 16 \times \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] \text V_\text b = 16\text V_\text s\end{aligned}$$

where V_s is the potential of a small drop. Therefore

$$\begin{aligned}64\ \text V = 16\text V_\text s \\[1em] \Rightarrow \quad \text V_\text s = 4\ \text V\end{aligned}$$

Hence, the potential of each small drop is 4 V.

Given,

• Number of electrons on the drop, n = 10, so q = 10 × 1.6 × 10^-19 = 1.6 × 10^-18 C
• Separation between the plates, d = 5 mm = 5 × 10^-3 m
• Mass of the drop, m = 3 × 10^-16 kg
• g = 10 m s^-2

For the drop to be balanced, the electric force on it must be equal to its weight,

$\text q\text E = \text{mg}$

Since the field between the plates is uniform, E = V/d, so

$$\begin{aligned}\text q\frac{\text V}{\text d} = \text{mg} \\[1em] \Rightarrow \quad \text V = \frac{\text{mgd}}{\text q}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text V = \frac{(3\times10^{-16})(10)(5\times10^{-3})}{1.6\times10^{-18}} \\[1em] = \frac{1.5\times10^{-17}}{1.6\times10^{-18}} \\[1em] \text V = 9.375\ \text V \approx 10\ \text V\end{aligned}$$

Hence, the voltage needed to balance the oil drop is about 10 V.

Given,

• Initial speed, u = 2.5 × 10⁶ m s^-1; final speed, v = 5 × 10⁶ m s^-1
• Mass of the α-particle, m = 6.64 × 10^-27 kg
• Charge on the α-particle, q = 3.2 × 10^-19 C
• Separation between the plates, d = 10 cm = 0.1 m

(a) Potential difference between the plates :

By the conservation of energy, the change in kinetic energy plus the change in potential energy is zero,

$$\begin{aligned}\Delta \text K + \text q\Delta \text V = 0 \\[1em] \Rightarrow \quad \Delta \text V = -\frac{\Delta \text K}{\text q} \\[1em] = -\frac{\frac{1}{2}\text m(\text v^2 - \text u^2)}{\text q}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text V_\text B - \text V_\text A = -\frac{\frac{1}{2}(6.64\times10^{-27})\left[(5\times10^6)^2 - (2.5\times10^6)^2\right]}{3.2\times10^{-19}} \\[1em] = -\frac{\frac{1}{2}(6.64\times10^{-27})(18.75\times10^{12})}{3.2\times10^{-19}} \\[1em] = -\frac{6.225\times10^{-14}}{3.2\times10^{-19}} \\[1em] \text V_\text B - \text V_\text A = -1.95\times10^5\ \text V\end{aligned}$$

(b) Electric field between the plates :

$$\begin{aligned}|\text E| = \frac{|\Delta \text V|}{\text d} \\[1em] = \frac{1.95\times10^5}{0.1} \\[1em] |\text E| = 1.95\times10^6\ \text V\text{-m}^{-1}\end{aligned}$$

(c) This principle is used in a particle accelerator, such as a linear accelerator (atom smasher).

The SI unit of electric potential in terms of the base units is kg m² s^-3 A^-1.

Explanation:

The electric potential is defined as

$\text V = \frac{\text W}{\text q_0}$

Hence

$$\begin{aligned}\text{Dimensions of electric potential} = \frac{\text{dimensions of work}}{\text{dimensions of charge}} \\[1em] = \frac{[\text M\text L^2\text T^{-2}]}{[\text A\text T]} \\[1em] = [\text M\text L^2\text T^{-3}\text A^{-1}]\end{aligned}$$

Given,

• q₁ = q₂ = 10 μC = 10 × 10^-6 C
• Initial separation, r₁ = 10 cm = 0.10 m
• Final separation, r₂ = 10 − 2 = 8 cm = 0.08 m

The electrostatic potential energy of a system of two point charges is

$\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r}$

The increase in potential energy is

$$\begin{aligned}\Delta \text U = \frac{1}{4\pi\epsilon_0}\text q_1\text q_2\left(\frac{1}{\text r_2} - \frac{1}{\text r_1}\right) \\[1em] = (9.0\times10^9)(10\times10^{-6})(10\times10^{-6})\left(\frac{1}{0.08} - \frac{1}{0.10}\right) \\[1em] = 0.9 \times (12.5 - 10) \\[1em] = 0.9 \times 2.5 \\[1em] = 2.25\ \text J\end{aligned}$$

Hence, the increase in the electrostatic potential energy is 2.25 J.

Given,

• q₁ = 3 μC, q₂ = 4 μC, q₃ = 7 μC
• Radius of the circle, r = 1.0 m

All the four charges lie on the circumference, so each is at the same distance r from the centre. Since potential is a scalar quantity,

$$\begin{aligned}\text V_\text C = \frac{1}{4\pi\epsilon_0}\frac{\text q_1 + \text q_2 + \text q_3 + \text q_4}{\text r} \\[1em] = 0\end{aligned}$$

This requires

$$\begin{aligned}\text q_1 + \text q_2 + \text q_3 + \text q_4 = 0 \\[1em] 3 + 4 + 7 + \text q_4 = 0 \\[1em] \Rightarrow \quad \text q_4 = -14\ \mu\text C\end{aligned}$$

Hence, the value of the charge q₄ is − 14 μC.

Given,

• Source charge, q = + 2e = 3.2 × 10^-19 C at the origin
• Initial position, r₁ = 3 cm = 0.03 m; final position, r₂ = 6 cm = 0.06 m
• Charge on the electron, q₀ = − 1.6 × 10^-19 C

The potentials at the two points are

$$\begin{aligned}\text V_1 = \frac{(9.0\times10^9)(3.2\times10^{-19})}{0.03} \\[1em] = \frac{2.88\times10^{-9}}{0.03} \\[1em] = 9.6\times10^{-8}\ \text V \\[1em] \text V_2 = \frac{(9.0\times10^9)(3.2\times10^{-19})}{0.06} \\[1em] = \frac{2.88\times10^{-9}}{0.06} \\[1em] = 4.8\times10^{-8}\ \text V\end{aligned}$$

The work done is

$$\begin{aligned}\text W = \text q_0(\text V_2 - \text V_1) \\[1em] = (-1.6\times10^{-19})(4.8\times10^{-8} - 9.6\times10^{-8}) \\[1em] = (-1.6\times10^{-19})(-4.8\times10^{-8}) \\[1em] = 7.68\times10^{-27}\ \text J\end{aligned}$$

Hence, the work done is 7.68 × 10^-27 J.

Given,

• Charge at the centre, Q = 5 C
• Radius of the circle, r = 10 cm
• Charge moved, q = 2 C

Both A and B lie on the circumference of the circle whose centre carries the charge of 5 C, so both are at the same distance r from it. Hence the circumference is an equipotential surface,

$\text V_\text A = \text V_\text B = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r}$

The work done is

$$\begin{aligned}\text W = \text q(\text V_\text B - \text V_\text A) \\[1em] = 0\end{aligned}$$

Hence, the work done in moving the charge from A to B is zero.

The work done in turning an electric dipole through 360° is zero.

Explanation:

The work done in rotating an electric dipole through an angle θ from the direction of the field is

$\text W = \text p\text E(1 - \cos\theta)$

For θ = 360°,

$$\begin{aligned}\text W = \text p\text E(1 - \cos 360°) \\[1em] = \text p\text E(1 - 1) \\[1em] = 0\end{aligned}$$

The dipole returns to its original position, so there is no change in its potential energy.

At a point where E = 0 the potential gradient is zero, so the electric potential there has a stationary value (a maximum, a minimum, or locally flat); it is not necessarily zero.

Explanation:

The electric field is the negative potential gradient,

$\text E = -\frac{\text{dV}}{\text{dr}}$

So E = 0 at a point tells us only that the potential gradient vanishes there; the potential is stationary at that point. It does not follow that V = 0, nor that V is constant throughout a region — the potential is uniform over a region only when E = 0 at every point of that region.

For example, inside a charged hollow metallic sphere the electric field is zero everywhere, so there the potential is indeed constant throughout, having the non-zero value $\frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R}$.

