Gauss' Theorem

100 Vm

Reason —

Given,

• Area of the frame, A = 10 m²
• Electric field, E = 20 N-C^-1
• Angle between the normal to the frame and the field, θ = 60°

The electric flux through a plane surface placed in a uniform electric field is

$\Phi_\text E = \text{EA}\cos\theta$

Substituting the values,

$$\begin{aligned}\Phi_\text E = 20 \times 10 \times \cos 60^\circ \\[1em] = 200 \times \frac{1}{2} \\[1em] \Phi_\text E = 100\ \text{Vm}\end{aligned}$$

zero

Reason —

Given,

• Number of electric dipoles located inside the closed surface = n

An electric dipole consists of two equal and opposite point-charges, + q and − q. The net charge of one dipole is therefore

$\text q + (-\text q) = 0$

For n such dipoles lying inside the closed surface, the algebraic sum of all the charges enclosed is again zero.

By Gauss' theorem, the electric flux through a closed surface is $\frac{1}{\epsilon_0}$ times the net charge enclosed by it,

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] = \frac{0}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

Hence the net electric flux emerging through the surface is zero.

14.4 × 10²⁰ Vm^-1

Reason —

Given,

• Radius of the nucleus of hydrogen atom, R = 10^-15 m
• Charge on the nucleus (a single proton), q = 1.6 × 10^-19 C
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

The nucleus is a uniformly charged sphere, so for a point on its surface the entire charge behaves as if concentrated at its centre,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2}$

Substituting the values,

$$\begin{aligned}\text E = \frac{(9.0 \times 10^9)(1.6 \times 10^{-19})}{(10^{-15})^2} \\[1em] = \frac{1.44 \times 10^{-9}}{10^{-30}} \\[1em] = 1.44 \times 10^{21} \\[1em] \text E = 14.4 \times 10^{20}\ \text{Vm}^{-1}\end{aligned}$$

$\text E = \left[\frac{1}{4\pi\epsilon_0}\right]\frac{2\lambda}{\text r}$

Reason —

Given,

• Linear charge density of the infinitely long linear charge X-X = λ
• Perpendicular distance of the point P from the linear charge = r

For the infinitely long linear charge, a coaxial cylindrical Gaussian surface of length l and radius r is drawn through the point P. The flux through the plane ends is zero, and over the curved surface E is constant and normal to the surface, so

$$\begin{aligned}\Phi_\text E = \int_\text A \vec{\text E} \cdot \text d\vec{\text A} \\[1em] \Phi_\text E = \int_\text A \text E\ \text{dA}\cos 0^\circ \\[1em] \Phi_\text E = \text E\int_\text A \text{dA} \\[1em] \Phi_\text E = \text E(2\pi \text{rl})\end{aligned}$$

The charge enclosed by this Gaussian surface is q = λl. Applying Gauss' theorem,

$$\begin{aligned}\text E(2\pi \text{rl}) = \frac{\lambda \text l}{\epsilon_0} \\[1em] \text E = \frac{\lambda}{2\pi\epsilon_0 \text r}\end{aligned}$$

Writing $\frac{1}{2\pi\epsilon_0} = \frac{1}{4\pi\epsilon_0} \times 2$,

$\text E = \left[\frac{1}{4\pi\epsilon_0}\right]\frac{2\lambda}{\text r}$

Thus the intensity is inversely proportional to the distance r from the linear charge, and not to r².

3 N/C

Reason —

Given,

• Surface charge density of each sheet, σ = 26.4 × 10^-12 C/m², of opposite signs
• ε₀ = 8.85 × 10^-12 C²/N-m²

For two parallel sheets carrying equal and opposite charge densities, the intensities due to the two sheets are in the same direction at a point between them. Hence

$$\begin{aligned}\text E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \\[1em] \text E = \frac{\sigma}{\epsilon_0}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text E = \frac{26.4 \times 10^{-12}}{8.85 \times 10^{-12}} \\[1em] \text E = 2.98 \approx 3\ \text{N/C}\end{aligned}$$

is the same for all the figures

Reason —

In each of the four figures the same charge + q lies inside the closed surface S, only the shape of the surface being different.

By Gauss' theorem,

$\Phi_\text E = \frac{\text q}{\epsilon_0}$

Gauss' theorem is valid for a closed surface no matter what its shape or size, and the flux passing through it depends only upon the net charge enclosed. It does not depend upon the shape of the surface or on the position of the charge inside it. Hence the electric flux is the same for all four figures.

E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q₂ and q₄ only.

Reason —

$\vec{\text E} \space$on S is the resultant field of all five charges. Only q₂ and q₄ lie inside S, so q = q₂ + q₄ on the right-hand side of Gauss' law.

the number of flux lines entering the surface must be equal to the number of flux lines leaving it

Reason — The electric flux is a measure of the number of lines of force passing through a surface. It is positive for the lines of force leaving the surface and negative for those entering the surface.

If

$\oint_\text S \vec{\text E} \cdot \text d\vec{\text S} = 0$

then the positive and the negative contributions cancel exactly, which means that the number of flux lines entering the surface is equal to the number of flux lines leaving it.

total flux through the surface of the sphere is $\frac{-\text Q}{\epsilon_0}$

Reason —

Given,

• Radius of the Gaussian surface = R
• Charges inside the surface : Q (at the centre) and − 2Q (at a distance $\frac{\text R}{2}$ from the centre)
• Charge outside the surface : 5Q (at a distance 2R from the centre)

The net charge enclosed is

$$\begin{aligned}\sum \text q = \text Q + (-2\text Q) \\[1em] \sum \text q = -\text Q\end{aligned}$$

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] \Phi_\text E = \frac{-\text Q}{\epsilon_0}\end{aligned}$$

$\frac{\text Q}{8\epsilon_0}$

Reason —

Given,

• Charge Q situated at the corner A of the cube

A charge placed at a corner of a cube is shared by eight such cubes meeting at that corner. To enclose the charge Q completely, eight identical cubes have to be placed around the corner A, so that the charge lies at the centre of the large cube so formed.

By Gauss' theorem, the total flux through this complete closed arrangement is $\frac{\text Q}{\epsilon_0}$. By symmetry this flux is shared equally by the eight cubes. Hence the flux through the given cube is

$$\begin{aligned}\Phi_\text E = \frac{1}{8}\left(\frac{\text Q}{\epsilon_0}\right) \\[1em] \Phi_\text E = \frac{\text Q}{8\epsilon_0}\end{aligned}$$

ERh

Reason —

Given,

• Base radius of the cone = R, height of the cone = h
• The cone stands with its base on a horizontal table, so its axis is vertical
• The electric field E is horizontal and hence parallel to the base

The cone together with its base forms a closed surface enclosing no charge, so the total flux through it is zero. As the base is horizontal and E is horizontal, the flux through the base is zero. Hence the flux entering the curved surface is equal in magnitude to the flux leaving it.

The flux entering the body is

$\Phi_\text E = \text E \times (\text{area of the cone projected on a plane perpendicular to } \vec{\text E})$

Viewed along the direction of the field, the cone projects as a triangle of base 2R and height h. Therefore

$$\begin{aligned}\text{projected area} = \frac{1}{2} \times 2\text R \times \text h \\[1em] \text{projected area} = \text{Rh} \\[1em] \Phi_\text E = \text{ERh}\end{aligned}$$

Note: The printed key selects option 2, $\frac{\text{ERh}}{2}$. The projected area is Rh, so the correct answer is option 3, ERh.

φ_A = − EπR², φ_B = 0, φ_C = EπR²

Reason —

Given,

• Radius of the cylinder = R, its axis parallel to the uniform electric field E
• A : left plane face, B : curved surface, C : right plane face

The area of each plane face is πR².

Surface A : The outward drawn normal points opposite to $\vec{\text E} \space$, so θ = 180°,

$$\begin{aligned}\phi_\text A = \text{EA}\cos 180^\circ \\[1em] \phi_\text A = -\text E\pi \text R^2\end{aligned}$$

Surface B : For every area element of the curved surface the outward drawn normal is at right angles to $\vec{\text E} \space$, so θ = 90°,

$$\begin{aligned}\phi_\text B = \text{EA}\cos 90^\circ \\[1em] \phi_\text B = 0\end{aligned}$$

Surface C : The outward drawn normal is along $\vec{\text E} \space$, so θ = 0,

$$\begin{aligned}\phi_\text C = \text{EA}\cos 0^\circ \\[1em] \phi_\text C = \text E\pi \text R^2\end{aligned}$$

The total flux through the closed cylinder is $\phi_\text A$ + $\phi_\text B$ + $\phi_\text C$ = 0, as it should be, since no charge is enclosed.

$\frac{\text q}{2\epsilon_0}$

Reason —

Given,

• Point-charge q at the centre of the circular rim of the hemisphere (from the figure)

This point is the centre of the sphere of which this hemisphere is one half.

The solid angle subtended by a complete closed surface at a point inside it is 4π steradian, and the total flux from the charge is $\frac{\text q}{\epsilon_0}$. The hemispherical surface subtends half of this solid angle, that is 2π steradian, at the charge.

Hence the flux through the hemispherical surface is half of the total flux,

$$\begin{aligned}\Phi_\text E = \frac{1}{2}\left(\frac{\text q}{\epsilon_0}\right) \\[1em] \Phi_\text E = \frac{\text q}{2\epsilon_0}\end{aligned}$$

$\frac{1}{2}\left(\frac{\text q}{\epsilon_0} - \phi\right)$

Reason —

Given,

• Charge enclosed within the hollow cylinder = q, placed centrally (from the figure)
• Electric flux through the curved surface B = φ
• A and C are the two plane surfaces

The hollow cylinder is a closed surface made up of the curved surface B and the two plane surfaces A and C.

By Gauss' theorem, the total flux through the closed cylinder is

$\phi_\text A + \phi_\text B + \phi_\text C = \frac{\text q}{\epsilon_0}$

The two plane surfaces A and C are situated symmetrically with respect to the charge, so the flux through each of them is the same, that is $\phi_\text A$ = $\phi_\text C$. Also $\phi_\text B$ = $\phi$ is given. Therefore

$$\begin{aligned}2\phi_\text A + \phi = \frac{\text q}{\epsilon_0} \\[1em] \phi_\text A = \frac{1}{2}\left(\frac{\text q}{\epsilon_0} - \phi\right)\end{aligned}$$

zero

Reason —

Given,

• Radius of the hemispherical body = R
• Uniform electric field E, parallel to the base

The curved surface together with the plane base forms a closed surface enclosing no charge. Hence by Gauss' theorem the total flux through it is zero,

$\phi_{\text{curved}} + \phi_{\text{base}} = 0$

The outward drawn normal to the plane base is perpendicular to the base and therefore perpendicular to $\vec{\text E} \space$, so

$$\begin{aligned}\phi_{\text{base}} = \text{EA}\cos 90^\circ \\[1em] \phi_{\text{base}} = 0 \\[1em] \therefore \quad \phi_{\text{curved}} = 0\end{aligned}$$

The flux linked with the curved surface is therefore zero.

$\frac{\sqrt{3}\lambda\text a}{\epsilon_0}$

Reason —

Given,

• Linear charge density of the string, λ
• Edge of the imaginary cube, a

By Gauss' theorem, the flux through the cube is $\frac{1}{\epsilon_0}$ times the charge enclosed by it. The charge enclosed is λ times the length of the string lying inside the cube. Hence the flux is maximum when this length is maximum.

The greatest length of a straight line that can lie within a cube of edge a is its body diagonal,

$$\begin{aligned}\text l_\text{max} = \sqrt{\text a^2 + \text a^2 + \text a^2} \\[1em] \text l_\text{max} = \sqrt{3}\ \text a\end{aligned}$$

The charge enclosed is then

$$\begin{aligned}\text q = \lambda \text l_\text{max} \\[1em] \text q = \sqrt{3}\ \lambda \text a \\[1em] \Phi_\text E = \frac{\text q}{\epsilon_0} \\[1em] \Phi_\text E = \frac{\sqrt{3}\lambda \text a}{\epsilon_0}\end{aligned}$$

This problem does not have spherical symmetry.

Reason — Gauss' theorem is mainly used for a symmetrical charge distribution. To obtain the electric field from it, the Gaussian surface is chosen to match the symmetry of the source, so that over the whole surface the field is everywhere normal to it and has the same magnitude. Only then can E be taken out of the integral $\oint \vec{\text E} \cdot \text d\vec{\text S}$.

An electric dipole consists of two equal and opposite point-charges separated by a small distance. Its field is not the same in magnitude at all points of a spherical surface drawn about it, and is not everywhere normal to that surface. The charge distribution has no spherical symmetry.

Gauss' theorem itself does not fail here; it remains valid for any closed surface. It is merely not convenient for finding E in this case.

All the charges

Reason —

The field $\vec{\text E} \space$at any point of a Gaussian surface is set up jointly by all the charges present, whether they lie inside the surface or outside it.

Hence in calculating the flux $\oint \vec{\text E} \cdot \text d\vec{\text S}$, the field $\vec{\text E} \space$appearing in the integral is the resultant field of + q₁, − q₁ and q₂, that is, of all the charges shown.

It should be noted, however, that the value of the flux depends only on the charges present inside the surface, since the net flux due to a charge outside a closed surface is zero.

$4\pi\epsilon_0 \text{Aa}^3$

Reason —

Given,

• Electric field acting radially outward, E = Ar
• Radius of the sphere centred at the origin = a

Since the field is radial and depends only on r, it has the same magnitude at every point of a sphere of radius a and is normal to it. At the surface of this sphere,

$\text E = \text{Aa}$

The electric flux through the spherical surface is

$$\begin{aligned}\Phi_\text E = \text E \times 4\pi \text a^2 \\[1em] = (\text{Aa})(4\pi \text a^2) \\[1em] \Phi_\text E = 4\pi \text A \text a^3\end{aligned}$$

By Gauss' theorem, $\Phi_\text E = \frac{\text q}{\epsilon_0}$. Therefore the charge contained in the sphere is

$$\begin{aligned}\text q = \epsilon_0 \Phi_\text E \\[1em] \text q = 4\pi\epsilon_0 \text{Aa}^3\end{aligned}$$

EπR²

Reason —

Given,

• Radius of the hemispherical body = R
• Uniform electric field E, perpendicular to the circular base (upward, as shown in the figure)

The curved surface together with the plane base forms a closed surface enclosing no charge, so by Gauss' theorem the total flux through it is zero. Hence the magnitude of the flux through the curved surface equals that through the plane base.