Given,

• Each charge, q = 5.0 μC = 5.0 × 10^-6 C
• Side of the triangle, a = 5.0 cm = 0.05 m

The number of pairs is $\frac{3 \times 2}{2} = 3$, and every pair is separated by the same distance a. Hence

$\text U = 3 \times \frac{1}{4\pi\epsilon_0}\frac{\text q^2}{\text a}$

Substituting the values,

$$\begin{aligned}= 3 \times \frac{(9.0\times10^9)(5.0\times10^{-6})^2}{0.05} \\[1em] = 3 \times \frac{(9.0\times10^9)(2.5\times10^{-11})}{0.05} \\[1em] = 3 \times \frac{0.225}{0.05} \\[1em] = 3 \times 4.5 \\[1em] = 13.5\ \text J\end{aligned}$$

Hence, the electrostatic potential energy of the system is 13.5 J.

Given,

• Charge on the electron, e = 1.6 × 10^-19 C
• Potential difference, V = 1000 V

The kinetic energy gained is equal to the work done on the electron,

$$\begin{aligned}\text K = \text q\text V \\[1em] = (1.6\times10^{-19})(1000) \\[1em] = 1.6\times10^{-16}\ \text J\end{aligned}$$

Hence, the kinetic energy gained by the electron is 1.6 × 10^-16 J.

Given,

• Radius of the orbit, r = 5 × 10^-11 m
• Charge on the hydrogen nucleus, Ze = 1.6 × 10^-19 C (Z = 1)
• Charge on the electron, e = 1.6 × 10^-19 C

(i) Intensity of the electric field of the nucleus :

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text{Ze}}{\text r^2} \\[1em] = \frac{(9\times10^9)(1)(1.6\times10^{-19})}{(5\times10^{-11})^2} \\[1em] = \frac{1.44\times10^{-9}}{2.5\times10^{-21}} \\[1em] = 5.76\times10^{11}\ \text N\text C^{-1}\end{aligned}$$

(ii) Electrostatic potential energy of the system :

$$\begin{aligned}\text U = \frac{1}{4\pi\epsilon_0}\frac{\text{Ze}\times(-\text e)}{\text r} \\[1em] = -\frac{(9\times10^9)(1.6\times10^{-19})(1.6\times10^{-19})}{5\times10^{-11}} \\[1em] = -\frac{2.304\times10^{-28}}{5\times10^{-11}} \\[1em] = -4.6\times10^{-18}\ \text J\end{aligned}$$

Hence, E = 5.76 × 10¹¹ N C^-1 and U = − 4.6 × 10^-18 J.

Note: The textbook answer key prints the unit of the answer to part (i) as joule. The intensity of the electric field is measured in N C^-1 (or V m^-1); the numerical value 5.76 × 10¹¹ is correct.

No work is done in this process.

Explanation:

The electrostatic force of attraction which furnishes the centripetal force always remains perpendicular to the direction of the velocity of the electron. Also, the circular orbit is an equipotential surface, so the potential difference between any two of its points is zero and

$$\begin{aligned}\text W = \text q \times \Delta \text V \\[1em] = 0 \qquad (\Delta \text V = 0)\end{aligned}$$

Given,

• Q₁ = 25 μC, Q₂ = 50 μC, Q₃ = 100 μC
• Each side of the equilateral triangle, a = 7.5 m

The total electrostatic potential energy is the algebraic sum of the energies of all the three pairs, each pair being separated by the same distance a,

$\text U = \frac{1}{4\pi\epsilon_0\text a}\left[\text Q_1\text Q_2 + \text Q_2\text Q_3 + \text Q_3\text Q_1\right]$

Substituting the values,

$$\begin{aligned}= \frac{9.0\times10^9}{7.5}\left[(25)(50) + (50)(100) + (100)(25)\right]\times10^{-12} \\[1em] = \frac{9.0\times10^9}{7.5}\left[1250 + 5000 + 2500\right]\times10^{-12} \\[1em] = \frac{(9.0\times10^9)(8750\times10^{-12})}{7.5} \\[1em] = \frac{78.75}{7.5} \\[1em] = 10.5\ \text J\end{aligned}$$

Hence, the total electrostatic potential energy of the system is 10.5 J.

Given,

• q₁ = q₂ = 10 C
• Separation, r = 3 m

The electrostatic potential energy of the system is

$$\begin{aligned}\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r} \\[1em] = \frac{(9.0\times10^9)(10)(10)}{3} \\[1em] = \frac{9.0\times10^{11}}{3} \\[1em] = 3\times10^{11}\ \text J\end{aligned}$$

Hence, the electrostatic potential energy is 3 × 10¹¹ J.

zero

Reason — The points M and N both lie on a circular arc of radius r drawn about the fixed charge Q, so each is at the same distance r from Q. Hence

$\text V_\text M = \text V_\text N = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r}$

The arc is thus part of an equipotential surface, and the work done is

$$\begin{aligned}\text W = \text q(\text V_\text N - \text V_\text M) \\[1em] = 0\end{aligned}$$

The length l of the arc is irrelevant, since no work is done in moving a charge along an equipotential surface.

9 × 10^-3 J

Reason —

Given, q₁ = q₂ = 1 μC = 1 × 10^-6 C and r = 1 m.

$$\begin{aligned}\text U = \frac{1}{4\pi\epsilon_0}\frac{\text q_1\text q_2}{\text r} \\[1em] = \frac{(9.0\times10^9)(1\times10^{-6})(1\times10^{-6})}{1} \\[1em] = 9.0\times10^9 \times 10^{-12} \\[1em] = 9\times10^{-3}\ \text J\end{aligned}$$

Let a positive point charge + Q be fixed at a point O in vacuum, and let P be the point, a distance r from O, whose potential is required. The potential at P is the work done per unit charge in carrying a positive test charge + q₀ from infinity up to P.

Imagine the test charge + q₀ momentarily at a point A, a distance x from O and lying beyond P (so that x > r). By Coulomb’s law the force on it has magnitude

$\text F = \frac{1}{4\pi\epsilon_0}\frac{\text Q\text q_0}{\text x^2}\ \text{newton} \qquad \dots(\text i)$

This force $\vec{\text F} \space$is directed radially away from O. When the charge is pushed inward through a small displacement dx to a neighbouring point B, the element of work done is

$\text{dW} = \text F(-\text{dx})$

the minus sign appearing because the displacement dx is opposite to the direction of the force.

Summing these elements as the charge moves from infinity (x = ∞) in to P (x = r) gives the total work

$$\begin{aligned}\text W = -\displaystyle\int_\infty^\text r \text F\ \text{dx} \\[1em] = -\frac{\text Q\text q_0}{4\pi\epsilon_0}\displaystyle\int_\infty^\text r \frac{1}{\text x^2}\ \text{dx} \\[1em] = -\frac{\text Q\text q_0}{4\pi\epsilon_0}\left[-\frac{1}{\text x}\right]_\infty^\text r \\[1em] = -\frac{\text Q\text q_0}{4\pi\epsilon_0}\left(-\frac{1}{\text r} + \frac{1}{\infty}\right) \\[1em] \text W = \frac{1}{4\pi\epsilon_0}\frac{\text Q\text q_0}{\text r}\end{aligned}$$

The potential at P is this work per unit test charge,

$$\begin{aligned}\text V = \frac{\text W}{\text q_0} \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r}\ \text{volt}\end{aligned}$$

Hence, the electric potential at a point distant r from the charge Q is $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r}$.

Any surface over which the electric potential is same everywhere is called an equipotential surface.

Explanation:

An equipotential surface may be the surface of a charged body or simply a surface in space; for example, the surface of a conductor is an equipotential surface. No work is done in moving a charge between any two points on such a surface, and the electric field is everywhere at right angles to it.

$\left(\frac{1}{4\pi\epsilon_0}\right)\frac{3\text q^2}{\text a}$

Reason — There are three pairs of charges, and each pair is separated by the same distance a. The potential energy of each pair is $\frac{1}{4\pi\epsilon_0}\frac{\text q^2}{\text a}$, and these add algebraically,

$$\begin{aligned}\text U = 3 \times \frac{1}{4\pi\epsilon_0}\frac{\text q^2}{\text a} \\[1em] = \left(\frac{1}{4\pi\epsilon_0}\right)\frac{3\text q^2}{\text a}\end{aligned}$$

The locus is the perpendicular bisector of the dipole, that is, the equatorial line of the dipole.

Explanation:

Every point on this line is equidistant from the charges + q and − q, so the potentials due to them are equal and opposite and cancel out. This also follows from $\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p\cos\theta}{\text r^2}$, which vanishes for θ = 90°.

Consider a dipole AB made up of charges − q and + q coulomb separated by a small distance 2l in vacuum, and let P lie on the axis of the dipole at a distance r from its mid-point O.

The point P is nearer the charge + q, at a distance (r − l) from it, and farther from the charge − q, at a distance (r + l) from it. The potential produced at P by the charge + q is therefore

$\text V_1 = \frac{1}{4\pi\epsilon_0}\frac{\text q}{(\text r - \text l)}$

while the charge − q contributes

$\text V_2 = -\frac{1}{4\pi\epsilon_0}\frac{\text q}{(\text r + \text l)}$

Potential is a scalar, so the resultant potential at P is the algebraic sum of the two,

$$\begin{aligned}\text V = \text V_1 + \text V_2 \\[1em] = \frac{\text q}{4\pi\epsilon_0}\left[\frac{1}{(\text r - \text l)} - \frac{1}{(\text r + \text l)}\right] \\[1em] = \frac{\text q}{4\pi\epsilon_0}\left[\frac{(\text r + \text l) - (\text r - \text l)}{\text r^2 - \text l^2}\right] \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{2\text q\text l}{\text r^2 - \text l^2}\end{aligned}$$

Writing the electric dipole moment p = 2ql, this becomes

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2 - \text l^2}$

For a point far from the dipole (r >> 2l), l² is negligible in comparison with r², and the potential reduces to the short-dipole result

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2}\ \text{volt}$

inversely proportional to square of the distance

Reason — For a short dipole (r >> 2l) the potential at a point on the axial line is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p}{\text r^2}$

Hence $\text V \propto \frac{1}{\text r^2}$. This is unlike the potential due to a single point charge, for which $\text V \propto \frac{1}{\text r}$.

electric potential is zero

Reason — The perpendicular bisector of a dipole is its equatorial line. Every point on it is equidistant from + q and − q, so the potentials due to the two charges are equal in magnitude but opposite in sign,

$$\begin{aligned}\text V = \frac{\text q}{4\pi\epsilon_0}\left(\frac{1}{\text{BP}} - \frac{1}{\text{AP}}\right) \\[1em] = 0 \qquad (\because\ \text{BP} = \text{AP})\end{aligned}$$

The electric field, however, is not zero there; it is directed antiparallel to the dipole axis. Hence the potential gradient along the axial direction is also not zero.

$-\frac{\text Q}{2}$

Reason —

Given, charges Q, Q and q at the vertices of an equilateral triangle of side a, with U = 0.

There are three pairs, all separated by the distance a,

$$\begin{aligned}\text U = \frac{1}{4\pi\epsilon_0\text a}\left[\text Q\text Q + \text Q\text q + \text q\text Q\right] \\[1em] = \frac{1}{4\pi\epsilon_0\text a}\left[\text Q^2 + 2\text{Qq}\right]\end{aligned}$$

Putting U = 0,

$$\begin{aligned}\text Q^2 + 2\text{Qq} = 0 \\[1em] \Rightarrow \quad \text Q(\text Q + 2\text q) = 0\end{aligned}$$

Since Q ≠ 0,

$$\begin{aligned}\text Q + 2\text q = 0 \\[1em] \Rightarrow \quad \text q = -\frac{\text Q}{2}\end{aligned}$$

Given,

• Charge, q = 50 μC = 50 × 10^-6 C
• Distance, r = 9 cm = 0.09 m

The potential due to a point charge is

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] = \frac{(9.0\times10^9)(50\times10^{-6})}{0.09} \\[1em] = \frac{4.5\times10^5}{0.09} \\[1em] = 5\times10^6\ \text V\end{aligned}$$

Hence, the electric potential at the point P is 5 × 10⁶ V.

Zero.

Explanation:

The potential is the same at every point of an equipotential surface, so V_A = V_B. Hence

$$\begin{aligned}\text W = \text q(\text V_\text B - \text V_\text A) \\[1em] = 0\end{aligned}$$

The result is independent of the distance AB, since the electric field is everywhere normal to the equipotential surface and has no component along it.

Given,

• Two charges, each Q, at A and B with AB = a
• The third charge q = Q is brought to C, where ABC is equilateral, so CA = CB = a

Since ABC is an equilateral triangle, the point C is at a distance a from each of the charges. As potential is a scalar, the potential at C due to the two charges is

$$\begin{aligned}\text V_\text C = \frac{1}{4\pi\epsilon_0}\left(\frac{\text Q}{\text a} + \frac{\text Q}{\text a}\right) \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{2\text Q}{\text a}\end{aligned}$$

The work done in bringing the charge q = Q from infinity (where the potential is zero) to C is

$$\begin{aligned}\text W = \text q\text V_\text C \\[1em] = \text Q \times \frac{1}{4\pi\epsilon_0}\frac{2\text Q}{\text a} \\[1em] = \frac{2\text Q^2}{4\pi\epsilon_0\text a} \\[1em] \text W = \frac{1}{2\pi\epsilon_0}\frac{\text Q^2}{\text a}\end{aligned}$$

Hence, the work done is $\frac{1}{2\pi\epsilon_0}\frac{\text Q^2}{\text a}$. The work is positive because the third charge is also positive and must be pushed against the repulsion of the two charges already present.

$-\left(6\hat{\text i} + 5\hat{\text j} + 2\hat{\text k}\right)$

Reason —

Given, V(x, y, z) = 6xy − y + 2yz.

The electric field is the negative gradient of the potential,

$\vec{\text E} = -\left(\frac{\partial \text V}{\partial \text x}\hat{\text i} + \frac{\partial \text V}{\partial \text y}\hat{\text j} + \frac{\partial \text V}{\partial \text z}\hat{\text k}\right)$

Taking the partial derivatives,

$\frac{\partial \text V}{\partial \text x} = 6\text y, \qquad \frac{\partial \text V}{\partial \text y} = 6\text x - 1 + 2\text z, \qquad \frac{\partial \text V}{\partial \text z} = 2\text y$

Therefore

$\vec{\text E} = -\left[6\text y\ \hat{\text i} + (6\text x - 1 + 2\text z)\hat{\text j} + 2\text y\ \hat{\text k}\right]$

At the point (1, 1, 0),

$$\begin{aligned}\vec{\text E} = -\left[6(1)\hat{\text i} + (6 - 1 + 0)\hat{\text j} + 2(1)\hat{\text k}\right] \\[1em] = -\left(6\hat{\text i} + 5\hat{\text j} + 2\hat{\text k}\right)\end{aligned}$$

In all the four cases the work done is the same.

Reason — The work done in moving a charge between two points in an electrostatic field is

$\text W = \text q\ \Delta \text V$

and depends only on the potential difference between the two points, not on the shape or spacing of the equipotential surfaces or on the path followed. In each of the four diagrams the point A lies on the 10 V equipotential and the point B on the 40 V equipotential, so ΔV = 30 V in every case. Hence the work done is the same in all four cases.

$\frac{\sigma}{\epsilon_0}\left[\frac{\text a^2 - \text b^2}{\text b} + \text c\right]$

Reason —

Given, three concentric shells of radii a, b and c with surface charge densities + σ, − σ and + σ respectively.

The charges on the shells are

$\text Q_\text A = \sigma(4\pi\text a^2), \qquad \text Q_\text B = -\sigma(4\pi\text b^2), \qquad \text Q_\text C = \sigma(4\pi\text c^2)$

The potential at the shell B is the sum of the potential due to the inner shell A (which behaves as a point charge at the centre), the potential due to its own charge, and the potential due to the outer shell C (which is the same at all interior points),

$\text V_\text B = \frac{1}{4\pi\epsilon_0}\left[\frac{\text Q_\text A}{\text b} + \frac{\text Q_\text B}{\text b} + \frac{\text Q_\text C}{\text c}\right]$

Substituting the charges,

$$\begin{aligned}= \frac{1}{4\pi\epsilon_0}\left[\frac{\sigma 4\pi\text a^2}{\text b} - \frac{\sigma 4\pi\text b^2}{\text b} + \frac{\sigma 4\pi\text c^2}{\text c}\right] \\[1em] = \frac{\sigma}{\epsilon_0}\left[\frac{\text a^2}{\text b} - \text b + \text c\right] \\[1em] = \frac{\sigma}{\epsilon_0}\left[\frac{\text a^2 - \text b^2}{\text b} + \text c\right]\end{aligned}$$

potential of a uniformly charged spherical shell

Reason — The graph is flat from r = 0 up to r = r₀ and then falls off smoothly with increasing r.