The base is a circle of area πR² whose normal is along $\vec{\text E} \space$. Therefore

$$\begin{aligned}\phi_{\text{curved}} = \text{EA}\cos 0^\circ \\[1em] = \text E \times \pi \text R^2 \\[1em] \phi_{\text{curved}} = \text E\pi \text R^2\end{aligned}$$

$\frac{\text q}{3\epsilon_0}$

Reason —

Given,

• Charge q distributed uniformly on a ring of radius r
• A sphere of the same radius r is constructed with its centre C lying on the periphery of the ring

Let O be the centre of the ring and C a point on the ring itself, taken as the centre of the sphere. Consider a point P on the ring such that ∠POC = θ. Since O is the centre of the ring, both P and C lie at a distance r from O, so the chord PC of the ring is

$\text{PC} = 2\text r\sin\frac{\theta}{2}$

The point P lies inside the sphere if PC < r, that is

$$\begin{aligned}2\text r\sin\frac{\theta}{2} \le \text r \quad \Rightarrow \quad \sin\frac{\theta}{2} \le \frac{1}{2} \quad \Rightarrow \quad \frac{\theta}{2} \le 30^\circ \\[1em] \therefore \quad \theta \le 60^\circ\end{aligned}$$

Measuring θ on either side of OC, the arc of the ring lying inside the sphere subtends an angle of 2 × 60° = 120° at the centre O. As the charge is uniformly distributed, the charge enclosed by the sphere is

$$\begin{aligned}\text q_\text{in} = \frac{120^\circ}{360^\circ}\text q \\[1em] \text q_\text{in} = \frac{\text q}{3}\end{aligned}$$

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\text q_\text{in}}{\epsilon_0} \\[1em] \Phi_\text E = \frac{\text q}{3\epsilon_0}\end{aligned}$$

A Gaussian surface is used to compute easily the electric field intensity due to certain charge systems or charged bodies, in cases where the use of Coulomb's law is quite difficult.

Explanation:
By choosing a closed surface having the same symmetry as the charge distribution, the field becomes normal to the surface and constant in magnitude over it, so that Gauss' theorem gives E at once.

The Gaussian surface should pass through the point at which the electric field is required.

Explanation:
Unless the surface passes through that point, the field obtained from Gauss' theorem would not be the field at the required location.

The field due to a charged conductor is twice that due to a sheet of charge.

Explanation:
Just outside a sheet of charge of surface charge density σ,

$\text E = \frac{\sigma}{2\epsilon_0}$

while just outside a charged conductor of surface charge density σ,

$\text E = \frac{\sigma}{\epsilon_0}$

The ratio of the electric flux through S₁ and S₂ is 1 : 1.

Explanation:
By Gauss' theorem, the flux through a closed surface is $\frac{1}{\epsilon_0}$ times the net charge enclosed by it. There is no additional charge between S₁ and S₂, so both the spheres enclose the same charge q.

$\Phi_1 = \frac{\text q}{\epsilon_0}, \qquad \Phi_2 = \frac{\text q}{\epsilon_0} \quad \Rightarrow \quad \frac{\Phi_1}{\Phi_2} = 1 : 1$

The flux does not depend upon the shape or the size of the surface enclosing the charge.

No, Gauss' law would not remain valid.

Explanation:
Gauss' theorem is valid for all vector fields obeying the inverse square law. If Coulomb's law involved a $\frac{1}{\text r^3}$ dependence, or any other power of r except 2, then the electric flux through a sphere of radius r would depend upon r also, instead of depending only on the charge enclosed. Hence Gauss' law would not hold.

Gauss' law in electrostatics : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface.

Mathematically,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

where ε₀ is the permittivity of free space. This is the integral form of Gauss' theorem.

A Gaussian surface is any closed surface, not passing through any point-charge, through which the flux of the electric field is calculated.

Explanation:
It is an arbitrary closed surface in three dimensional space. It may be of any shape and may or may not enclose any charge. The surface of a sphere, a cylinder or a cube are examples of valid Gaussian surfaces, whereas a disc surface or a square surface cannot be used as a Gaussian surface since these are not closed.

(i) The electric flux Φ_E will not be affected. Gauss' theorem is valid for a closed surface no matter what its shape or size, so the flux does not depend upon the shape or the volume of the surface enclosing the charge.

(ii) Φ_E will not be affected. The flux passing through a closed surface depends only upon the net charge enclosed by the surface and not upon the position of the charge inside it.

(iii) Φ_E will change. Since $\Phi_\text E = \frac{\text q}{\epsilon_0}$, where q is the net charge, taking the sign of the charge into account, the flux is an additive quantity and alters when a second charge is placed inside.

(iv) Φ_E will become zero. An electric dipole consists of two equal and opposite charges, so that the net charge enclosed by the surface is zero.

No, the electric intensity $\vec{\text E} \space$is not necessarily zero at all points on the surface.

When the net charge q inside a Gaussian surface is zero, Gauss' law gives

$$\begin{aligned}\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

This does not mean that $\vec{\text E} \space$is zero at every point of the surface. For example, if $\vec{\text E} \space$is non-zero but at right angles to the area vector $\text d\vec{\text A}$at all points of the surface, the value of $\oint \vec{\text E} \cdot \text d\vec{\text A}$is still zero.

The converse is, however, true. If $\vec{\text E} \space$is zero at every point on the Gaussian surface, then the flux through the surface is zero, and Gauss' law then predicts that there must be no net charge inside the Gaussian surface.

Gauss' theorem : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

Variation of electric field with distance :

In a charged metal ball the entire charge resides on the surface, so the ball behaves just like a charged thin spherical shell.

(i) Inside the ball (r < R) : A Gaussian surface drawn inside encloses no charge, so E = 0 everywhere.

(ii) At the surface (r = R) : The field is maximum,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2}$

(iii) Outside the ball (r > R) : The ball behaves as if its whole charge were concentrated at the centre, so

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2} \quad \text{that is} \quad \text E \propto \frac{1}{\text r^2}$

Hence the graph is zero from the centre up to the surface, rises abruptly to its maximum value at the surface and then decreases rapidly as $\frac{1}{\text r^2}$ at outside points.

The cylinder is immersed in the uniform field $\vec{\text E} \space$with its axis parallel to the field. The flux through the whole cylinder is the sum of the fluxes through the left plane face, the right plane face and the curved surface,

$\Phi_\text E = \int_{\text{left plane}} \vec{\text E} \cdot \text d\vec{\text A} + \int_{\text{right plane}} \vec{\text E} \cdot \text d\vec{\text A} + \int_{\text{curved surface}} \vec{\text E} \cdot \text d\vec{\text A}$

Drawing the outward normals, the angle between $\vec{\text E} \space$and $\text d\vec{\text A}$is 180° for all area elements on the left face, 0 for all elements on the right face and 90° for all elements on the curved surface. Therefore

$$\begin{aligned}\Phi_\text E = \int_{\text{left plane}} \text E\ \text{dA}\cos 180^\circ + \int_{\text{right plane}} \text E\ \text{dA}\cos 0^\circ + \int_{\text{curved surface}} \text E\ \text{dA}\cos 90^\circ \\[1em] = -\text E\int_{\text{left plane}} \text{dA} + \text E\int_{\text{right plane}} \text{dA} + 0 \\[1em] \Phi_\text E = 0\end{aligned}$$

Hence the electric flux through the cylinder is zero. This is expected from Gauss' theorem also, since the cylinder encloses no charge.

The electric field at a point distant r from the infinite line charge is

$\text E = \frac{\lambda}{2\pi\epsilon_0 \text r}$

directed radially outwards.

(i) The electric flux is zero for the two plane faces of the cylinder, because for these faces the angle between the field vector $\vec{\text E} \space$and the area vector $\text d\vec{\text A}$is 90°.

(ii) $\vec{\text E} \space$is not zero over any of the surfaces. It cannot be zero at a finite distance from the line charge.

(iii) $|\vec{\text E}|$ is constant over the curved surface of the cylinder, because every point of the curved surface is at the same perpendicular distance from the infinite line charge.

(iv) $|\vec{\text E}|$ varies over the plane faces of the cylinder, since different points of a plane face are at different distances from the line charge, and E decreases as the distance increases.

Given,

• Charge enclosed in the hollow cylinder = q
• Electric flux through the curved surface of the cylinder = Φ_C

Let the electric flux through each plane surface be Φ_P. The hollow cylinder is a closed surface made up of the curved surface and the two plane surfaces, so the total flux through the cylinder is

$\Phi_\text C + \Phi_\text P + \Phi_\text P = \Phi_\text C + 2\Phi_\text P$

Hence by Gauss' theorem,

$$\begin{aligned}\Phi_\text C + 2\Phi_\text P = \frac{\text q}{\epsilon_0} \\[1em] 2\Phi_\text P = \frac{\text q}{\epsilon_0} - \Phi_\text C \\[1em] \Phi_\text P = \frac{1}{2}\left(\frac{\text q}{\epsilon_0} - \Phi_\text C\right)\end{aligned}$$

Hence the electric flux through either plane surface of the cylinder is $\frac{1}{2}\left(\frac{\text q}{\epsilon_0} - \Phi_\text C\right)$.

Gauss' law states that the electric flux through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the net charge within the surface,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

Let a Gaussian surface be drawn entirely within the material of a charged conductor. The charge given to a conductor does not reside inside it; the entire charge lies on its surface. Hence the charge enclosed by this Gaussian surface is q = 0, and

$\oint \vec{\text E} \cdot \text d\vec{\text A} = 0$

Since such a Gaussian surface may be drawn about every interior point of the conductor, the field must vanish at each of them.

Hence the electric field does not exist inside a charged conductor, that is, $\vec{\text E} \space$= 0.

Since the electric field inside a charged conductor is zero, electric lines of force cannot get into the empty space enclosed by a hollow conductor. A conductor can therefore act as an electrostatic shield.

When the sensitive instrument is covered fully by a metallic surface, the lines of force of the external fields pass along the outside of the surface and do not enter inside it. The field inside the metallic covering remains zero, so the instrument placed within it remains safe from the external electric fields.

This phenomenon is known as electrostatic shielding.

Given,

• Inner radius of the spherical conducting shell = r₁, outer radius = r₂
• Charge on the shell = Q
• Charge placed at the centre of the shell = q

The charge Q given to the conducting shell resides on its outer surface. The charge q placed at the centre induces a charge − q on the inner surface of the shell and a charge + q on its outer surface.

Inner surface : The charge on the inner surface is − q and its area is $4\pi \text r_1^2$. Hence the surface charge density is

$\sigma_1 = \frac{-\text q}{4\pi \text r_1^2}$

Outer surface : The total charge on the outer surface is (Q + q) and its area is $4\pi \text r_2^2$. Hence

$\sigma_2 = \frac{\text Q + \text q}{4\pi \text r_2^2}$

Hence the surface charge densities on the inner and the outer surfaces of the shell are $-\frac{\text q}{4\pi \text r_1^2}$ and $\frac{\text Q + \text q}{4\pi \text r_2^2}$ respectively.

Given,

• Mass of the oil-drop = m
• Area of each conducting plate = A metre²
• Charges on the plates = + q and − q coulomb

Let the charge on the drop be q′ coulomb. The drop is suspended in equilibrium, so the upward electric force on it balances its weight,

$\text q'\text E = \text{mg} \qquad \dots(\text i)$

where E is the electric field between the plates.

The surface charge density on each plate is $\sigma = \frac{\text q}{\text A}$. The field between two parallel conducting plates carrying equal and opposite charges is

$$\begin{aligned}\text E = \frac{\sigma}{\epsilon_0} \\[1em] \text E = \frac{\text q}{\text A\epsilon_0} \qquad \dots(\text{ii})\end{aligned}$$

Substituting equation (ii) in equation (i),

$$\begin{aligned}\text q' = \frac{\text{mg}}{\text E} \\[1em] = \frac{\text{mg}}{\left(\frac{\text q}{\text A\epsilon_0}\right)} \\[1em] \text q' = \frac{\text{mgA}\epsilon_0}{\text q}\end{aligned}$$

Hence the charge on the drop is $\frac{\text{mgA}\epsilon_0}{\text q}$ coulomb.

The field will remain unchanged, that is, it will still be E.

Explanation:
Let the surface area of the plate be A.

Non-conducting plate : The charge q is distributed uniformly over its surface, so the surface charge density is $\sigma = \frac{\text q}{\text A}$. The field just outside a sheet of charge is

$$\begin{aligned}\text E = \frac{\sigma}{2\epsilon_0} \\[1em] \text E = \frac{\text q}{2\epsilon_0 \text A}\end{aligned}$$

Metallic plate : The charge q now spreads over both the surfaces of the plate, so the surface charge density is $\sigma = \frac{\text q}{2\text A}$. The field just outside a charged conductor is

$$\begin{aligned}\text E = \frac{\sigma}{\epsilon_0} \\[1em] \text E = \frac{\text q}{2\epsilon_0 \text A}\end{aligned}$$

which is the same as before.

Given,

• Area of the plane surface, A = 10 m², lying in the X-Z plane
• Electric field, $\vec{\text E} = (2\hat{\text i} + 4\hat{\text j} + 7\hat{\text k})$ N-C^-1

The surface lies in the X-Z plane, so the normal to the surface is along the Y-axis. Hence the area vector is

$\vec{\text A} = 10\ \hat{\text j}\ \text m^2$

The electric flux through the surface is

$\Phi_\text E = \vec{\text E} \cdot \vec{\text A}$

Substituting the values,

$\Phi_\text E = (2\hat{\text i} + 4\hat{\text j} + 7\hat{\text k}) \cdot (10\ \hat{\text j})$

Since $\hat{\text i} \cdot \hat{\text j} = 0$, $\hat{\text j} \cdot \hat{\text j} = 1$ and $\hat{\text k} \cdot \hat{\text j} = 0$,

$$\begin{aligned}\Phi_\text E = 4 \times 10 \\[1em] \Phi_\text E = 40\ \text{V-m}\end{aligned}$$

Hence the electric flux passing through the surface is 40 V-m.