Due to a charged spherical shell, the potential at any interior point is the same as that at its surface,

$\text V_{\text{inside}} = \text V_{\text{surface}} = \frac{\text{KQ}}{\text r_0}$

which explains the constant portion of the graph. Outside the shell,

$\text V_{\text{outside}} = \frac{\text{KQ}}{\text r} \quad \Rightarrow \quad \text V \propto \frac{1}{\text r}$

which explains the falling portion. The electric field of a shell would be zero inside and then drop as $\frac{1}{\text r^2}$, while for a uniformly charged solid sphere the potential rises to $\frac{3}{2}\frac{\text{KQ}}{\text R}$ at the centre and is not constant inside. Hence the graph represents the potential of a uniformly charged spherical shell.

$\sqrt{\frac{2}{\text m}}\left(\frac{2}{15}\frac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}$

Reason —

Given, radius of the ring = 3a, charge on the ring = q, moving charge = q of mass m, initial position z = 4a.

The potential on the axis of a charged ring at an axial distance z is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\sqrt{\text R^2 + \text z^2}}$

At z = 4a :

$$\begin{aligned}\text V_1 = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\sqrt{(3\text a)^2 + (4\text a)^2}} \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{\text q}{5\text a}\end{aligned}$$

At the centre (z = 0) :

$\text V_2 = \frac{1}{4\pi\epsilon_0}\frac{\text q}{3\text a}$

To just cross the origin, the whole of the kinetic energy must be used up in increasing the potential energy,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text q(\text V_2 - \text V_1) \\[1em] = \frac{\text q^2}{4\pi\epsilon_0}\left(\frac{1}{3\text a} - \frac{1}{5\text a}\right) \\[1em] = \frac{\text q^2}{4\pi\epsilon_0}\cdot\frac{2}{15\text a}\end{aligned}$$

Therefore

$$\begin{aligned}\text v^2 = \frac{2}{\text m}\left(\frac{2}{15}\frac{\text q^2}{4\pi\epsilon_0\text a}\right) \\[1em] \Rightarrow \quad \text v = \sqrt{\frac{2}{\text m}}\left(\frac{2}{15}\frac{\text q^2}{4\pi\epsilon_0\text a}\right)^{1/2}\end{aligned}$$

$\frac{\text Q^2}{4\pi\epsilon_0}\left(1 + \frac{1}{\sqrt5}\right)$

Reason —

Given, four charges Q at (0, 2), (4, 2), (4, − 2) and (0, − 2).

The distances of these points from the origin are

$\text r_1 = 2, \qquad \text r_2 = 2, \qquad \text r_3 = \sqrt{4^2 + 2^2} = \sqrt{20}, \qquad \text r_4 = \sqrt{20}$

Since potential is a scalar, the potential at the origin is

$$\begin{aligned}\text V = \text{KQ}\left[\frac{1}{2} + \frac{1}{2} + \frac{1}{\sqrt{20}} + \frac{1}{\sqrt{20}}\right] \\[1em] = \text{KQ}\left[1 + \frac{2}{2\sqrt5}\right] \\[1em] = \text{KQ}\left[1 + \frac{1}{\sqrt5}\right]\end{aligned}$$

The work required to bring the fifth charge Q from infinity to the origin is

$$\begin{aligned}\text W = \text Q\text V \\[1em] = \frac{\text Q^2}{4\pi\epsilon_0}\left(1 + \frac{1}{\sqrt5}\right)\end{aligned}$$

$\text v \propto \sqrt{\log\left(\frac{\text r}{\text r_0}\right)}$

Reason —

The electric field due to an infinitely long uniformly charged wire of linear charge density λ at a distance r is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\epsilon_0\text r} \\[1em] = -\frac{\text{dV}}{\text{dr}}\end{aligned}$$

Integrating between r₀ and r,

$$\begin{aligned}\Delta \text V = -\displaystyle\int_{\text r_0}^{\text r}\frac{\lambda}{2\pi\epsilon_0\text r}\ \text{dr} \\[1em] = -\frac{\lambda}{2\pi\epsilon_0}\log\left(\frac{\text r}{\text r_0}\right)\end{aligned}$$

The charge is released from rest, so the loss in potential energy appears as kinetic energy,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text q(\text V_{\text r_0} - \text V_\text r) \\[1em] = \frac{\text q\lambda}{2\pi\epsilon_0}\log\left(\frac{\text r}{\text r_0}\right) \\[1em] \Rightarrow \quad \text v^2 \propto \log\left(\frac{\text r}{\text r_0}\right) \\[1em] \Rightarrow \quad \text v \propto \sqrt{\log\left(\frac{\text r}{\text r_0}\right)}\end{aligned}$$

−7 × 10^-27 J

Reason —

Given,

• Electric field, E = 1000 V m^-1
• Angle, θ = 45°
• Dipole moment, p = 10^-29 C-m

The potential energy of a dipole in a uniform electric field is

$$\begin{aligned}\text U = -\vec{\text p}\cdot\vec{\text E} \\[1em] = -\text p\text E\cos\theta\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text U = -(10^{-29})(1000)\cos 45° \\[1em] = -(10^{-26})(0.707) \\[1em] \text U \approx -7\times10^{-27}\ \text J\end{aligned}$$

The graph in which the speed increases from zero at R₀ and then flattens out, approaching a constant value as R(t) becomes large.

Reason —

Let the radius of the distribution at any instant t be R, the initial radius being R₀. By the conservation of energy,

$0 + \frac{\text{KQ}^2}{2\text R_0} = \frac{1}{2}\text m\text v^2 + \frac{\text{KQ}^2}{2\text R}$

Differentiating this equation with respect to R,

$$\begin{aligned}\frac{1}{2}\text m(2\text v)\frac{\text{dv}}{\text{dR}} - \frac{\text{KQ}^2}{2\text R^2} = 0 \\[1em] \Rightarrow \quad \frac{\text{dv}}{\text{dR}} = \frac{\text{KQ}^2}{2\text{mv}\text R^2}\end{aligned}$$

Here $\frac{\text{dv}}{\text{dR}}$ is the slope of the v-R graph. It decreases as both v and R increase, and the slope tends to zero as R → ∞. Hence the curve starts at v = 0 when R = R₀, rises steeply at first and then becomes almost horizontal, the speed approaching a limiting value.

$\frac{\text Q(\text a + \text b + \text c)}{4\pi\epsilon_0(\text a^2 + \text b^2 + \text c^2)}$

Reason —

Given, three concentric shells of radii a, b and c with equal surface charge densities, carrying a total charge Q.

Since $\sigma_\text a = \sigma_\text b = \sigma_\text c$,

$$\begin{aligned}\frac{\text Q_\text a}{4\pi\text a^2} = \frac{\text Q_\text b}{4\pi\text b^2} = \frac{\text Q_\text c}{4\pi\text c^2} \\[1em] \Rightarrow \quad \text Q_\text a : \text Q_\text b : \text Q_\text c = \text a^2 : \text b^2 : \text c^2\end{aligned}$$

Also Q_a + Q_b + Q_c = Q, so

$\text Q_\text a = \frac{\text a^2}{\text a^2 + \text b^2 + \text c^2}\text Q, \quad \text Q_\text b = \frac{\text b^2}{\text a^2 + \text b^2 + \text c^2}\text Q, \quad \text Q_\text c = \frac{\text c^2}{\text a^2 + \text b^2 + \text c^2}\text Q$

For a point with r < a, the potential due to each shell equals the potential at its own surface. Hence

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\left[\frac{\text Q_\text a}{\text a} + \frac{\text Q_\text b}{\text b} + \frac{\text Q_\text c}{\text c}\right] \\[1em] = \frac{1}{4\pi\epsilon_0}\cdot\frac{\text Q(\text a + \text b + \text c)}{\text a^2 + \text b^2 + \text c^2}\end{aligned}$$

180 V

Reason —

Given, $\vec{\text E} = (20\text x + 10)\hat{\text i}$ N C^-1.