Given,

• Charges enclosed by S₁ : q, − 2q and 3q
• Charges enclosed by S₂ : q, − 2q, 3q, Q and − q
• Φ₂ = 4Φ₁

Flux through S₁ :

$$\begin{aligned}\Phi_1 = \frac{\text q + (-2\text q) + 3\text q}{\epsilon_0} \\[1em] \Phi_1 = \frac{2\text q}{\epsilon_0}\end{aligned}$$

Flux through S₂ :

$$\begin{aligned}\Phi_2 = \frac{\text q + (-2\text q) + 3\text q + \text Q + (-\text q)}{\epsilon_0} \\[1em] \Phi_2 = \frac{\text q + \text Q}{\epsilon_0}\end{aligned}$$

Using Φ₂ = 4Φ₁,

$$\begin{aligned}\frac{\text q + \text Q}{\epsilon_0} = 4\left(\frac{2\text q}{\epsilon_0}\right) \\[1em] \text q + \text Q = 8\text q \\[1em] \text Q = 7\text q\end{aligned}$$

Hence the required relation is Q = 7q.

Electric flux : The electric flux is a measure of the number of lines of force passing through some surface held in the electric field. It is denoted by Φ_E.

If $\text d\vec{\text A}$is a small area element of a surface placed in a field $\vec{\text E} \space$, the scalar product $\vec{\text E} \cdot \text d\vec{\text A}$is the electric flux through the element, and for the entire surface

$\Phi_\text E = \int_\text A \vec{\text E} \cdot \text d\vec{\text A}$

The electric flux linked with a surface is therefore obtained by integrating the normal component of the electric field over that surface. For a plane surface of area A held in a uniform field, the normal to it making an angle θ with the field,

$\Phi_\text E = \text{EA}\cos\theta$

Electric flux density : In an electric field, the ratio of the electric flux Φ_E through a surface to the area A of the surface is called the electric flux density at the location of the surface,

$\text{electric flux density} = \frac{\Phi_\text E}{\text A}$

For a plane surface normal to the electric field, Φ_E = EA, so that

$$\begin{aligned}\text{electric flux density} = \frac{\text{EA}}{\text A} \\[1em] \text{electric flux density} = \text E\end{aligned}$$

Its unit is the same as that of the electric field. Electric flux is a scalar, but the electric flux density is a vector quantity.

Negative flux : The electric flux is positive for the lines of force pointing outwards, that is, leaving the surface, and negative for those pointing inwards, that is, entering the surface.

For field lines entering a plane surface normally, θ = 180° and

$$\begin{aligned}\Phi_\text E = \text{EA}\cos 180^\circ \\[1em] \Phi_\text E = -\text{EA}\end{aligned}$$

The negative sign therefore shows that the lines of force are entering the surface.

SI unit of electric flux : From Φ_E = EA cos θ,

$\frac{\text N}{\text C} \times \text m^2 = \text{N-m}^2\text C^{-1}$

As E is also expressed in V/m, another SI unit of Φ_E is

$\frac{\text V}{\text m} \times \text m^2 = \text{V-m}$

Dimensions of electric flux :

$$\begin{aligned}\text{dimensions of } \Phi_\text E = \text{dimensions of E} \times \text{dimensions of A} \\[1em] = \left[\frac{\text{MLT}^{-2}}{\text{AT}}\right] \times [\text L^2] \\[1em] \text{dimensions of } \Phi_\text E = [\text{ML}^3\text T^{-3}\text A^{-1}]\end{aligned}$$

Consider an infinitely long, uniformly charged wire (taken as positive) whose linear charge density, that is, charge per unit length, is λ coulomb/metre. The electric intensity $\vec{\text E} \space$is to be found at a point P at a perpendicular distance r metre from the wire.

Through P, take a coaxial cylindrical Gaussian surface of radius r and length l. Every point of its curved surface is at the same distance r from the wire, so by symmetry the field has the same magnitude E at all these points and points radially outwards. The area vector $\text d\vec{\text A}$of any element dA of the curved surface is also radially outward, so $\vec{\text E} \space$and $\text d\vec{\text A}$point the same way. Therefore

$$\begin{aligned}\vec{\text E} \cdot \text d\vec{\text A} = \text E\ \text{dA}\cos 0 \\[1em] \vec{\text E} \cdot \text d\vec{\text A} = \text E\ \text{dA}\end{aligned}$$

Hence the electric flux through the curved surface of the Gaussian surface is

$$\begin{aligned}\Phi_\text E = \int_\text A \vec{\text E} \cdot \text d\vec{\text A} \\[1em] = \int_\text A \text E\ \text{dA} \\[1em] = \text E\int_\text A \text{dA} \qquad (\because \text{E is same everywhere}) \\[1em] \Phi_\text E = \text E(2\pi \text{rl})\end{aligned}$$

On each plane end the area vector is parallel to the wire while $\vec{\text E} \space$is perpendicular to it, so $\vec{\text E} \cdot \text d\vec{\text A} = 0$ there and the plane ends contribute no flux. The total flux through the Gaussian surface is therefore

$\Phi_\text E = \text E(2\pi \text{rl}) \qquad \dots(\text i)$

But by Gauss' theorem this must be $\frac{\text q}{\epsilon_0}$, where q is the net charge enclosed by the Gaussian surface. Here q = λl, so that

$\Phi_\text E = \frac{\lambda \text l}{\epsilon_0} \qquad \dots(\text{ii})$

Comparing equations (i) and (ii),

$$\begin{aligned}\text E(2\pi \text{rl}) = \frac{\lambda \text l}{\epsilon_0} \\[1em] \text E = \frac{\lambda}{2\pi\epsilon_0 \text r}\end{aligned}$$

In vector notation,

$\vec{\text E} = \frac{\lambda}{2\pi\epsilon_0 \text r}\hat{\text r}$

where $\hat{\text r}$ is the unit vector in the direction of r. The direction of $\vec{\text E} \space$is radially outwards for a positive charge.

Hence the electric field of a linear charge is inversely proportional to the perpendicular distance r from it and, for a positive charge distribution, points perpendicularly away from the linear charge.

Gauss' theorem : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface.

Mathematical expression :

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

where ε₀ is the permittivity of free space.

Electric field due to an infinite plane sheet of charge :

Consider a thin sheet of charge of infinite extent, with surface charge density (charge per unit area) σ. The electric intensity is required at a point P at a distance r from the sheet.

Take P′ as the mirror image of P on the other side of the sheet. The Gaussian surface is a cylinder that passes through the sheet, its two plane ends, each of area A, lying parallel to the sheet and passing through P and P′.

As the sheet extends to infinity, $\vec{\text E} \space$near it on either side is perpendicular to the sheet, pointing outwards for a positively charged sheet. It is therefore normal to the plane ends of the cylinder and parallel to its curved surface, and it has equal magnitude at P and P′. The flux through the two plane ends is then

$\Phi_\text E = \oint_\text A \vec{\text E} \cdot \text d\vec{\text A} + \oint_\text A \vec{\text E} \cdot \text d\vec{\text A}$

Since $\vec{\text E} \space$and $\text d\vec{\text A}$are parallel, $\vec{\text E} \cdot \text d\vec{\text A} = \text E\ \text{dA}$, so that

$\Phi_\text E = \oint_\text A \text E\ \text{dA} + \oint_\text A \text E\ \text{dA}$

But E is the same for all points on a plane end. Therefore

$$\begin{aligned}\Phi_\text E = \text E\oint \text{dA} + \text E\oint \text{dA} \\[1em] = \text{EA} + \text{EA} \\[1em] \Phi_\text E = 2\text{EA}\end{aligned}$$

The flux through the curved surface of the Gaussian cylinder is zero, because $\vec{\text E} \space$and $\text d\vec{\text A}$are at right angles everywhere on the curved surface. Hence the total flux through the Gaussian cylinder is

$\Phi_\text E = 2\text{EA} \qquad \dots(\text i)$

By Gauss' theorem, $\Phi_\text E = \frac{\text q}{\epsilon_0}$, where q is the net charge enclosed by the Gaussian cylinder. Here q = σA, and so

$\Phi_\text E = \frac{\sigma \text A}{\epsilon_0} \qquad \dots(\text{ii})$

Comparing equations (i) and (ii),

$$\begin{aligned}2\text{EA} = \frac{\sigma \text A}{\epsilon_0} \\[1em] \text E = \frac{\sigma}{2\epsilon_0}\end{aligned}$$

The result does not contain r, so the field has the same value at all points 'close' to the sheet, for any sheet of charge. Moving farther from the sheet brings more of its charge into view, and this makes up for the weakening of the field with distance.

$\vec{\text E} \space$at any point is directed away from the sheet of positive charge. If the sheet be of negative charge, then $\vec{\text E} \space$would be directed towards the sheet.

Gauss' law connects the electric field just outside a charged conductor in equilibrium with the charge distributed over its surface.

Consider a positively charged conductor in vacuum or in air. All the charge given to a conductor resides on its surface, and in equilibrium the electric field at the surface is normal to it. Because the conductor is irregular in shape, the surface charge density varies from place to place.

To find the field at a point P just outside the conductor, where the surface charge density is σ, a small closed cylinder is taken as the Gaussian surface. One plane face, of area A, passes through P; the other, also of area A, lies inside the conductor, parallel to its surface.

The inner plane face contributes no flux because the electric field inside a charged conductor is zero, and the curved surface contributes none because $\vec{\text E} \space$and $\text d\vec{\text A}$are perpendicular over it. Only the outer plane face through P contributes, so the flux through the Gaussian cylinder is

$$\begin{aligned}\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} \\[1em] = \oint \text E \cdot \text{dA}\cos 0^\circ \\[1em] = \text E\oint \text{dA} \\[1em] \Phi_\text E = \text{EA} \qquad \dots(\text i)\end{aligned}$$

But from Gauss' law, $\Phi_\text E = \frac{\text q}{\epsilon_0}$, where q (= σA) is the net charge enclosed by the Gaussian surface, and so

$\Phi_\text E = \frac{\sigma \text A}{\epsilon_0} \qquad \dots(\text{ii})$

Now, from equations (i) and (ii),

$$\begin{aligned}\text{EA} = \frac{\sigma \text A}{\epsilon_0} \\[1em] \text E = \frac{\sigma}{\epsilon_0}\end{aligned}$$

This field is directed away from the conductor. If the conductor were negatively charged, E would be directed towards the conductor.

Gaussian surface : A Gaussian surface is an arbitrary closed surface in three dimensional space through which the flux of a vector field (the gravitational field, the electric field or the magnetic field) is calculated. The surface of a sphere, a cylinder or a cube are examples of valid Gaussian surfaces, while a disc surface or a square surface cannot be used as a Gaussian surface.

Is it always a closed surface? Yes. It is an essential property of a Gaussian surface that it should be a closed surface, so that a clear distinction can be made between points that are inside the surface, on the surface and outside the surface.

Deduction of Coulomb's law from Gauss' law :

A point charge + q is placed at O, and its field $\vec{\text E} \space$is required at a point P at a distance r from O. All points at the same distance from O have the same field magnitude, so the Gaussian surface chosen is a sphere of radius r centred at O and passing through P. For a small element of this sphere around P, with area vector $\text d\vec{\text A}$, the electric flux is

$\text d\phi_\text E = \vec{\text E} \cdot \text d\vec{\text A}$

Therefore, the electric flux through the entire Gaussian surface is

$$\begin{aligned}\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} \\[1em] = \oint \text E\ \text{dA}\cos 0 \\[1em] \Phi_\text E = \text{EA}\end{aligned}$$

But A = 4πr² = surface area of the spherical Gaussian surface,

$\Phi_\text E = \text E \cdot 4\pi \text r^2 \qquad \dots(\text i)$

But from Gauss' theorem,

$\Phi_\text E = \frac{\text q}{\epsilon_0} \qquad \dots(\text{ii})$

From equations (i) and (ii),

$$\begin{aligned}\text E \cdot 4\pi \text r^2 = \frac{\text q}{\epsilon_0} \\[1em] \text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2}\end{aligned}$$

This is the expression for the electric intensity at P due to the point charge placed at O.

Now place a test charge q₀ at P. The electric force acting on it is

$$\begin{aligned}\text F = \text q_0 \text E \\[1em] \text F = \frac{1}{4\pi\epsilon_0}\frac{\text{qq}_0}{\text r^2}\end{aligned}$$

This represents Coulomb's law.

Gauss' theorem in electrostatics : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface. That is,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

where ε₀ is the permittivity of free space.

Proof for an asymmetric curved surface :

Consider a point-charge + q at O, enclosed by a closed surface A of arbitrary shape. Take a small area element dA around a point P of the surface, with OP = r, and represent it by the vector $\text d\vec{\text A}$along its outward normal.

The field $\vec{\text E} \space$at P due to + q is directed along OP, so the normal electric flux through the element dA is

$$\begin{aligned}\text d\Phi_\text E = \vec{\text E} \cdot \text d\vec{\text A} \\[1em] \text d\Phi_\text E = \text E\ \text{dA}\cos\theta\end{aligned}$$

where θ is the angle between the vectors $\vec{\text E} \space$and $\text d\vec{\text A}$.

Now,

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2} \\[1em] \therefore \quad \text d\Phi_\text E = \frac{\text q}{4\pi\epsilon_0}\frac{\text{dA}\cos\theta}{\text r^2}\end{aligned}$$

The quantity $\frac{\text{dA}\cos\theta}{\text r^2}$ is the solid angle dω that the element dA subtends at O. Hence

$\text d\Phi_\text E = \frac{\text q}{4\pi\epsilon_0}\text d\omega$

Integrating over the whole closed surface A gives the total normal electric flux Φ_E,

$\Phi_\text E = \frac{\text q}{4\pi\epsilon_0}\oint \text d\omega$

The whole closed surface A subtends a total solid angle $\oint \text d\omega = 4\pi$ steradian at O, and this is true for a closed surface of any shape enclosing the point.

$$\begin{aligned}\therefore \quad \Phi_\text E = \frac{\text q}{4\pi\epsilon_0} \times 4\pi \\[1em] \Phi_\text E = \frac{\text q}{\epsilon_0}\end{aligned}$$

Hence Gauss' theorem is proved. The shape of the surface never enters the argument, so the theorem holds for an asymmetric curved surface as well.

(i) The response of the student is correct.

Coulomb's law is an equivalent way of describing the relation between charge and electric field in static condition. Gauss' theorem and Coulomb's law are complementary to each other; in fact Coulomb's law can be deduced from Gauss' law and vice versa.

The difference is that Gauss' law can be applied conveniently to a uniform charge distribution having symmetry only, whereas Coulomb's law can be applied even if there is no symmetry.

(ii) The response of the student is incorrect.

The electric flux through a Gaussian surface is

$\phi = \frac{\text q_\text{in}}{\epsilon_0}$

The flux depends only upon the charge enclosed, and does not depend upon the shape or the size of the Gaussian surface. Halving the radius of the spherical Gaussian surface therefore leaves the flux unchanged.

To make the outgoing flux double, the charge enclosed must be doubled.

(iii) The response of the student is correct.

Given,

• Area of the plane surface, A = 10^-2 m², parallel to the xy plane
• Electric field, $\vec{\text E} = (4\hat{\text i} + 5\hat{\text j} + 8\hat{\text k})$ N/C

The surface is parallel to the xy plane, so its area vector is along the z-axis,

$$\begin{aligned}\vec{\text A} = (10^{-2})\hat{\text k}\ \text m^2 \\[1em] \phi = \vec{\text E} \cdot \vec{\text A} \\[1em] = (4\hat{\text i} + 5\hat{\text j} + 8\hat{\text k}) \cdot (10^{-2}\hat{\text k}) \\[1em] \phi = 8 \times 10^{-2}\ \text{V-m}\end{aligned}$$

which is exactly what the student has written.

Gauss' theorem in electrostatics : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

where ε₀ is the permittivity of free space.

Electric intensity due to an infinite sheet of charge :

Consider a thin sheet of charge of infinite extent with surface charge density σ, and a point P at a distance r from it where the electric intensity is required.

With P′ as the mirror image of P on the other side of the sheet, the Gaussian surface taken is a cylinder passing through the sheet, whose plane ends, each of area A, are parallel to the sheet and pass through P and P′.

Because the sheet is infinite, $\vec{\text E} \space$close to it on either side is perpendicular to the sheet, pointing outwards for a positively charged sheet. Hence $\vec{\text E} \space$is normal to the plane ends, parallel to the curved surface, and equal in magnitude at P and P′. The flux through the two plane ends is therefore

$$\begin{aligned}\Phi_\text E = \oint_\text A \vec{\text E} \cdot \text d\vec{\text A} + \oint_\text A \vec{\text E} \cdot \text d\vec{\text A} \\[1em] = \text E\oint \text{dA} + \text E\oint \text{dA} \\[1em] \Phi_\text E = 2\text{EA}\end{aligned}$$

The flux through the curved surface is zero, because there $\vec{\text E} \space$and $\text d\vec{\text A}$are at right angles. Hence the total flux through the Gaussian cylinder is

$\Phi_\text E = 2\text{EA} \qquad \dots(\text i)$

By Gauss' theorem, with the enclosed charge q = σA,

$\Phi_\text E = \frac{\sigma \text A}{\epsilon_0} \qquad \dots(\text{ii})$

Comparing equations (i) and (ii),

$$\begin{aligned}2\text{EA} = \frac{\sigma \text A}{\epsilon_0} \\[1em] \text E = \frac{\sigma}{2\epsilon_0}\end{aligned}$$

This is independent of r, so the electric intensity is the same at all points close to the sheet.

Electric intensity due to two parallel oppositely charged thin infinite sheets :

Let us now consider two sheets 1 and 2 carrying charges of equal magnitude and opposite sign, of charge densities σ₁ = + σ and σ₂ = − σ. Let $\vec{\text E}_1$ and $\vec{\text E}_2$ be the electric intensities at any point due to sheets 1 and 2 respectively, each of magnitude $\frac{\sigma}{2\epsilon_0}$.

(i) At a point P in between the sheets :

Here $\vec{\text E}_1$ is directed away from sheet 1 and $\vec{\text E}_2$ is directed towards sheet 2, so the two are in the same direction. The resultant intensity is

$$\begin{aligned}\text E = \text E_1 + \text E_2 \\[1em] = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \\[1em] \text E = \frac{\sigma}{\epsilon_0}\end{aligned}$$

The magnitude of E is free from the position of the point P between the sheets. Hence the electric field between the sheets, except near the edges, is uniform everywhere and independent of the separation between the sheets. It points from the positive towards the negative sheet.

(ii) At a point P′ outside the sheets :

Here $\vec{\text E}_1$ is directed away from sheet 1 and $\vec{\text E}_2$ is directed towards sheet 2, so the two are oppositely directed. The resultant intensity is

$$\begin{aligned}\text E = \text E_1 - \text E_2 \\[1em] = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} \\[1em] \text E = 0\end{aligned}$$

Thus a uniform electric field $\frac{\sigma}{\epsilon_0}$ is confined to the region between the sheets, and the field outside them is zero. Such a pair of sheets is therefore used to obtain a uniform field in a limited region of space, provided the sheets are much larger in linear dimensions than their separation.

Consider a thin, isolated spherical shell of centre O and radius R, with a charge + q spread uniformly over its surface.

(i) At an external point :

The electric intensity is required at a point P at a distance r (> R) from the centre O.

The Gaussian surface taken is a sphere of radius r, concentric with the shell and passing through P. Every point of this sphere is at the same distance from the charged shell, so by spherical symmetry the field has the same magnitude E all over it and is directed radially outwards.

For an area element dA around P, both $\vec{\text E} \space$and the area vector $\text d\vec{\text A}$point radially outwards, so the angle between them is zero and the electric flux through the element is

$$\begin{aligned}\text d\Phi_\text E = \vec{\text E} \cdot \text d\vec{\text A} \\[1em] = \text E\ \text{dA}\cos 0 \\[1em] \text d\Phi_\text E = \text E\ \text{dA}\end{aligned}$$

The flux through the entire Gaussian surface is

$$\begin{aligned}\Phi_\text E = \oint \text E\ \text{dA} \\[1em] = \text E\oint \text{dA} \\[1em] = \text{EA} \\[1em] \Phi_\text E = \text E(4\pi \text r^2)\end{aligned}$$

But, by Gauss' theorem, $\Phi_\text E = \frac{\text q}{\epsilon_0}$, where q is the total charge enclosed by the Gaussian surface. Therefore

$$\begin{aligned}\text E(4\pi \text r^2) = \frac{\text q}{\epsilon_0} \\[1em] \text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2} \qquad (\text{for r } \gt \text{ R}) \qquad \dots(\text i)\end{aligned}$$

This has the same form as the electric intensity of a point-charge q at a distance r. Hence, for points 'outside' it, the charged spherical shell behaves as if its whole charge were concentrated at its centre.

If σ be the uniform charge density at the surface of the shell, then q = 4πR²σ. Substituting this value of q in equation (i),

$\text E = \frac{\sigma}{\epsilon_0}\frac{\text R^2}{\text r^2} \qquad \dots(\text{ii})$

(ii) At the surface :

If the point P lies just on the surface of the shell, then r = R. Equations (i) and (ii) now give

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2} \\[1em] \text E = \frac{\sigma}{\epsilon_0}\end{aligned}$$

(iii) At an internal point :

For points inside the shell, such as P′, the Gaussian surface through P′ does not enclose any charge, since the entire charge lies on the surface of the shell. Hence, according to Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \text E(4\pi \text r^2) \\[1em] \text E(4\pi \text r^2) = 0 \\[1em] \text E = 0\end{aligned}$$

The electric field intensity is zero everywhere inside the charged shell.

Graph :

The electric field is zero everywhere inside the shell (from r = 0 to r = R), is maximum at the surface of the shell where it equals $\frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2}$, and then decreases rapidly outside the shell as $\text E \propto \frac{1}{\text r^2}$.

In case of a metallic (conducting) sphere, the entire charge will lie on the surface and it will behave just like a shell.

Given,

• Magnitude of the surface charge density on the inner faces of the plates, σ = 17.7 × 10^-22 Cm^-2, of opposite signs
• ε₀ = 8.85 × 10^-12 C²/N-m²

From the figure, plate A carries + σ and plate B carries − σ. The point M lies in the outer region of plate A and the point P lies between the plates.

The magnitude of the electric field due to each plate alone is

$\text E_1 = \text E_2 = \frac{\sigma}{2\epsilon_0}$

Substituting the values,

$$\begin{aligned}\text E_1 = \text E_2 = \frac{17.7 \times 10^{-22}}{2 \times 8.85 \times 10^{-12}} \\[1em] = \frac{17.7 \times 10^{-22}}{17.7 \times 10^{-12}} \\[1em] \text E_1 = \text E_2 = 1 \times 10^{-10}\ \text{N/C}\end{aligned}$$

(i) Resultant field at the point M :

At a point outside the two plates, the field due to the positively charged plate A is directed away from A while the field due to the negatively charged plate B is directed towards B. The two are therefore oppositely directed and equal in magnitude, so

$$\begin{aligned}\text E_\text M = \text E_1 - \text E_2 \\[1em] = 1 \times 10^{-10} - 1 \times 10^{-10} \\[1em] \text E_\text M = 0\end{aligned}$$

Hence the resultant electric field at the point M is zero.

(ii) Field at the point P due to plate B :

Plate B carries a negative charge, so the field due to it is directed towards plate B.

$$\begin{aligned}\text E_2 = \frac{\sigma}{2\epsilon_0} \\[1em] \text E_2 = 1 \times 10^{-10}\ \text{N/C}\end{aligned}$$

Hence the electric field at P due to plate B is 1 × 10^-10 N/C, directed from A towards B.

(iii) Resultant field at the point P :

At a point between the plates, the field due to A is directed away from A (that is, from A towards B) and the field due to B is directed towards B (again from A towards B). The two are in the same direction, so

$$\begin{aligned}\text E_\text P = \text E_1 + \text E_2 \\[1em] = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \\[1em] \text E_\text P = \frac{\sigma}{\epsilon_0} \\[1em] = 1 \times 10^{-10} + 1 \times 10^{-10} \\[1em] \text E_\text P = 2 \times 10^{-10}\ \text{N/C}\end{aligned}$$

Hence the magnitude of the resultant field at the point P is 2 × 10^-10 N/C, directed from plate A towards plate B.

(iv) For a thin infinitely long straight uniformly charged wire, the field is the same in magnitude at all points equidistant from the wire and is directed radially outwards. The Gaussian surface must have a shape according to this symmetry, so the convenient shape is a cylindrical Gaussian surface drawn coaxially with the wire.

(i) Copper

Explanation:
A Faraday cage is an enclosure made of a conducting material. Copper is a conductor.

(ii) car

Explanation:
The metal body of a car acts as a big hollow conductor. The electric field within the enclosure is zero, so the occupants are shielded from external electric fields.

(iii) Zero

Explanation:
Since there is no electric field inside a charged conductor, electric lines of force cannot get into the empty space enclosed by a hollow conductor. Hence the field inside the cage is zero and the electrical shocks pass harmlessly around the outside of the cage. This is electrostatic shielding.

(iv) 2.26 × 10⁵ Nm²/C leaving the surface

Explanation:
Given,

• Point charge at the centre of the cube, q = 2 μC = 2 × 10^-6 C
• Edge of the cube = 9 cm
• ε₀ = 8.85 × 10^-12 C²/N-m²

By Gauss' theorem, the electric flux through the closed surface of the cube is $\frac{1}{\epsilon_0}$ times the charge enclosed by it,

$\Phi_\text E = \frac{\text q}{\epsilon_0}$

Substituting the values,

$$\begin{aligned}\Phi_\text E = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}} \\[1em] \Phi_\text E = 2.26 \times 10^5\ \text{Nm}^2\text{/C}\end{aligned}$$

Since the enclosed charge is positive, the flux is directed outwards, that is, leaving the surface. The edge of the cube does not enter the calculation, as the flux does not depend upon the shape or the size of the closed surface.

Given,

• Point-charge at the centre of the cube, q = 1.8 μC = 1.8 × 10^-6 C
• Side of the cubical Gaussian surface = 55 cm
• ε₀ = 8.85 × 10^-12 C²/N-m²

By Gauss' theorem, the electric flux through the closed surface of the cube is $\frac{1}{\epsilon_0}$ times the charge q enclosed by the cube,

$\Phi_\text E = \frac{\text q}{\epsilon_0}$

Substituting the values,

$\Phi_\text E = \frac{1.8 \times 10^{-6}}{8.85 \times 10^{-12}}$

$$\begin{aligned}= 0.2034 \times 10^6 \\[1em] = 2.0 \times 10^5\ \text{N-m}^2\ \text C^{-1}\end{aligned}$$

The side of the cube does not enter the calculation, since the flux through a closed surface does not depend upon its shape or size.

Hence the net electric flux through the surface is 2.0 × 10⁵ N-m² C^-1.

Given,

• Net outward electric flux through the surface of the box, Φ_E = 8.0 × 10³ N m² C^-1
• ε₀ = 8.85 × 10^-12 C²/N-m²

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\text q}{\epsilon_0} \\[1em] \Rightarrow \text q = \epsilon_0 \Phi_\text E\end{aligned}$$

Substituting the values,

$\text q = (8.85 \times 10^{-12})(8.0 \times 10^3)$

$$\begin{aligned}= 70.8 \times 10^{-9}\ \text C \\[1em] = 70.8\ \text{nC}\end{aligned}$$

The flux is outward, so the net charge inside the box is positive.

Hence the net charge inside the box is 70.8 nC (nano-coulomb).

Given,

• Electric flux through the spherical Gaussian surface, Φ_E = − 750 N m² C^-1
• Radius of the Gaussian surface = 10 cm, with centre at the charge
• ε₀ = 8.85 × 10^-12 C²/N-m²

(i) By Gauss' theorem, the flux through a closed surface depends only upon the net charge enclosed by the surface, and not upon the shape or the size of the surface.

When the radius of the Gaussian surface is doubled, the charge enclosed remains the same. Hence the flux passing through the surface is still

$\Phi_\text E = -750\ \text N\ \text m^2\ \text C^{-1}$

(ii) From Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\text q}{\epsilon_0} \\[1em] \Rightarrow \text q = \epsilon_0 \Phi_\text E\end{aligned}$$

Substituting the values,

$\text q = (8.85 \times 10^{-12})(-750)$

$$\begin{aligned}= -6637.5 \times 10^{-12} \\[1em] = -6.64 \times 10^{-9}\ \text C\end{aligned}$$

The negative sign shows that the charge is negative, as is also indicated by the inward (negative) flux.

Hence (i) the flux remains − 750 N m² C^-1 and (ii) the magnitude of the point-charge is 6.64 × 10^-9 C.

Note: The printed answer gives − 6.64 × 10^-9 C, which is the signed charge; since the question asks for magnitude, the answer is 6.64 × 10^-9 C.