The potential difference is obtained from

$$\begin{aligned}\Delta \text V = -\displaystyle\int_{\text r_1}^{\text r_2}\vec{\text E}\cdot\text d\vec{\text r} \\[1em] \text V_1 - \text V_2 = -\displaystyle\int_{-5}^{1}(20\text x + 10)\ \text{dx}\end{aligned}$$

Integrating term by term,

$$\begin{aligned}= -\left[\frac{20\text x^2}{2} + 10\text x\right]_{-5}^{1} \\[1em] = -\left[10\text x^2 + 10\text x\right]_{-5}^{1}\end{aligned}$$

Evaluating the limits,

$$\begin{aligned}= -\left[\left(10(1)^2 + 10(1)\right) - \left(10(-5)^2 + 10(-5)\right)\right] \\[1em] = -\left[(10 + 10) - (250 - 50)\right] \\[1em] = -\left[20 - 200\right] \\[1em] = 180\ \text V\end{aligned}$$

$0,\ \frac{-\vec{\text p}}{4\pi\epsilon_0\text d^3}$

Reason —

Given, $\vec{\text p} = -\text p_0\hat{\text x}$ placed at the origin, and the field point A lies on the Y-axis at a distance d.

Since the dipole axis is along the X-axis, the point A on the Y-axis lies on the equatorial line of the dipole. Hence

$$\begin{aligned}\text V_\text A = \frac{1}{4\pi\epsilon_0}\frac{\text p\cos 90°}{\text d^2} \\[1em] = 0\end{aligned}$$

The electric field at a point on the equatorial line is directed antiparallel to the dipole moment,

$$\begin{aligned}\vec{\text E}_\text A = -\frac{1}{4\pi\epsilon_0}\frac{\vec{\text p}}{\text d^3} \\[1em] = -\frac{1}{4\pi\epsilon_0}\frac{(-\text p_0\hat{\text x})}{\text d^3} \\[1em] = \frac{\text p_0}{4\pi\epsilon_0\text d^3}\hat{\text x}\end{aligned}$$

Note: The textbook answer key gives option 2 for this question, but the worked hint given in the textbook itself obtains $\vec{\text E}_\text A = \frac{\text p_0}{4\pi\epsilon_0\text d^3}\hat{\text x}$, which is option 4. The correct option is therefore 4.

V

Reason —

Let r₁ be the radius of the inner solid sphere and r₂ that of the outer surface of the shell, and let $\text k = \frac{1}{4\pi\epsilon_0}$.

In the first condition, the outer shell is uncharged, so both potentials are due to the charge Q alone,

$$\begin{aligned}\Delta \text V = \text V_{\text{in}} - \text V_{\text{out}} \\[1em] = \frac{\text{kQ}}{\text r_1} - \frac{\text{kQ}}{\text r_2} \\[1em] = \text{kQ}\left(\frac{1}{\text r_1} - \frac{1}{\text r_2}\right) \\[1em] = \text V\end{aligned}$$

In the second condition, the shell carries a charge − 4Q as well, and this charge contributes the same amount $-\frac{4\text{kQ}}{\text r_2}$ to both the potentials,

$$\begin{aligned}\Delta \text V' = \left(\frac{\text{kQ}}{\text r_1} - \frac{4\text{kQ}}{\text r_2}\right) - \left(\frac{\text{kQ}}{\text r_2} - \frac{4\text{kQ}}{\text r_2}\right) \\[1em] = \frac{\text{kQ}}{\text r_1} - \frac{\text{kQ}}{\text r_2} \\[1em] = \text V\end{aligned}$$

Hence the potential difference does not depend on the charge given to the outer shell and remains the same.

2 : 1

Reason —

For a charge accelerated from rest through a potential difference V,

$$\begin{aligned}\frac{1}{2}\text m\text v^2 = \text q\text V \\[1em] \Rightarrow \quad \text v = \sqrt{\frac{2\text q\text V}{\text m}}\end{aligned}$$

The hydrogen ion and the singly ionized helium atom carry the same charge, but the mass of the helium ion is four times that of the hydrogen ion. Hence

$$\begin{aligned}\frac{\text v_\text H}{\text v_{\text{He}}} = \sqrt{\frac{\text q_\text H \text m_{\text{He}}}{\text m_\text H \text q_{\text{He}}}} \\[1em] = \sqrt{\frac{4\text m_\text H}{\text m_\text H}} \\[1em] = 2\end{aligned}$$

that is, the ratio of the final speeds is 2 : 1.

$\frac{3\text Q_1}{16\pi\epsilon_0\text R}$

Reason —

Given, spheres of radii R and 4R with charges Q₁ and Q₂, having equal surface charge densities.

$$\begin{aligned}\frac{\text Q_1}{4\pi\text R^2} = \frac{\text Q_2}{4\pi(4\text R)^2} \\[1em] \Rightarrow \quad \text Q_2 = 16\text Q_1\end{aligned}$$

Writing $\text k = \frac{1}{4\pi\epsilon_0}$, the potentials at the two surfaces are

$\text V(\text R) = \frac{\text{kQ}_1}{\text R} + \frac{\text{kQ}_2}{4\text R}, \qquad \text V(4\text R) = \frac{\text{kQ}_1}{4\text R} + \frac{\text{kQ}_2}{4\text R}$

Therefore

$$\begin{aligned}\text V(\text R) - \text V(4\text R) = \frac{\text{kQ}_1}{\text R} - \frac{\text{kQ}_1}{4\text R} \\[1em] = \frac{3}{4}\cdot\frac{\text{kQ}_1}{\text R} \\[1em] = \frac{3\text Q_1}{16\pi\epsilon_0\text R}\end{aligned}$$

decrease by $\frac{4\text q^2}{3\pi\epsilon_0\text d}$

Reason —

Given, 4q at $\text x = -\frac{\text d}{2}$ and − q at $\text x = \frac{\text d}{2}$, and the charge q is taken from the origin to x = d.

Writing $\text k = \frac{1}{4\pi\epsilon_0}$ :

At the origin, the distance from each fixed charge is $\frac{\text d}{2}$,

$$\begin{aligned}\text U_1 = \text{kq}\left[\frac{4\text q}{\text d/2} - \frac{\text q}{\text d/2}\right] \\[1em] = \text{kq}^2\left[\frac{8}{\text d} - \frac{2}{\text d}\right] \\[1em] = \frac{6\text{kq}^2}{\text d}\end{aligned}$$

At x = d, the distance from 4q is $\frac{3\text d}{2}$ and that from − q is $\frac{\text d}{2}$,

$$\begin{aligned}\text U_2 = \text{kq}\left[\frac{4\text q}{3\text d/2} - \frac{\text q}{\text d/2}\right] \\[1em] = \text{kq}^2\left[\frac{8}{3\text d} - \frac{2}{\text d}\right] \\[1em] = \frac{2\text{kq}^2}{3\text d}\end{aligned}$$

The change in the energy of the charge is

$$\begin{aligned}\Delta \text U = \text U_2 - \text U_1 \\[1em] = \frac{2\text{kq}^2}{3\text d} - \frac{6\text{kq}^2}{\text d} \\[1em] = -\frac{16\text{kq}^2}{3\text d}\end{aligned}$$

Since U₁ > U₂, there is a decrease, of magnitude

$\frac{16}{3\text d}\cdot\frac{\text q^2}{4\pi\epsilon_0} = \frac{4\text q^2}{3\pi\epsilon_0\text d}$

The path being a semicircle does not matter, since the electrostatic field is conservative.

$\frac{\text p}{\text a}\sqrt{\frac{1}{2\pi\epsilon_0\text{ma}}}$

Reason —

The electric field due to the first dipole at an axial distance a is

$\vec{\text E} = \frac{2\vec{\text p}}{4\pi\epsilon_0\text a^3}$

The second dipole is located in this field, so its potential energy is

$$\begin{aligned}\text U = -\vec{\text p}_2\cdot\vec{\text E} \\[1em] = -(-\text p)\left(\frac{2\text p}{4\pi\epsilon_0\text a^3}\right) \\[1em] = \frac{2\text p^2}{4\pi\epsilon_0\text a^3}\end{aligned}$$

When the dipoles are infinitely far apart, U' = 0. Hence the loss in potential energy appears as the kinetic energy of the two dipoles,

$$\begin{aligned}\Delta \text U = \text U - \text U' \\[1em] = \frac{2\text p^2}{4\pi\epsilon_0\text a^3} \\[1em] = 2 \times \frac{1}{2}\text m\text v^2 \\[1em] \text m\text v^2 = \frac{2\text p^2}{4\pi\epsilon_0\text a^3} \\[1em] = \frac{\text p^2}{2\pi\epsilon_0\text a^3} \\[1em] \text v^2 = \frac{\text p^2}{2\pi\epsilon_0\text{ma}^3} \\[1em] \Rightarrow \quad \text v = \frac{\text p}{\text a}\sqrt{\frac{1}{2\pi\epsilon_0\text{ma}}}\end{aligned}$$

$\frac{1}{4\pi\epsilon_0}\frac{(\text R + \text r)}{(\text R^2 + \text r^2)}\text Q$

Reason —

Given, two concentric shells of radii r and R carrying charges q₁ and q₂ with equal surface charge densities, and q₁ + q₂ = Q.