Given,

• Net charge on the isolated conductor = + 10 μC
• Point-charge inside the cavity = + 3.0 μC

Charge on the cavity wall :

Let a Gaussian surface be drawn within the material of the conductor, enclosing the cavity. The electric field inside the material of a conductor is zero, so the flux through this Gaussian surface is zero, and by Gauss' theorem the net charge enclosed by it must also be zero.

$$\begin{aligned}\oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0} \\[1em] = 0 \\[1em] \Rightarrow \text q = 0\end{aligned}$$

The charge enclosed is the sum of the point-charge in the cavity and the charge induced on the cavity wall. Therefore

$$\begin{aligned}3.0\ \mu\text C + \text q_{\text{wall}} = 0 \\[1em] \text q_{\text{wall}} = -3.0\ \mu\text C\end{aligned}$$

Charge on the outer surface :

The conductor as a whole carries a net charge of + 10 μC. Since − 3.0 μC of it appears on the cavity wall, the charge on the outer surface is

$$\begin{aligned}\text q_{\text{outer}} = 10\ \mu\text C - (-3.0\ \mu\text C) \\[1em] = +13\ \mu\text C\end{aligned}$$

Hence the charge on the cavity wall is − 3.0 μC and that on the outer surface of the conductor is + 13 μC.

Given,

• Uniform electric field, $\vec{\text E} = 5 \times 10^3\ \hat{\text i}$ NC^-1
• Side of the square = 10 cm = 0.10 m

The area of the square is

$$\begin{aligned}\text A = 0.10 \times 0.10 \\[1em] = 1 \times 10^{-2}\ \text m^2\end{aligned}$$

Case (i) : Plane of the square parallel to the YZ-plane

When the plane of the square is parallel to the YZ-plane, the normal to the square is along the X-axis, that is, along $\vec{\text E} \space$. Hence

$$\begin{aligned}\Phi_\text E = \vec{\text E} \cdot \vec{\text A} \\[1em] = (5 \times 10^3\ \hat{\text i}) \cdot (1 \times 10^{-2}\ \hat{\text i}) \\[1em] = 50\ \text{N-m}^2\text C^{-1}\end{aligned}$$

Case (ii) : Plane of the square making an angle of 30° with the X-axis

If the plane of the square makes an angle of 30° with the X-axis, then the normal to the square makes an angle of

$$\begin{aligned}\theta = 90^\circ - 30^\circ \\[1em] = 60^\circ\end{aligned}$$

with the X-axis, that is, with the direction of $\vec{\text E} \space$. Hence

$$\begin{aligned}\Phi_\text E = \text{EA}\cos\theta \\[1em] = (5 \times 10^3)(1 \times 10^{-2})\cos 60^\circ \\[1em] = 50 \times \frac{1}{2} \\[1em] = 25\ \text{N-m}^2\text C^{-1}\end{aligned}$$

Hence the electric flux is 50 N-m² C^-1 in the first case and 25 N-m² C^-1 in the second case.

Gauss' law in electrostatics : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

Given,

• Side of the cube = a
• Electric field, $\vec{\text E} = \text{Kx}\hat{\text i}$, where K is a positive dimensional constant

From the figure, the cube is placed with the face ABCD parallel to the Y-Z plane at x = a and the opposite face at x = 2a. The area of each face is a².

The field $\vec{\text E} \space$is directed along the X-axis, so the flux is zero through the four faces whose outward normals are perpendicular to the X-axis, that is, through the faces parallel to the X-Z and X-Y planes, since for each of these the angle between the area vector and the electric field is 90°.

(i) Electric flux through the cube :

Flux through the face at x = a : Here E_x = Ka and the outward drawn normal is along the negative X-axis, so θ = 180°,

$$\begin{aligned}\Phi_1 = \text E_x \text A\cos 180^\circ \\[1em] = (\text{Ka})(\text a^2)(-1) \\[1em] = -\text{Ka}^3\end{aligned}$$

Flux through the face at x = 2a : Here E_x = K(2a) and the outward drawn normal is along the positive X-axis, so θ = 0,

$$\begin{aligned}\Phi_2 = \text E_x \text A\cos 0^\circ \\[1em] = \text K(2\text a)(\text a^2)(1) \\[1em] = 2\text{Ka}^3\end{aligned}$$

The net flux associated with the cube is

$$\begin{aligned}\Phi = \Phi_1 + \Phi_2 \\[1em] = -\text{Ka}^3 + 2\text{Ka}^3 \\[1em] \Phi = \text{Ka}^3\ \text{Nm}^2\text C^{-1}\end{aligned}$$

(ii) Net charge inside the cube :

According to Gauss' theorem,

$$\begin{aligned}\text q_\text{in} = \epsilon_0 \Phi \\[1em] = \epsilon_0 \text{Ka}^3\ \text{coulomb}\end{aligned}$$

Hence the electric flux through the cube is Ka³ Nm²C^-1 and the net charge inside the cube is $\epsilon_0 \text a^3 \text K$ coulomb.

Given,

• Length of the hollow cylindrical box = 1 m
• Area of cross-section, A = 25 cm² = 25 × 10^-4 m²
• Electric field, $\vec{\text E} = 50\text x\hat{\text i}$ NC^-1, with x in metres
• ε₀ = 8.85 × 10^-12 C²/N-m²

From the figure, the axis of the cylinder lies along the X-axis, the left plane face L being at x = 1 m and the right plane face R at x = 2 m.

(i) Net flux through the cylinder :

The flux through the curved surface is zero, because there the area vector is at right angles to $\vec{\text E} \space$.

Flux through the left face (x = 1 m) : Here E = 50 × 1 = 50 NC^-1 and the outward drawn normal is along the negative X-axis, so θ = 180°,

$$\begin{aligned}\Phi_\text L = \text{EA}\cos 180^\circ \\[1em] = -(50)(25 \times 10^{-4}) \\[1em] = -0.125\ \text{Nm}^2\text C^{-1}\end{aligned}$$

Flux through the right face (x = 2 m) : Here E = 50 × 2 = 100 NC^-1 and the outward drawn normal is along the positive X-axis, so θ = 0,

$$\begin{aligned}\Phi_\text R = \text{EA}\cos 0^\circ \\[1em] = (100)(25 \times 10^{-4}) \\[1em] = 0.250\ \text{Nm}^2\text C^{-1}\end{aligned}$$

The net flux through the cylinder is

$$\begin{aligned}\Phi_\text{net} = \Phi_\text L + \Phi_\text R \\[1em] = -0.125 + 0.250 \\[1em] \Phi_\text{net} = 0.125\ \text{Nm}^2\text C^{-1}\end{aligned}$$

(ii) Charge enclosed by the cylinder :

By Gauss' theorem,

$$\begin{aligned}\text q = \epsilon_0 \Phi_\text{net} \\[1em] = (8.85 \times 10^{-12})(0.125) \\[1em] = 1.106 \times 10^{-12}\ \text C\end{aligned}$$

Hence the net flux through the cylinder is 0.125 Nm²C^-1 and the charge enclosed by it is 1.106 × 10^-12 C.

Given,

• Side of the cube = a
• Charges at the eight corners of the cube = 1 μC, 2 μC, 3 μC, 4 μC, 5 μC, 6 μC, 7 μC and 8 μC
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

From the figure, charges of 1 μC, 2 μC, 3 μC, 4 μC, 5 μC, 6 μC, 7 μC and 8 μC are placed at the eight corners of the cube of side a.

A charge placed at a corner of a cube is shared by eight such cubes meeting at that corner. Hence only $\frac{1}{8}$ of the flux due to each corner charge passes through the given cube.

The sum of all the charges at the corners is

$$\begin{aligned}\sum \text q = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)\ \mu\text C \\[1em] = 36\ \mu\text C \\[1em] = 36 \times 10^{-6}\ \text C\end{aligned}$$

The effective charge enclosed by the cube is therefore

$$\begin{aligned}\text q_\text{in} = \frac{1}{8}\sum \text q \\[1em] = \frac{36 \times 10^{-6}}{8} \\[1em] = 4.5 \times 10^{-6}\ \text C\end{aligned}$$

By Gauss' theorem, the total electric flux associated with the cube is

$\Phi_\text E = \frac{\text q_\text{in}}{\epsilon_0}$

Using $\frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9$, we have $\frac{1}{\epsilon_0} = 4\pi \times 9.0 \times 10^9 = 36\pi \times 10^9$. Therefore

$$\begin{aligned}\Phi_\text E = (4.5 \times 10^{-6})(36\pi \times 10^9) \\[1em] = 162\pi \times 10^3\ \text{Nm}^2\text{/C}\end{aligned}$$

Hence the total electric flux associated with the cube is $162\pi \times 10^3$ Nm²/C.

Given,

• Electric field due to the infinite line charge, E = 9.0 × 10⁴ N C^-1
• Distance from the line charge, r = 2.0 cm = 2.0 × 10^-2 m
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

The electric intensity due to an infinitely long line charge of linear charge density λ, at a point distant r from it, is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\epsilon_0 \text r} \\[1em] \therefore \quad \lambda = \text E \times 2\pi\epsilon_0 \text r \\[1em] = \frac{\text E \times \text r}{2 \times \left(\frac{1}{4\pi\epsilon_0}\right)}\end{aligned}$$

Substituting the values,

$\lambda = \frac{(9.0 \times 10^4)(2.0 \times 10^{-2})}{2 \times 9.0 \times 10^9}$

$$\begin{aligned}= \frac{1.8 \times 10^3}{1.8 \times 10^{10}} \\[1em] = 1 \times 10^{-7}\ \text{C/m} \\[1em] = 0.1\ \mu\text{C/m}\end{aligned}$$

Hence the linear charge density is 0.1 μC/m.

Given,

• Surface charge density on the inner faces, σ = 17.0 × 10^-12 C m^-2, equal and opposite
• ε₀ = 8.85 × 10^-12 C²/N-m²

The magnitude of the field due to each plate alone is $\frac{\sigma}{2\epsilon_0}$.

To the left of the plates and to the right of the plates :

At a point outside the two plates the fields due to the positively and the negatively charged plates are equal in magnitude but oppositely directed. Therefore

$$\begin{aligned}\text E = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} \\[1em] = 0\end{aligned}$$

In between the plates :

At a point between the plates the two fields are in the same direction, so they add up,

$$\begin{aligned}\text E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \\[1em] = \frac{\sigma}{\epsilon_0}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text E = \frac{17.0 \times 10^{-12}}{8.85 \times 10^{-12}} \\[1em] = 1.92 \approx 2.0\ \text N\ \text C^{-1}\end{aligned}$$

Hence the electric field is zero to the left of the plates, zero to the right of the plates and 2.0 N C^-1 in between the plates.

Given,

• Area of each plate, A = 1.0 m²
• Separation between the plates = 5.0 cm
• Electric field between the plates, E = 55 N C^-1
• ε₀ = 8.85 × 10^-12 C²/N-m²

For two large metal plates carrying equal and opposite charges on their inner surfaces, the field between the plates is

$\text E = \frac{\sigma}{\epsilon_0}$

where $\sigma = \frac{\text q}{\text A}$ is the surface charge density on the inner surfaces. Therefore

$$\begin{aligned}\text E = \frac{\text q}{\epsilon_0 \text A} \\[1em] \Rightarrow \text q = \epsilon_0 \text{EA}\end{aligned}$$

Substituting the values,

$\text q = (8.85 \times 10^{-12})(55)(1.0)$

$$\begin{aligned}= 486.75 \times 10^{-12} \\[1em] = 4.9 \times 10^{-10}\ \text C\end{aligned}$$

The separation between the plates does not enter the calculation, since the field between the plates is independent of the separation.

Hence the charges on the plates are + 4.9 × 10^-10 C and − 4.9 × 10^-10 C.

Given,

• Radius of the conducting sphere, R = 1.2 m
• Surface charge density, σ = 80.0 μC/m² = 80.0 × 10^-6 C/m²
• ε₀ = 8.85 × 10^-12 C²/N-m²

Charge on the sphere :

The charge resides on the surface of the conducting sphere, whose area is 4πR². Hence

$\text q = \sigma \times 4\pi \text R^2$

Substituting the values,

$\text q = (80.0 \times 10^{-6}) \times 4 \times 3.14 \times (1.2)^2$

$$\begin{aligned}= (80.0 \times 10^{-6}) \times 4 \times 3.14 \times 1.44 \\[1em] = (80.0 \times 10^{-6}) \times 18.09 \\[1em] = 1.45 \times 10^{-3}\ \text C\end{aligned}$$

Total electric flux leaving the surface :

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\text q}{\epsilon_0} \\[1em] = \frac{1.45 \times 10^{-3}}{8.85 \times 10^{-12}} \\[1em] = 1.64 \times 10^8\ \text{N-m}^2\ \text C^{-1}\end{aligned}$$

Hence the charge on the sphere is 1.45 × 10^-3 C and the total electric flux leaving its surface is 1.64 × 10⁸ N-m² C^-1.

Given,

• Radius of the charged conducting sphere = 10 cm
• Distance of the point from the centre, r = 20 cm = 0.20 m
• Electric field at that point, E = 1.5 × 10³ N C^-1, pointing radially inwards
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

The point lies outside the sphere, since 20 cm > 10 cm. For an external point the charged conducting sphere behaves as if its entire charge were concentrated at its centre,

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

Substituting the values,

$\text q = \frac{(1.5 \times 10^3)(0.20)^2}{9.0 \times 10^9}$

$$\begin{aligned}= \frac{(1.5 \times 10^3)(0.04)}{9.0 \times 10^9} \\[1em] = \frac{60}{9.0 \times 10^9} \\[1em] = 6.67 \times 10^{-9}\ \text C\end{aligned}$$

As the field points radially inwards, the charge on the sphere is negative.

Hence the charge on the sphere is − 6.67 nC.

Given,

• Radius of the metal sphere, R = 12 cm = 0.12 m
• Charge distributed uniformly on its surface, q = 1.6 × 10^-7 C
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

In a metal sphere the entire charge lies on the surface, so it behaves just like a charged thin spherical shell.

(i) At a point inside the sphere :

A Gaussian surface drawn inside the sphere encloses no charge. Hence, by Gauss' theorem,

$$\begin{aligned}\text E(4\pi \text r^2) = 0 \\[1em] \Rightarrow \text E = 0\end{aligned}$$

The electric field intensity inside the sphere is zero.

(ii) At a point just outside the sphere :

Here r = R = 0.12 m,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2}$

Substituting the values,

$\text E = \frac{(9.0 \times 10^9)(1.6 \times 10^{-7})}{(0.12)^2}$

$$\begin{aligned}= \frac{1440}{0.0144} \\[1em] = 1.0 \times 10^5\ \text N\ \text C^{-1}\end{aligned}$$

(iii) At 18 cm from the centre of the sphere :

Here r = 18 cm = 0.18 m, which is an external point,

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

Substituting the values,

$\text E = \frac{(9.0 \times 10^9)(1.6 \times 10^{-7})}{(0.18)^2}$

$$\begin{aligned}= \frac{1440}{0.0324} \\[1em] = 4.4 \times 10^4\ \text N\ \text C^{-1}\end{aligned}$$

Hence the electric field intensity is (i) zero inside the sphere, (ii) 1.0 × 10⁵ N C^-1 just outside the sphere and (iii) 4.4 × 10⁴ N C^-1 at 18 cm from the centre.

Gaussian surface : A Gaussian surface is any closed surface, not passing through any point-charge, through which the flux of the electric field is calculated. It may be of any shape and may or may not enclose any charge.

Given,

• Point-charge at the centre of the cube, q = 2.0 μC = 2.0 × 10^-6 C
• Side of the cubic Gaussian surface = 1 m
• ε₀ = 8.85 × 10^-12 C²/N-m²

By Gauss' theorem, the electric flux through the closed surface of the cube is $\frac{1}{\epsilon_0}$ times the charge enclosed by it,

$\Phi_\text E = \frac{\text q}{\epsilon_0}$

Substituting the values,

$$\begin{aligned}\Phi_\text E = \frac{2.0 \times 10^{-6}}{8.85 \times 10^{-12}} \\[1em] = 2.26 \times 10^5\ \text{N-m}^2\text{/C}\end{aligned}$$

Hence the net electric flux through the surface is 2.26 × 10⁵ N-m²/C.

Gauss' theorem : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

Variation of electric field with distance :

In a charged metal ball the entire charge resides on its surface, so the ball behaves just like a charged thin spherical shell.

(i) From the centre up to the surface (r < R) : A Gaussian surface drawn inside the ball encloses no charge, so by Gauss' theorem E = 0 at every interior point.

(ii) At the surface (r = R) : The field rises abruptly to its maximum value,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text R^2}$

(iii) At a far-off outside point (r > R) : The ball behaves as if its whole charge were concentrated at its centre,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text q}{\text r^2} \quad \text{that is} \quad \text E \propto \frac{1}{\text r^2}$

Hence the graph is zero from the centre up to the surface, is maximum at the surface and then decreases rapidly as $\frac{1}{\text r^2}$ outside the ball.

The Gaussian surface should pass through the point at which the electric field is required.

Explanation:
Unless the chosen closed surface passes through that point, the electric field obtained from Gauss' theorem would not be the field at the required location.

By definition, the electric flux through a plane surface held in a uniform electric field is

$\Phi_\text E = \text{EA}\cos\theta$

So the unit of electric flux is the unit of E multiplied by the unit of area.

The unit of the electrostatic field E is volt/metre and the unit of area is metre². Therefore

$$\begin{aligned}\text{unit of } \Phi_\text E = \frac{\text V}{\text m} \times \text m^2 \\[1em] \text{unit of } \Phi_\text E = \text{V-m}\end{aligned}$$

Hence the unit of electric flux is volt-metre (V-m).

Given,

• Electric field due to the infinite line charge, E = 9.0 × 10⁴ N C^-1
• Distance from the line charge, r = 2.0 cm = 2.0 × 10^-2 m
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

The electric intensity due to an infinitely long line charge having linear charge density λ, at a point distant r from it, is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\epsilon_0 \text r} \\[1em] \therefore \quad \lambda = \text E \times 2\pi\epsilon_0 \text r \\[1em] \lambda = \frac{\text E \times \text r}{2 \times \left(\frac{1}{4\pi\epsilon_0}\right)}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\lambda = \frac{(9.0 \times 10^4)(2.0 \times 10^{-2})}{2 \times 9.0 \times 10^9} \\[1em] = \frac{1.8 \times 10^3}{1.8 \times 10^{10}} \\[1em] = 1 \times 10^{-7}\ \text{C/m} \\[1em] = 0.1\ \mu\text{C/m}\end{aligned}$$

Hence the linear charge density is 0.1 μC/m.

zero

Reason — The two point-charges kept inside the metallic cube are + e and − e. The net charge enclosed by the cube is therefore

$$\begin{aligned}\sum \text q = (+\text e) + (-\text e) \\[1em] \sum \text q = 0\end{aligned}$$

By Gauss' theorem, the electric flux through a closed surface is $\frac{1}{\epsilon_0}$ times the net charge enclosed by it,

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] \Phi_\text E = \frac{0}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

Hence the electric flux emerging out of the cube is zero.

Given,

• Positive charge at the centre of the hollow metallic sphere, q = 35.4 μC = 35.4 × 10^-6 C
• ε₀ = 8.85 × 10^-12 C²/N-m²

By Gauss' theorem, the electric flux emanating from the closed spherical surface is $\frac{1}{\epsilon_0}$ times the charge enclosed by it,

$\Phi_\text E = \frac{\text q}{\epsilon_0}$

Substituting the values,

$$\begin{aligned}\Phi_\text E = \frac{35.4 \times 10^{-6}}{8.85 \times 10^{-12}} \\[1em] = 4 \times 10^6\ \frac{\text{Nm}^2}{\text C}\end{aligned}$$

Hence the electric flux emanating from the sphere is 4 × 10⁶ Nm²/C.

zero

Reason —

A short electric dipole consists of two point-charges, + q and − q, of equal magnitude and opposite sign. Both these charges lie inside the cube, since the dipole is placed at the centre O.

The net charge enclosed by the cube is therefore

$$\begin{aligned}\sum \text q = (+\text q) + (-\text q) \\[1em] \sum \text q = 0\end{aligned}$$

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

Hence the electric flux emanating through the cube is zero, whatever the length L of the cube.

Gauss' theorem of electrostatics : The electric flux Φ_E through any closed surface is equal to $\frac{1}{\epsilon_0}$ times the 'net' charge q enclosed by the surface.

Mathematically,

$\Phi_\text E = \oint \vec{\text E} \cdot \text d\vec{\text A} = \frac{\text q}{\epsilon_0}$

where ε₀ is the permittivity of free space.

$\frac{2\text q}{\epsilon_0}$

Reason — The charges enclosed by the closed surface are − q and + 3q. The net charge enclosed is the algebraic sum of all the charges within the surface,

$$\begin{aligned}\sum \text q = (-\text q) + (3\text q) \\[1em] \sum \text q = 2\text q\end{aligned}$$

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] \Phi_\text E = \frac{2\text q}{\epsilon_0}\end{aligned}$$

$\text E = \left(\frac{1}{4\pi\epsilon_0}\right)\frac{2\lambda}{\text r}$

Reason — A coaxial cylindrical Gaussian surface of radius r and length l is drawn through the given point. The flux through the plane ends is zero, while over the curved surface E is constant in magnitude and directed radially outwards, so

$\Phi_\text E = \text E(2\pi \text{rl})$

The charge enclosed by this Gaussian surface is q = λl. Applying Gauss' theorem,

$$\begin{aligned}\text E(2\pi \text{rl}) = \frac{\lambda \text l}{\epsilon_0} \\[1em] \text E = \frac{\lambda}{2\pi\epsilon_0 \text r}\end{aligned}$$

Writing $\frac{1}{2\pi\epsilon_0} = \frac{1}{4\pi\epsilon_0} \times 2$,

$\text E = \left(\frac{1}{4\pi\epsilon_0}\right)\frac{2\lambda}{\text r}$

Thus the intensity varies inversely as the first power of the distance r, and not as r².

remains Φ

Reason — By Gauss' theorem, the electric flux through a closed surface is

$\Phi_\text E = \frac{\text q}{\epsilon_0}$

The electric flux passing through a closed surface depends only upon the net charge enclosed by the surface. Gauss' theorem is valid for a closed surface no matter what its shape or size, so the flux does not depend upon the radius of the sphere.

Here the charge enclosed is unchanged, so on making the radius four times, the electric flux still remains Φ.

2 × 10⁶ NC^-1 (not among the given options)

Reason —

Given,

• Surface charge density of the large conducting sheet, σ = 17.7 × 10^-6 Cm^-2
• ε₀ = 8.85 × 10^-12 C²/N-m²

The electric field intensity just outside a charged conductor, at a place where its surface charge density is σ, is

$\text E = \frac{\sigma}{\epsilon_0}$

The same result holds for a large conducting sheet of finite thickness. Substituting the values,

$$\begin{aligned}\text E = \frac{17.7 \times 10^{-6}}{8.85 \times 10^{-12}} \\[1em] \text E = 2 \times 10^6\ \text{NC}^{-1}\end{aligned}$$

Note: The printed key gives option 4, 1 × 10⁶ NC^-1, which is the value of $\frac{\sigma}{2\epsilon_0}$ for a thin sheet of charge; for a conducting sheet the chapter gives $\text E = \frac{\sigma}{\epsilon_0}$, so none of the options is correct.

become 2Φ

Reason — By Gauss' theorem, the electric flux emanating from the closed spherical surface is

$\Phi = \frac{\text Q}{\epsilon_0}$

The flux depends only upon the net charge enclosed and not upon the shape or the size of the surface. Hence doubling the radius of the sphere produces no change in the flux.

When the charge is doubled, that is Q′ = 2Q, the new flux is

$$\begin{aligned}\Phi' = \frac{2\text Q}{\epsilon_0} \\[1em] \Phi' = 2\Phi\end{aligned}$$

Hence the electric flux emanating from the sphere becomes 2Φ.

Given,

• Intensity of electric field at the point P, E = 2.7 × 10⁴ Vm^-1
• Perpendicular distance from the infinite line charge, r = 5 cm = 5 × 10^-2 m
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

The electric intensity due to an infinitely long line charge having linear charge density λ, at a point distant r from it, is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\epsilon_0 \text r} \\[1em] \therefore \quad \lambda = \text E \times 2\pi\epsilon_0 \text r \\[1em] \lambda = \frac{\text E \times \text r}{2 \times \left(\frac{1}{4\pi\epsilon_0}\right)}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\lambda = \frac{(2.7 \times 10^4)(5 \times 10^{-2})}{2 \times 9.0 \times 10^9} \\[1em] = \frac{1.35 \times 10^3}{1.8 \times 10^{10}} \\[1em] = 7.5 \times 10^{-8}\ \text{C/m}\end{aligned}$$

Hence the linear charge density on the line charge is 7.5 × 10^-8 C/m.

(i) The electric flux emanating from the sphere is, by Gauss' theorem,

$\text X = \frac{\text q}{\epsilon_0}$

(a) When the radius of the sphere is doubled : There is no change in the electric flux.

The electric flux passing through a closed surface depends only upon the net charge enclosed by the surface. Gauss' theorem is valid for a closed surface no matter what its shape or size, so the flux remains X.

(b) When the charge q is replaced by an electric dipole : The electric flux becomes zero.

An electric dipole consists of two equal and opposite charges, so the net charge enclosed by the sphere is zero, and hence

$$\begin{aligned}\Phi_\text E = \frac{0}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

OR

(ii) The intensity of electric field at a point distant r from an infinite line charge of linear charge density λ is

$\text E = \frac{\lambda}{2\pi\epsilon_0 \text r}$

(a) When the charge on it is doubled : The intensity of electric field is doubled.

Since $\text E \propto \lambda$, replacing λ by 2λ gives

$$\begin{aligned}\text E' = \frac{2\lambda}{2\pi\epsilon_0 \text r} \\[1em] \text E' = 2\text E\end{aligned}$$

(b) When the distance of the point is halved : The intensity of electric field is doubled.

Since $\text E \propto \frac{1}{\text r}$, replacing r by $\frac{\text r}{2}$ gives

$$\begin{aligned}\text E' = \frac{\lambda}{2\pi\epsilon_0 \left(\frac{\text r}{2}\right)} \\[1em] \text E' = 2\text E\end{aligned}$$

Given,

• Q₁ = 100 μC, Q₂ = − 200 μC, Q₃ = 17.7 μC, Q₄ = − 8.85 μC
• ε₀ = 8.85 × 10^-12 C²/N-m²

From the figure, the charges Q₃ and Q₄ lie inside the hollow sphere S and Q₂ lies outside it, while Q₁ is drawn on the surface S. A Gaussian surface should not pass through any discrete charge, because the electric field at the location of the charge is not well defined; so, strictly, Gauss' theorem cannot be applied to S as drawn.

Taking Q₁ to lie just outside S, as the printed answer does, Q₁ and Q₂ make no contribution, since the net electric flux through a closed surface due to a charge outside the surface is zero. The net charge enclosed by the sphere S is then

$$\begin{aligned}\sum \text q = \text Q_3 + \text Q_4 \\[1em] \sum \text q = 17.7\ \mu\text C + (-8.85\ \mu\text C) \\[1em] \sum \text q = 8.85\ \mu\text C \\[1em] \sum \text q = 8.85 \times 10^{-6}\ \text C\end{aligned}$$

By Gauss' theorem,

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] \Phi_\text E = \frac{8.85 \times 10^{-6}}{8.85 \times 10^{-12}} \\[1em] \Phi_\text E = 1 \times 10^6\ \text{V-m}\end{aligned}$$

Hence, taking Q₁ to lie just outside S, the electric flux emanating from the hollow sphere S is 1 × 10⁶ V-m.

Note: The printed figure places Q₁ on the surface S, where Gauss' theorem cannot be applied; the printed answer 1 × 10⁶ V-m is obtained only by treating Q₁ as lying outside S.