Since the surface charge densities are equal,

$\frac{\text q_1}{4\pi\text r^2} = \frac{\text q_2}{4\pi\text R^2}$

Solving these two relations together,

$\text q_1 = \frac{\text r^2}{\text R^2 + \text r^2}\text Q, \qquad \text q_2 = \frac{\text R^2}{\text R^2 + \text r^2}\text Q$

The potential inside a shell equals that at its surface, so the potential at the common centre is

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\left(\frac{\text q_1}{\text r} + \frac{\text q_2}{\text R}\right) \\[1em] = \frac{1}{4\pi\epsilon_0}\left(\frac{\text r\text Q}{\text R^2 + \text r^2} + \frac{\text R\text Q}{\text R^2 + \text r^2}\right) \\[1em] = \frac{1}{4\pi\epsilon_0}\cdot\frac{(\text R + \text r)}{(\text R^2 + \text r^2)}\text Q\end{aligned}$$

1980 V

Reason —

Given, n = 27 and the potential of each small drop is 220 V.

If r and q are the radius and charge of a small drop, and R and 27q those of the bigger drop, then from the conservation of volume,

$$\begin{aligned}\frac{4}{3}\pi\text R^3 = 27 \times \frac{4}{3}\pi\text r^3 \\[1em] \Rightarrow \quad \text R = 3\text r\end{aligned}$$

For each small drop,

$220 = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r}$

For the bigger drop,

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{27\text q}{3\text r} \\[1em] = 9 \times \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r} \\[1em] = 9 \times 220 \\[1em] = 1980\ \text V\end{aligned}$$

$\frac{\text Q}{2\pi\epsilon_0}\left[\frac{1}{\text a} - \frac{1}{\sqrt{\text s^2 + \text a^2}}\right]$

Reason —

Given, two coaxial rings A and B, each of radius a, carrying charges + Q and − Q, separated by a distance s.

Every point of a ring is at a distance a from its own centre and at a distance $\sqrt{\text s^2 + \text a^2}$ from the centre of the other ring. Writing $\text K = \frac{1}{4\pi\epsilon_0}$,

$$\begin{aligned}\text V_\text A = \frac{\text{KQ}}{\text a} - \frac{\text{KQ}}{\sqrt{\text a^2 + \text s^2}} \\[1em] \text V_\text B = -\frac{\text{KQ}}{\text a} + \frac{\text{KQ}}{\sqrt{\text a^2 + \text s^2}}\end{aligned}$$

Therefore

$$\begin{aligned}\text V_\text A - \text V_\text B = \frac{2\text{KQ}}{\text a} - \frac{2\text{KQ}}{\sqrt{\text a^2 + \text s^2}} \\[1em] = \frac{\text Q}{2\pi\epsilon_0}\left[\frac{1}{\text a} - \frac{1}{\sqrt{\text s^2 + \text a^2}}\right]\end{aligned}$$

2 : 1

Reason —

Given, r_A = 5 mm and r_B = 10 mm.

When two conducting spheres are connected by a conducting wire, charge flows from the sphere at the higher potential to the one at the lower potential till both acquire the same potential,

$$\begin{aligned}\text V_\text A = \text V_\text B \\[1em] \Rightarrow \quad \frac{1}{4\pi\epsilon_0}\frac{\text q_\text A}{\text r_\text A} = \frac{1}{4\pi\epsilon_0}\frac{\text q_\text B}{\text r_\text B} \\[1em] \Rightarrow \quad \frac{\text q_\text A}{\text q_\text B} = \frac{\text r_\text A}{\text r_\text B} \\[1em] = \frac{5}{10} \\[1em] = \frac{1}{2}\end{aligned}$$

The ratio of the electric fields at the surfaces is

$$\begin{aligned}\frac{\text E_\text A}{\text E_\text B} = \frac{\text K\frac{\text q_\text A}{\text r_\text A^2}}{\text K\frac{\text q_\text B}{\text r_\text B^2}} \\[1em] = \frac{\text q_\text A}{\text q_\text B} \times \frac{\text r_\text B^2}{\text r_\text A^2} \\[1em] = \frac{1}{2}\left(\frac{10}{5}\right)^2 \\[1em] = \frac{1}{2} \times 4 \\[1em] = \frac{2}{1}\end{aligned}$$

Hence the ratio is 2 : 1.

Both statement I and statement II are correct.

Reason — The electric field is zero at every point inside a conductor, so

$$\begin{aligned}\text E = -\frac{\text{dV}}{\text{dr}} \\[1em] = 0\end{aligned}$$

there is no potential gradient within the conductor and the potential is constant throughout its volume and over its surface. Hence Statement I is correct.

The surface of a conductor is therefore an equipotential surface. Since the electric field is always at right angles to an equipotential surface, the field just outside a charged conductor is perpendicular to the surface at every point. Hence Statement II is also correct.

4 : 1

Reason —

Given, n = 64, r = 0.02 m and q = 5 μC.

From the conservation of volume,

$$\begin{aligned}\text n\left(\frac{4}{3}\pi\text r^3\right) = \frac{4}{3}\pi\text R^3 \\[1em] \Rightarrow \quad \text R = (64)^{1/3}\text r \\[1em] = 4 \times 0.02 \\[1em] = 0.08\ \text m\end{aligned}$$

The surface charge densities of the small and the big drop are

$\sigma_\text r = \frac{\text q}{4\pi\text r^2}, \qquad \sigma_\text R = \frac{\text{nq}}{4\pi\text R^2}$

Therefore

$$\begin{aligned}\frac{\sigma_\text R}{\sigma_\text r} = \frac{\text{nq}}{4\pi\text R^2} \times \frac{4\pi\text r^2}{\text q} \\[1em] = \text n\frac{\text r^2}{\text R^2} \\[1em] = 64 \times \frac{(0.02)^2}{(0.08)^2} \\[1em] = 64 \times \frac{1}{16} \\[1em] = \frac{4}{1}\end{aligned}$$

Hence the ratio is 4 : 1.

6 V m^-1, directed along negative x-axis

Reason —

Given, V = 3x² volt.

The potential depends only on x, so the field has only an x-component,

$$\begin{aligned}\vec{\text E} = -\frac{\text{dV}}{\text{dx}}\hat{\text i} \\[1em] = -\frac{\text d}{\text{dx}}(3\text x^2)\hat{\text i} \\[1em] = -3(2\text x)\hat{\text i} \\[1em] = -6\text x\ \hat{\text i}\end{aligned}$$

At the point (1, 0, 3), x = 1 m,

$\vec{\text E} = -6\ \hat{\text i}\ \text V\text m^{-1}$

The magnitude is 6 V m^-1 and the negative sign shows that it is directed along the negative x-axis.

Options 1, 2 and 3 are correct.

Reason —

The distance of the centre O of a regular hexagon of side a from any of its sides (the apothem) is $\frac{\sqrt3}{2}\text a$. Since the perpendicular drawn from each charge to the nearest side passes through O and is bisected by that side, the distance of each charge from O is

$$\begin{aligned}\text d = 2 \times \frac{\sqrt3}{2}\text a \\[1em] = \sqrt3\ \text a\end{aligned}$$

Option 1 : When x = q, all the six charges are equal and are symmetrically placed about O. The fields due to diametrically opposite charges cancel in pairs, so the resultant electric field at O is zero. Correct.

Option 2 : When x = − q, we may write the sixth charge as q + (− 2q). The six equal charges q give zero field at O, so only the extra charge − 2q contributes,

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{2\text q}{(\sqrt3 \text a)^2} \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{2\text q}{3\text a^2} \\[1em] = \frac{\text q}{6\pi\epsilon_0\text a^2}\end{aligned}$$

Correct.

Option 3 : When x = 2q, the total charge is 5q + 2q = 7q, and every charge is at the same distance $\sqrt3 \text a$ from O. Since potential is a scalar,

$$\begin{aligned}\text V_\text O = \frac{1}{4\pi\epsilon_0}\frac{7\text q}{\sqrt3 \text a} \\[1em] = \frac{7\text q}{4\sqrt3 \pi\epsilon_0\text a}\end{aligned}$$

Correct.