$\frac{\text a}{2}\log\left(\frac{1}{1 - \frac{\text Q}{2\pi \text{aA}}}\right)$

Reason —

Given,

• Volume charge density, $\rho(\text r) = \frac{\text A}{\text r^2}\text e^{\frac{-2\text r}{\text a}}$
• Total charge of the distribution = Q, radius of the sphere = R

Consider a thin spherical shell of radius r and thickness dr inside the distribution. Its volume is

$\text{dV} = 4\pi \text r^2\ \text{dr}$

The charge on this shell element is

$$\begin{aligned}\text{dq} = \rho\ \text{dV} \\[1em] \text{dq} = \frac{\text A}{\text r^2}\text e^{\frac{-2\text r}{\text a}}(4\pi \text r^2\ \text{dr}) \\[1em] \text{dq} = 4\pi \text A\ \text e^{\frac{-2\text r}{\text a}}\ \text{dr}\end{aligned}$$

Integrating from r = 0 to r = R to obtain the total charge,

$$\begin{aligned}\text Q = \int_0^\text R 4\pi \text A\ \text e^{\frac{-2\text r}{\text a}}\ \text{dr} \\[1em] \text Q = 4\pi \text A\int_0^\text R \text e^{\frac{-2\text r}{\text a}}\ \text{dr} \\[1em] = 4\pi \text A\left[\frac{\text e^{\frac{-2\text r}{\text a}}}{-\frac{2}{\text a}}\right]_0^\text R \\[1em] = 4\pi \text A \times \left(\frac{-\text a}{2}\right)\left(\text e^{\frac{-2\text R}{\text a}} - \text e^0\right) \\[1em] \text Q = 2\pi \text{aA}\left(1 - \text e^{\frac{-2\text R}{\text a}}\right)\end{aligned}$$

Rearranging for the exponential term,

$\text e^{\frac{-2\text R}{\text a}} = 1 - \frac{\text Q}{2\pi \text{aA}}$

Taking logarithms on both sides,

$$\begin{aligned}\frac{-2\text R}{\text a} = \log\left(1 - \frac{\text Q}{2\pi \text{aA}}\right) \\[1em] \text R = -\frac{\text a}{2}\log\left(1 - \frac{\text Q}{2\pi \text{aA}}\right) \\[1em] \text R = \frac{\text a}{2}\log\left(\frac{1}{1 - \frac{\text Q}{2\pi \text{aA}}}\right)\end{aligned}$$

electric field outside the shell is the same as that of a point-charge at the centre of the shell

Reason —

The dipole at the centre has zero net charge, so the total charge induced on the inner surface of the shell is zero and the net charge on the outer surface is Q. For any point outside the shell, the resultant electric field is therefore due to Q uniformly distributed on the outer surface only, which behaves as if it were concentrated at the centre,

$\text E = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r^2}$

which is the field of a point-charge placed at the centre of the shell.

a = 8^-1/4 R

Reason —

Given,

• Total charge distributed in the sphere of radius R = 2Q
• Charge density, ρ(r) = kr
• Two charges of − Q each placed at diametrically opposite points, at a distance a from the centre

Step 1 — Determination of k :

Consider a thin shell of radius r and thickness dr. Its volume is 4πr² dr, so the charge on it is

$$\begin{aligned}\text{dq} = \rho\ \text{dV} \\[1em] \text{dq} = (\text{kr})(4\pi \text r^2\ \text{dr}) \\[1em] \text{dq} = 4\pi \text k \text r^3\ \text{dr}\end{aligned}$$

Integrating over the whole sphere,

$$\begin{aligned}2\text Q = \int_0^\text R 4\pi \text k \text r^3\ \text{dr} \\[1em] 2\text Q = 4\pi \text k\left[\frac{\text r^4}{4}\right]_0^\text R \\[1em] 2\text Q = \pi \text k \text R^4 \\[1em] \therefore \quad \text k = \frac{2\text Q}{\pi \text R^4}\end{aligned}$$

Step 2 — Charge enclosed within radius a :

Since the two charges experience no force, they must lie inside the sphere, so a < R. The charge enclosed within a sphere of radius a is

$$\begin{aligned}\text q_{in} = \int_0^\text a 4\pi \text k \text r^3\ \text{dr} \\[1em] \text q_{in} = \pi \text k \text a^4 \\[1em] \text q_{in} = \pi \times \frac{2\text Q}{\pi \text R^4} \times \text a^4 \\[1em] \text q_{in} = \frac{2\text Q \text a^4}{\text R^4}\end{aligned}$$

Step 3 — Condition for zero net force :

The force on the charge at A due to the enclosed charge of the sphere is attractive,

$$\begin{aligned}\text F_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\text q_{in}\text Q}{\text a^2} \\[1em] \text F_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{2\text Q^2 \text a^2}{\text R^4}\end{aligned}$$

The two charges A and B are of the same sign and are separated by a distance 2a, so the force between them is repulsive,

$$\begin{aligned}\text F_{\text{AB}} = \frac{1}{4\pi\epsilon_0}\frac{\text Q^2}{(2\text a)^2} \\[1em] \text F_{\text{AB}} = \frac{1}{4\pi\epsilon_0}\frac{\text Q^2}{4\text a^2}\end{aligned}$$

For no net force on A and B, these two must be equal,

$$\begin{aligned}\frac{2\text Q^2 \text a^2}{\text R^4} = \frac{\text Q^2}{4\text a^2} \\[1em] 8\text a^4 = \text R^4 \\[1em] \text a = \frac{\text R}{8^{1/4}} \\[1em] \text a = 8^{-1/4}\ \text R\end{aligned}$$

zero as r increases for r < R, decreases as r increases for r > R

Reason —

In a hollow metal sphere the entire charge Q resides on its surface, so it behaves just like a uniformly charged thin spherical shell.

For r < R (inside the sphere) : A Gaussian surface of radius r drawn inside encloses no charge. Hence by Gauss' theorem,

$$\begin{aligned}\text E(4\pi \text r^2) = \frac{0}{\epsilon_0} \\[1em] \text E = 0\end{aligned}$$

The field remains zero as r increases, right up to the surface.

For r > R (outside the sphere) : The Gaussian surface now encloses the whole charge Q, so

$$\begin{aligned}\text E(4\pi \text r^2) = \frac{\text Q}{\epsilon_0} \\[1em] \text E =\frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r^2}\end{aligned}$$

Since $\text E \propto \frac{1}{\text r^2}$, the field decreases as r increases.

$\frac{\lambda}{\pi\epsilon_0 \text R}$ N/C

Reason —

Given,

• Linear charge densities of the two parallel infinite line charges, + λ C/m and − λ C/m
• Separation between them = 2R

The mid-point lies at a perpendicular distance R from each line charge. The intensity due to an infinite line charge at a perpendicular distance R is

$\text E_1 = \text E_2 = \frac{\lambda}{2\pi\epsilon_0 \text R}$

The field due to the positive line charge is directed away from it, that is, towards the negative line charge. The field due to the negative line charge is directed towards it, again in the same direction. Hence the two fields are in the same direction and they add up,

$$\begin{aligned}\text E = \text E_1 + \text E_2 \\[1em] \text E = \frac{\lambda}{2\pi\epsilon_0 \text R} + \frac{\lambda}{2\pi\epsilon_0 \text R} \\[1em] \text E = \frac{2\lambda}{2\pi\epsilon_0 \text R} \\[1em] \text E = \frac{\lambda}{\pi\epsilon_0 \text R}\ \text{N/C}\end{aligned}$$

Statement I is true but Statement II is false.

Reason —

Statement I is true : An electric dipole consists of two equal and opposite charges, so the net charge enclosed by the hollow sphere is zero, and by Gauss' theorem the flux through the sphere is zero,

$$\begin{aligned}\Phi_\text E = \frac{0}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

However, in a Gaussian surface, flux $\phi$ = 0 does not imply E = 0. The two charges of the dipole produce fields which do not cancel at any point of the sphere, so the electric field is not zero anywhere in the sphere.

Statement II is false : In a solid metallic sphere the entire charge Q resides on its outer surface. A spherical surface of radius r (< R) drawn inside the metal encloses no charge, so the electric field at every point on it is zero. This part of the statement is correct.

But then, by Gauss' theorem, the flux through this closed spherical surface is

$$\begin{aligned}\Phi_\text E = \frac{\text q_{in}}{\epsilon_0} \\[1em] \Phi_\text E = \frac{0}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

So the electric flux through the surface of radius r is also zero, and not non-zero as claimed. This follows from the property that E = 0 implies $\phi$ = 0.

$\frac{\text q}{24\epsilon_0}$

Reason —

The charge q is placed at one corner of the cube. A charge at a corner is shared by eight such cubes, so the flux through the whole of the given cube is

$$\begin{aligned}\Phi_{cube} = \frac{1}{8}\left(\frac{\text q}{\epsilon_0}\right) \\[1em] \Phi_{cube} = \frac{\text q}{8\epsilon_0}\end{aligned}$$

The cube has six faces. The three faces which meet at the corner where the charge is placed contain the charge in their own planes, so for every area element of these faces the electric field is parallel to the surface and the flux through them is zero.

Hence the whole of the flux $\frac{\text q}{8\epsilon_0}$ passes through the remaining three faces, which are symmetrically situated with respect to the charge. The flux through each of these faces is therefore

$$\begin{aligned}\Phi_{face} = \frac{1}{3} \times \frac{\text q}{8\epsilon_0} \\[1em] \Phi_{face} = \frac{\text q}{24\epsilon_0}\end{aligned}$$

The shaded area shown in the figure is made up of halves of two such faces, so the flux through it is

$$\begin{aligned}\Phi = \frac{1}{2}\left(\frac{\text q}{24\epsilon_0}\right) + \frac{1}{2}\left(\frac{\text q}{24\epsilon_0}\right) \\[1em] \Phi = \frac{\text q}{24\epsilon_0}\end{aligned}$$

0.424 nCm^-2

Reason —

Given,

• Linear charge density of the line charge along the z-axis, λ = 8 nC/m = 8 × 10^-9 C/m
• Distance of the point of intersection from the z-axis, r = 3 m
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C², ε₀ = 8.85 × 10^-12 C²/N-m²

The electric field at a perpendicular distance r from an infinite line charge is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\epsilon_0 \text r} \\[1em] \text E = \frac{2\lambda}{4\pi\epsilon_0 \text r}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text E = \frac{2 \times (9.0 \times 10^9)(8 \times 10^{-9})}{3} \\[1em] = \frac{144}{3} \\[1em] \text E = 48\ \text{N/C}\end{aligned}$$

The surface charge density at this location is obtained from the electric flux density, which for a surface normal to the field is

$\sigma = \epsilon_0 \text E$

Substituting the values,

$$\begin{aligned}\sigma = (8.85 \times 10^{-12})(48) \\[1em] = 4.248 \times 10^{-10}\ \text C/\text m^2 \\[1em] = 0.424\ \text{nCm}^{-2}\end{aligned}$$

the number of flux lines entering the surface must be equal to the number of flux lines leaving it

Reason — The electric flux is a measure of the number of lines of force passing through a surface. It is positive for the lines of force leaving the surface and negative for those entering the surface.

If

$\oint \vec{\text E} \cdot \text d\vec{\text S} = 0$

then the positive and the negative contributions cancel exactly. This means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

$\text T = 2\pi \text r\sqrt{\frac{\text m}{2\text k\lambda \text q}}$

Reason —

Given,

• Charge on the particle = − q, mass = m
• Radius of the circular path = r
• Linear charge density of the infinitely long line charge = + λ
• k is Coulomb's constant, so $\text k = \frac{1}{4\pi\epsilon_0}$

The electric field at a perpendicular distance r from the infinite line charge is

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\epsilon_0 \text r} \\[1em] \text E = \frac{2\text k\lambda}{\text r}\end{aligned}$$

The line charge is positive and the particle is negative, so the electrostatic force on the particle is directed towards the line charge and provides the necessary centripetal force,

$\text{qE} = \frac{\text{mv}^2}{\text r}$

Substituting the value of E,

$$\begin{aligned}\text q\left(\frac{2\text k\lambda}{\text r}\right) = \frac{\text{mv}^2}{\text r} \\[1em] \text v^2 = \frac{2\text k\lambda \text q}{\text m} \\[1em] \text v =\sqrt{\frac{2\text k\lambda \text q}{\text m}}\end{aligned}$$

The speed is independent of r. The time period of the circular motion is

$$\begin{aligned}\text T = \frac{2\pi \text r}{\text v} \\[1em] \text T = \frac{2\pi \text r}{\sqrt{\frac{2\text k\lambda \text q}{\text m}}} \\[1em] \text T = 2\pi \text r\sqrt{\frac{\text m}{2\text k\lambda \text q}}\end{aligned}$$

The dipole will undergo small oscillations at any finite value of r > R.

The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{\sigma \text P_0}{100\ \epsilon_0 \text I}}$ at r = 10R.

Reason —

Given,

• Dipole moment = P₀, moment of inertia about its centre = I
• Radius of the spherical shell = R, uniform surface charge density = σ
• Distance of the dipole from the centre of the shell = r

The total charge on the shell is

$\text Q = \sigma \times 4\pi \text R^2$

Field due to the shell : For an external point (r > R) the shell behaves as if its entire charge were concentrated at its centre,

$$\begin{aligned}\text E = \frac{1}{4\pi\epsilon_0}\frac{\text Q}{\text r^2} \\[1em] \text E = \frac{1}{4\pi\epsilon_0}\frac{4\pi \text R^2\sigma}{\text r^2} \\[1em] \text E = \frac{\sigma \text R^2}{\epsilon_0 \text r^2}\end{aligned}$$

For an internal point (r < R) the field is zero, so no torque acts on the dipole and it cannot oscillate. Hence the first option is incorrect and the second option is correct — oscillations occur for any finite value of r > R.

Angular frequency of small oscillations : The restoring torque on a dipole inclined at a small angle θ to the field is

$$\begin{aligned}\tau = -\text P_0 \text E\sin\theta \approx -\text P_0 \text E\ \theta \\[1em] \text I\frac{\text d^2\theta}{\text{dt}^2} = -\text P_0 \text E\ \theta \\[1em] \omega =\sqrt{\frac{\text P_0 \text E}{\text I}}\end{aligned}$$

At r = 2R :

$$\begin{aligned}\text E = \frac{\sigma \text R^2}{\epsilon_0 (2\text R)^2} \\[1em] \text E = \frac{\sigma}{4\epsilon_0} \\[1em] \omega =\sqrt{\frac{\sigma \text P_0}{4\epsilon_0 \text I}}\end{aligned}$$

This does not agree with the third option, so that option is incorrect.

At r = 10R :

$$\begin{aligned}\text E = \frac{\sigma \text R^2}{\epsilon_0 (10\text R)^2} \\[1em] \text E = \frac{\sigma}{100\epsilon_0} \\[1em] \omega =\sqrt{\frac{\sigma \text P_0}{100\ \epsilon_0 \text I}}\end{aligned}$$

which agrees with the fourth option.

Note: This question has more than one correct option; the printed answer key gives both option 2 and option 4.

$\frac{5\text Q}{\epsilon_0}$

Reason —

Given,

• Charge 5Q at the point (3a, 0) and charge − 2Q at the point (− 5a, 0)
• Radius of the sphere = 4a, with its centre at the origin

The distance of each charge from the centre of the sphere is measured to decide whether it lies inside the sphere.

For the charge 5Q : Its distance from the origin is 3a. Since 3a < 4a, this charge lies inside the sphere.