Option 4 : When x = − 3q, the total charge is 5q − 3q = 2q, so

$$\begin{aligned}\text V_\text O = \frac{1}{4\pi\epsilon_0}\frac{2\text q}{\sqrt3 \text a} \\[1em] = \frac{\text q}{2\sqrt3 \pi\epsilon_0\text a}\end{aligned}$$

which is positive and not equal to the value stated. Not correct.

90°

Reason — There is no potential gradient along any direction parallel to an equipotential surface, so the electric field can have no component parallel to the surface,

$$\begin{aligned}\text E_{\text{parallel}} = -\frac{\text{dV}}{\text{dr}} \\[1em] = 0\end{aligned}$$

Hence the electric field, and with it the lines of force, are always at right angles to the equipotential surface.

more on smaller sphere

Reason — The potential of a charged conducting sphere is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R} \quad \Rightarrow \quad \text V \propto \frac{1}{\text R}$

for a given charge q. Since R₁ >> R₂, we get V₂ >> V₁, that is, the potential is more on the smaller sphere. The potential does not depend on the material of the sphere.

$\frac{3}{8}\text q\text K$

Reason —

From the figure, the two charges − q and + q are 3 cm on either side of the centre O, so the half-length of the dipole is l = 3 cm = 3 × 10^-2 m, and the point P lies on the axis at r = 5 cm = 5 × 10^-2 m from O.

The potential at an axial point of a dipole is

$$\begin{aligned}\text V = \text K\frac{\text p}{\text r^2 - \text l^2} \\[1em] = \text K\frac{\text q(2\text l)}{\text r^2 - \text l^2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}= \frac{\text{Kq}(6\times10^{-2})}{(5^2 - 3^2)\times10^{-4}} \\[1em] = \frac{6\text{Kq}\times10^{-2}}{16\times10^{-4}} \\[1em] = \frac{6\text{Kq}\times10^2}{16} \\[1em] = \frac{3}{8}\text q\text K \times 10^2\ \text V\end{aligned}$$

Hence, in units of 10² V, the potential at P is $\frac{3}{8}\text q\text K$.

$\frac{\lambda}{2\epsilon_0}$

Reason —

Given, two concentric half rings of radii R₁ and R₂ having the same linear charge density λ.

The length of a half ring of radius R is πR, so the charges on the two half rings are

$\text Q_1 = \lambda(\pi\text R_1), \qquad \text Q_2 = \lambda(\pi\text R_2)$

Every element of a half ring is at the same distance from its centre, so the potential at O is

$$\begin{aligned}\text V = \frac{1}{4\pi\epsilon_0}\frac{\text Q_1}{\text R_1} + \frac{1}{4\pi\epsilon_0}\frac{\text Q_2}{\text R_2} \\[1em] = \frac{1}{4\pi\epsilon_0}\frac{\lambda\pi\text R_1}{\text R_1} + \frac{1}{4\pi\epsilon_0}\frac{\lambda\pi\text R_2}{\text R_2} \\[1em] = \frac{\lambda\pi}{4\pi\epsilon_0}(1 + 1) \\[1em] = \frac{\lambda}{2\epsilon_0}\end{aligned}$$

zero

Reason —

Given, a thin charged spherical shell of radius 3 cm carrying a charge of 1 μC, with C at the centre and P on the surface.

The electric field is zero at every point inside a charged spherical shell, so there is no potential gradient inside it. Hence the potential at every interior point is the same as that at the surface,

$$\begin{aligned}\text V_\text C = \text V_\text P = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R} \\[1em] \Rightarrow \quad \text V_\text C - \text V_\text P = 0\end{aligned}$$

A is true, but R is false.

Reason —

Explanation

Assertion (A) is correct : The potential at any point on the axis of a short dipole is

$\text V = \pm\frac{1}{4\pi\epsilon_0}\frac{|\vec{\text P}|}{\text r^2}$

Substituting P = 4 × 10^-6 C-m and r = 2 m,

$$\begin{aligned}\text V = \pm\frac{(9\times10^9)(4\times10^{-6})}{(2)^2} \\[1em] = \pm\frac{3.6\times10^4}{4} \\[1em] = \pm 9\times10^3\ \text V\end{aligned}$$

which is exactly the value stated.

Reason (R) is incorrect : The expression given carries an extra factor of 2. The correct relation for the potential at an axial point of a short dipole is $\text V = \pm\frac{\text P}{4\pi\epsilon_0\text r^2}$, without the factor 2. The factor 2 belongs to the expression for the electric intensity at an axial point, $\text E = \frac{2\text P}{4\pi\epsilon_0\text r^3}$.

Therefore, Assertion is true but Reason is false.

Given,

• Potential difference, V = 0.1 V
• Charge on the electron, e = 1.6 × 10^-19 C
• Boltzmann constant, K_B = 1.38 × 10^-23 J/K

The average translational kinetic energy of a gas molecule is

$\text{KE} = \frac{3}{2}\text K_\text B\text T$

and the kinetic energy of the electron accelerated through a potential difference V is eV. Equating the two,

$\frac{3}{2}\text K_\text B\text T = \text e\text V$

Substituting the values,

$$\begin{aligned}\frac{3}{2}(1.38\times10^{-23})\text T = (1.6\times10^{-19})(0.1) \\[1em] 2.07\times10^{-23}\ \text T = 1.6\times10^{-20} \\[1em] \text T = \frac{1.6\times10^{-20}}{2.07\times10^{-23}} \\[1em] = 773\ \text K\end{aligned}$$

Converting to the Celsius scale,

$$\begin{aligned}\text t = 773 - 273 \\[1em] = 500\ °\text C\end{aligned}$$

Hence, the required temperature is 500 °C.

Given,

• Number of drops, n = 27
• Potential of each small drop, V₁ = 10 V

Let q₁ and r₁ be the charge and radius of a small drop, and Q and R those of the bigger drop. Since the volume remains constant,

$$\begin{aligned}\frac{4}{3}\pi\text R^3 = 27 \times \frac{4}{3}\pi\text r_1^3 \\[1em] \Rightarrow \quad \text R = 3\text r_1\end{aligned}$$

Also, the charge is conserved, so Q = 27q₁.

The electrostatic potential energy of a charged spherical drop is

$\text U = \frac{3}{5}\frac{\text{kq}^2}{\text r}$

For a small drop :

$\text U_\text s = \frac{3}{5}\frac{\text{kq}_1^2}{\text r_1} \qquad \dots(\text i)$

For the bigger drop :

$$\begin{aligned}\text U_\text B = \frac{3}{5}\frac{\text k(27\text q_1)^2}{3\text r_1} \\[1em] = \frac{3}{5}\frac{\text k(27)^2\text q_1^2}{3\text r_1} \qquad \dots(\text{ii})\end{aligned}$$

Dividing Eq. (ii) by Eq. (i),

$$\begin{aligned}\frac{\text U_\text B}{\text U_\text s} = \frac{(27)^2}{3} \\[1em] = \frac{729}{3} \\[1em] = 243\end{aligned}$$

Hence, the potential energy of the bigger drop is 243 times that of a smaller drop.

Given,

• Number of drops, n = 512
• Potential of each drop, V = 2 V

Let q and r be the charge and radius of a small drop, and Q and R those of the single big drop. Since the volume remains constant when the drops join,

$$\begin{aligned}512 \times \frac{4}{3}\pi\text r^3 = \frac{4}{3}\pi\text R^3 \\[1em] \Rightarrow \quad \text R = (512)^{1/3}\text r \\[1em] = 8\text r\end{aligned}$$

For a small drop,

$$\begin{aligned}\text V = \frac{\text{kq}}{\text r} \\[1em] = 2 \qquad \dots(\text i)\end{aligned}$$

For the big drop, whose charge is Q = 512q,

$\text V_{\text{net}} = \frac{\text k(512\text q)}{8\text r} \qquad \dots(\text{ii})$

Dividing Eq. (ii) by Eq. (i),

$$\begin{aligned}\frac{\text V_{\text{net}}}{\text V} = \frac{512}{8} \\[1em] = 64 \\[1em] \text V_{\text{net}} = 64 \times 2 \\[1em] = 128\ \text V\end{aligned}$$

Hence, the potential of the single big drop is 128 V.

Given,

• Charge − Q at the origin (0, 0)
• Charge $+\frac{\text Q}{\sqrt3}$ at the point (2, 0)

Let (x, y) be any point of the zero-potential circle. Since potential is a scalar,

$$\begin{aligned}\text V(\text x, \text y) = \frac{1}{4\pi\epsilon_0}\left(-\frac{\text Q}{\sqrt{\text x^2 + \text y^2}} + \frac{\text Q}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}}\right) \\[1em] = 0 \\[1em] \Rightarrow \quad \frac{1}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}} = \frac{1}{\sqrt{\text x^2 + \text y^2}}\end{aligned}$$

Squaring both sides and cross-multiplying,

$$\begin{aligned}\text x^2 + \text y^2 = 3\left[(\text x - 2)^2 + \text y^2\right] \\[1em] \text x^2 + \text y^2 = 3\text x^2 - 12\text x + 12 + 3\text y^2 \\[1em] 2\text x^2 + 2\text y^2 - 12\text x + 12 = 0\end{aligned}$$

Dividing throughout by 2,

$\text x^2 + \text y^2 - 6\text x + 6 = 0$

Completing the square in x,

$$\begin{aligned}(\text x - 3)^2 + \text y^2 = 9 - 6 \\[1em] = 3 \\[1em] (\text x - 3)^2 + \text y^2 = \left(\sqrt3\right)^2\end{aligned}$$

Comparing with the standard form (x − h)² + (y − k)² = R²,

$$\begin{aligned}\text R = \sqrt3 \\[1em] = 1.73\ \text m\end{aligned}$$

Hence, the value of R is 1.73 metre.

Given,

• Charge − Q at the origin (0, 0)
• Charge $+\frac{\text Q}{\sqrt3}$ at the point (2, 0)
• The zero-potential circle has its centre at (b, 0)

Let (x, y) be any point lying on the circle of zero potential. Its distances from the two charges are

$\text r_1 = \sqrt{\text x^2 + \text y^2}, \qquad \text r_2 = \sqrt{(\text x - 2)^2 + \text y^2}$

Since potential is a scalar quantity, the potentials due to the two charges add algebraically,

$$\begin{aligned}\text V(\text x, \text y) = \frac{1}{4\pi\epsilon_0}\left(-\frac{\text Q}{\sqrt{\text x^2 + \text y^2}} + \frac{\text Q}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}}\right) \\[1em] = 0\end{aligned}$$

Cancelling $\frac{\text Q}{4\pi\epsilon_0}$ and transposing,

$$\begin{aligned}\frac{1}{\sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}} = \frac{1}{\sqrt{\text x^2 + \text y^2}} \\[1em] \Rightarrow \quad \sqrt{\text x^2 + \text y^2} = \sqrt3 \sqrt{(\text x - 2)^2 + \text y^2}\end{aligned}$$

Squaring both sides,

$\text x^2 + \text y^2 = 3\left[(\text x - 2)^2 + \text y^2\right]$

Expanding the right-hand side,

$\text x^2 + \text y^2 = 3\text x^2 - 12\text x + 12 + 3\text y^2$

Bringing all the terms to one side,

$2\text x^2 + 2\text y^2 - 12\text x + 12 = 0$

Dividing throughout by 2,

$\text x^2 + \text y^2 - 6\text x + 6 = 0$

Completing the square in x,

$$\begin{aligned}(\text x^2 - 6\text x + 9) + \text y^2 = 9 - 6 \\[1em] (\text x - 3)^2 + \text y^2 = \left(\sqrt3\right)^2\end{aligned}$$

Comparing this with the standard equation of a circle (x − h)² + (y − k)² = R², the centre of the circle is (3, 0).

Since the centre is given as (b, 0),

$\text b = 3.00\ \text m$

Hence, the value of b is 3.00 metre.

Given,

• The charges − q and + q are separated by 2l, forming a dipole of moment p₁ = q(2l)
• The charges + 2q and − 2q are separated by 4l, forming a dipole of moment p₂ = 2q(4l) = 8ql
• The point P is at a distance r from the centre O, and the line OP makes an angle of 60° with the dipole axis

From the figure, the two dipole moments are oppositely directed. Hence the net dipole moment is

$$\begin{aligned}\text p = \text p_2 - \text p_1 \\[1em] = 8\text q\text l - 2\text q\text l \\[1em] = 6\text q\text l\end{aligned}$$

directed along the direction of $\vec{\text p}_2$, that is, opposite to $\vec{\text p}_1$. Therefore the angle between the net dipole moment and the line OP is

$$\begin{aligned}\theta = 180° - 60° \\[1em] = 120°\end{aligned}$$

The potential at a general point due to a short dipole is

$\text V = \frac{1}{4\pi\epsilon_0}\frac{\text p\cos\theta}{\text r^2}$

Substituting the values,

$$\begin{aligned}\text V = (9\times10^9)\frac{(6\text q\text l)\cos 120°}{\text r^2} \\[1em] = (9\times10^9)\frac{(6\text q\text l)\left(-\frac{1}{2}\right)}{\text r^2} \\[1em] = -27\left[\frac{\text q\text l}{\text r^2}\right]\times10^9\ \text V\end{aligned}$$

Comparing with $\text V = -\alpha\left[\frac{\text q\text l}{\text r^2}\right]\times10^9$ V,

$\alpha = 27$

Hence, the value of α is 27.

Note: The value 9 × 10⁹ N m² C^-2 is the standard value of $\frac{1}{4\pi\epsilon_0}$, and it is this constant that has been used above to obtain the printed answer α = 27. The constant printed in the question as $\frac{1}{2\pi\epsilon_0}$ appears to be a misprint for $\frac{1}{4\pi\epsilon_0}$.

Given,

• Radius of the half ring, R = 10 cm = 0.1 m
• Linear charge density, λ = 4 n C m^-1 = 4 × 10^-9 C m^-1

Every element of the half ring is at the same distance R from the centre. Hence

$$\begin{aligned}\text V = \displaystyle\int \text K\frac{\text{dQ}}{\text R} \\[1em] = \text K\frac{\text Q}{\text R}\end{aligned}$$

The total charge on the half ring of length πR is

$\text Q = \lambda(\pi\text R)$

Therefore

$$\begin{aligned}\text V = \text K\frac{\lambda\pi\text R}{\text R} \\[1em] = \text K\lambda\pi\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text V = (9\times10^9)(4\times10^{-9})\pi \\[1em] = 36\pi\ \text{volt}\end{aligned}$$

Comparing with V = xπ volt,

$\text x = 36$

Hence, the value of x is 36.

Given,

• Linear charge density of the wire, λ = 5 n C/m = 5 × 10^-9 C/m
• Charge on the shell, Q = 10 nC = 10 × 10^-9 C, radius of the shell = 1 m
• P is at 0.5 m from the centre and R is at 2 m from the centre
• ln 2 = 0.7

Contribution of the wire : The field of an infinitely long wire is $\text E = \frac{2\text k\lambda}{\text x}$, so

$$\begin{aligned}\displaystyle\int_{\text V_\text P}^{\text V_\text R}\text{dV} = -\displaystyle\int_{0.5}^{2}\frac{2\text k\lambda}{\text x}\ \text{dx} \\[1em] \text V_\text R - \text V_\text P = -2\text k\lambda\ln\frac{2}{0.5} \\[1em] = -2\text k\lambda\ln 4 \\[1em] = -2\text k\lambda(2\ln 2)\end{aligned}$$

Substituting the values,

$$\begin{aligned}= -2(9\times10^9)(5\times10^{-9})(2 \times 0.7) \\[1em] = -90 \times 1.4 \\[1em] = -126\ \text V\end{aligned}$$

Contribution of the shell : The point P lies inside the shell, so the potential there equals that at the surface; the point R lies outside at 2 m,

$$\begin{aligned}\text V_\text R - \text V_\text P = \frac{\text{kQ}}{2} - \frac{\text{kQ}}{1} \\[1em] = -\frac{\text{kQ}}{2} \\[1em] = -\frac{(9\times10^9)(10\times10^{-9})}{2} \\[1em] = -45\ \text V\end{aligned}$$

Total :

$$\begin{aligned}\text V_\text R - \text V_\text P = -126 - 45 \\[1em] = -171\ \text V \\[1em] \Rightarrow \quad \text V_\text P - \text V_\text R = 171\ \text V\end{aligned}$$

Hence, the magnitude of the potential difference between P and R is 171 V.