For the charge − 2Q : Its distance from the origin is 5a. Since 5a > 4a, this charge lies outside the sphere. The net electric flux through a closed surface due to a charge outside the surface is zero, so it makes no contribution.

Hence the net charge enclosed by the sphere is 5Q, and by Gauss' theorem

$$\begin{aligned}\Phi_\text E = \frac{\sum \text q}{\epsilon_0} \\[1em] \Phi_\text E = \frac{5\text Q}{\epsilon_0}\end{aligned}$$

Given,

• Incremental volume located at the origin, dV = 2 × 10^-9 m³
• Electric flux density, $\text D = \text e^{-\text x}\sin \text y\ \hat{\text i} - \text e^{-\text x}\cos \text y\ \hat{\text j} + 2\text z\hat{\text k}$ C/m²

The volume charge density is obtained from the differential form of Gauss' theorem,

$\rho = \nabla \cdot \text D = \frac{\partial \text D_x}{\partial \text x} + \frac{\partial \text D_y}{\partial \text y} + \frac{\partial \text D_z}{\partial \text z}$

Differentiating each component :

$$\begin{aligned}\frac{\partial \text D_x}{\partial \text x} = \frac{\partial}{\partial \text x}\left(\text e^{-\text x}\sin \text y\right) \\[1em] \frac{\partial \text D_x}{\partial \text x} = -\text e^{-\text x}\sin \text y \\[1em] \frac{\partial \text D_y}{\partial \text y} = \frac{\partial}{\partial \text y}\left(-\text e^{-\text x}\cos \text y\right) \\[1em] \frac{\partial \text D_y}{\partial \text y} = -\text e^{-\text x}(-\sin \text y) \\[1em] \frac{\partial \text D_y}{\partial \text y} = \text e^{-\text x}\sin \text y \\[1em] \frac{\partial \text D_z}{\partial \text z} = \frac{\partial}{\partial \text z}(2\text z) \\[1em] \frac{\partial \text D_z}{\partial \text z} = 2\end{aligned}$$

Adding these,

$$\begin{aligned}\rho = -\text e^{-\text x}\sin \text y + \text e^{-\text x}\sin \text y + 2 \\[1em] \rho = 2\ \text C/\text m^3\end{aligned}$$

The charge density is a constant, so its value at the origin is also 2 C/m³. The charge enclosed in the incremental volume is

$$\begin{aligned}\text q = \rho\ \text{dV} \\[1em] \text q = 2 \times (2 \times 10^{-9}) \\[1em] = 4 \times 10^{-9}\ \text C \\[1em] \text q = 4\ \text{nC}\end{aligned}$$

Hence the total charge enclosed is 4 nC.

Given,

• Electric field, $\text E = \frac{2}{5}\text E_0\ \hat{\text i} + \frac{3}{5}\text E_0\ \hat{\text j}$, with E₀ = 4.0 × 10³ N/C
• Area of the rectangular surface, A = 0.4 m², parallel to the yz-plane

Since the surface is parallel to the yz-plane, the normal to it is along the X-axis. Hence the area vector is

$\vec{\text A} = 0.4\ \hat{\text i}\ \text m^2$

Only the x-component of the field contributes to the flux,

$$\begin{aligned}\Phi_\text E = \vec{\text E} \cdot \vec{\text A} \\[1em] \Phi_\text E = \left(\frac{2}{5}\text E_0\ \hat{\text i} + \frac{3}{5}\text E_0\ \hat{\text j}\right) \cdot (0.4\ \hat{\text i}) \\[1em] \Phi_\text E = \frac{2}{5}\text E_0 \times 0.4\end{aligned}$$

Substituting E₀ = 4.0 × 10³ N/C,

$$\begin{aligned}\Phi_\text E = \frac{2}{5} \times (4.0 \times 10^3) \times 0.4 \\[1em] = (1.6 \times 10^3) \times 0.4 \\[1em] = 640\ \text{N-m}^2\text C^{-1}\end{aligned}$$

Hence the flux of the field through the rectangular surface is 640 N-m²C^-1.

Given,

• Electric field, $\vec{\text E} = \left(\frac{3}{5}\text E_0\ \hat{\text i} + \frac{4}{5}\text E_0\ \hat{\text j}\right)$ N/C
• Rectangular surface of area 0.2 m², parallel to the YZ-plane
• Rectangular surface of area 0.3 m², parallel to the XZ-plane

Flux through the surface parallel to the YZ-plane :

The normal to this surface is along the X-axis, so its area vector is $\vec{\text A}_1 = 0.2\ \hat{\text i}$ m². Only the x-component of the field contributes,

$$\begin{aligned}\Phi_1 = \left(\frac{3}{5}\text E_0\ \hat{\text i} + \frac{4}{5}\text E_0\ \hat{\text j}\right) \cdot (0.2\ \hat{\text i}) \\[1em] \Phi_1 = \frac{3}{5}\text E_0 \times 0.2 \\[1em] \Phi_1 = 0.12\ \text E_0\end{aligned}$$

Flux through the surface parallel to the XZ-plane :

The normal to this surface is along the Y-axis, so its area vector is $\vec{\text A}_2 = 0.3\ \hat{\text j}$ m². Only the y-component of the field contributes,

$$\begin{aligned}\Phi_2 = \left(\frac{3}{5}\text E_0\ \hat{\text i} + \frac{4}{5}\text E_0\ \hat{\text j}\right) \cdot (0.3\ \hat{\text j}) \\[1em] \Phi_2 = \frac{4}{5}\text E_0 \times 0.3 \\[1em] \Phi_2 = 0.24\ \text E_0\end{aligned}$$

Ratio of the two fluxes :

$$\begin{aligned}\frac{\Phi_1}{\Phi_2} = \frac{0.12\ \text E_0}{0.24\ \text E_0} \\[1em] \frac{\Phi_1}{\Phi_2} = \frac{1}{2}\end{aligned}$$

Comparing with the given ratio a : b, we get a = 1 and b = 2.

Hence the value of a is 1.

Given,

• Point-charge, q = + 12 μC = 12 × 10^-6 C
• Perpendicular distance of the charge above the centre of the square = 6 cm
• Side of the square = 12 cm
• ε₀ = 8.85 × 10^-12 C²/N-m²

The charge lies at a height of 6 cm above the centre of a square of side 12 cm. Since 6 cm is exactly half of 12 cm, the square can be regarded as one face of a cube of edge 12 cm having the charge q at its centre.

By Gauss' theorem, the total flux through the closed surface of this cube is

$\Phi_{total} = \frac{\text q}{\epsilon_0}$

All the six faces of the cube are symmetrically situated with respect to the point-charge at the centre, so the flux through each face is $\frac{1}{6}$ of the total flux,

$\Phi = \frac{1}{6}\left(\frac{\text q}{\epsilon_0}\right)$

Substituting the values,

$$\begin{aligned}\Phi = \frac{1}{6} \times \frac{12 \times 10^{-6}}{8.85 \times 10^{-12}} \\[1em] = \frac{1}{6} \times (1.3559 \times 10^6) \\[1em] = 0.226 \times 10^6 \\[1em] = 226 \times 10^3\ \text{N-m}^2\text{/C}\end{aligned}$$

Hence the magnitude of the electric flux through the square is 226 × 10³ N-m²/C.

Given,

• Charge q surrounded by a closed surface consisting of an inverted cone of height h and base radius R, and a hemisphere of radius R
• Electric flux through the conical surface = $\frac{\text{nq}}{6\epsilon_0}$

From the figure, the charge q lies at the centre of the circular rim of radius R, which is the common boundary of the hemisphere and the cone. This point is therefore the centre of the sphere of which the given hemisphere is one half.

The cone and the hemisphere together form a closed surface enclosing the charge q. By Gauss' theorem, the total flux through this closed surface is

$\Phi_{total} = \frac{\text q}{\epsilon_0}$

The solid angle subtended by a closed surface at a point inside it is 4π steradian. The hemispherical surface subtends half of this, that is 2π steradian, at the charge. Hence the flux through the hemispherical surface is half of the total flux,

$$\begin{aligned}\Phi_{hemisphere} = \frac{1}{2}\left(\frac{\text q}{\epsilon_0}\right) \\[1em] \Phi_{hemisphere} = \frac{\text q}{2\epsilon_0}\end{aligned}$$

The flux through the conical surface is therefore

$$\begin{aligned}\Phi_{cone} = \Phi_{total} - \Phi_{hemisphere} \\[1em] \Phi_{cone} = \frac{\text q}{\epsilon_0} - \frac{\text q}{2\epsilon_0} \\[1em] \Phi_{cone} = \frac{\text q}{2\epsilon_0} \\[1em] \Phi_{cone} = \frac{3\text q}{6\epsilon_0}\end{aligned}$$

Comparing with the given form $\frac{\text{nq}}{6\epsilon_0}$,

$\text n = 3$

Hence the value of n is 3.

Given,

• Electric field, $\vec{\text E} = (2\text x^2\ \hat{\text i} - 4\text y\ \hat{\text j} + 6\hat{\text k})$ N/C
• Dimensions of the cuboid = 1 × 2 × 3 m³
• Magnitude of the charge within the cuboid = nε₀ C

From the figure, the cuboid extends from x = 0 to x = 1 m, from y = 0 to y = 2 m and from z = 0 to z = 3 m.

Flux through the faces perpendicular to the X-axis :

Area of each such face = 2 × 3 = 6 m².

At x = 1, E_x = 2(1)² = 2 N/C and the outward normal is along + x,

$$\begin{aligned}\Phi_{x_2} = (2)(6) \\[1em] \Phi_{x_2} = 12\ \text{N-m}^2\text C^{-1}\end{aligned}$$

At x = 0, E_x = 2(0)² = 0, so

$\Phi_{x_1} = 0$

Flux through the faces perpendicular to the Y-axis :

Area of each such face = 1 × 3 = 3 m².

At y = 2, E_y = − 4(2) = − 8 N/C and the outward normal is along + y,

$$\begin{aligned}\Phi_{y_2} = (-8)(3) \\[1em] \Phi_{y_2} = -24\ \text{N-m}^2\text C^{-1}\end{aligned}$$

At y = 0, E_y = − 4(0) = 0, so

$\Phi_{y_1} = 0$

Flux through the faces perpendicular to the Z-axis :

Area of each such face = 1 × 2 = 2 m². Here E_z = 6 N/C is the same at both faces, while their outward normals are oppositely directed,

$$\begin{aligned}\Phi_{z_2} + \Phi_{z_1} = (6)(2) + (-6)(2) \\[1em] \Phi_{z_2} + \Phi_{z_1} = 0\end{aligned}$$

Net flux through the cuboid :

$$\begin{aligned}\Phi_{net} = 12 + 0 - 24 + 0 + 0 \\[1em] \Phi_{net} = -12\ \text{N-m}^2\text C^{-1}\end{aligned}$$

By Gauss' theorem, the charge within the cuboid is

$$\begin{aligned}\text q = \epsilon_0 \Phi_{net} \\[1em] \text q = -12\epsilon_0\ \text C\end{aligned}$$

Its magnitude is 12ε₀ C. Comparing with nε₀ C,

$\text n = 12$

Hence the value of n is 12.

Note: The printed solution writes φ = 2(1)² − 4(2) + 6(3) = 12, adding field components without face areas or outward-normal signs; the net flux is actually − 12 N-m²C^-1 (charge − 12ε₀ C), although its magnitude still gives n = 12.

Given,

• Uniform linear charge density of the infinite cylindrical wire, λ = 2 × 10^-8 Cm^-1
• Mass of the electron, m = 9 × 10^-31 kg
• Charge on the electron, e = 1.6 × 10^-19 C
• $\frac{1}{4\pi\epsilon_0}$ = 9.0 × 10⁹ N-m²/C²

The electric field at a perpendicular distance r from the infinite line charge is

$\text E = \frac{\lambda}{2\pi\epsilon_0 \text r}$

The wire is positively charged and the electron is negative, so the electrostatic force on the electron is directed towards the wire and supplies the necessary centripetal force for the circular path,

$\text{eE} = \frac{\text{mv}^2}{\text r}$

Substituting the value of E,

$\text e\left(\frac{\lambda}{2\pi\epsilon_0 \text r}\right) = \frac{\text{mv}^2}{\text r}$

The radius r cancels from both sides, so

$$\begin{aligned}\text v^2 = \frac{\text e\lambda}{2\pi\epsilon_0 \text m} \\[1em] \text v^2 = \frac{\text e\lambda}{\text m} \times 2 \times \left(\frac{1}{4\pi\epsilon_0}\right)\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v^2 = \frac{(1.6 \times 10^{-19})(2 \times 10^{-8}) \times 2 \times (9.0 \times 10^9)}{9 \times 10^{-31}} \\[1em] = \frac{(3.2 \times 10^{-27})(1.8 \times 10^{10})}{9 \times 10^{-31}} \\[1em] = \frac{5.76 \times 10^{-17}}{9 \times 10^{-31}} \\[1em] = 6.4 \times 10^{13}\ \text m^2\text s^{-2} \\[1em] \text v = \sqrt{6.4 \times 10^{13}} \\[1em] \text v = \sqrt{64 \times 10^{12}} \\[1em] \text v = 8 \times 10^6\ \text{ms}^{-1}\end{aligned}$$

Hence the velocity of the electron is 8 × 10⁶ ms^-1.

Given,

• Surface charge density of the uniformly charged infinite plane sheet S = + σ
• Initial distance of the electron from S at t = 0 is 1 m
• Initial speed of the electron, u = 1 m/s
• Time taken to strike the sheet, t = 1 s
• Maximum value of σ is $\alpha\left[\frac{\text m\epsilon_0}{\text e}\right]$ C/m²

The electric field due to a uniformly charged infinite plane sheet is

$\text E = \frac{\sigma}{2\epsilon_0}$

The sheet is positively charged and the electron carries a negative charge, so the force on the electron is directed towards the sheet. The magnitude of its acceleration is

$$\begin{aligned}\text a = \frac{\text{eE}}{\text m} \\[1em] \text a = \frac{\text e\sigma}{2\epsilon_0 \text m}\end{aligned}$$

This acceleration is independent of the distance from the sheet, so the motion is uniformly accelerated.

For σ to be maximum, the electron must take the longest possible route before striking the sheet. This happens when the electron initially moves away from the sheet with the speed of 1 m/s, is retarded, returns and then strikes the sheet at t = 1 s.

Taking the direction away from the sheet as positive, the displacement in 1 s must be − 1 m. Using

$$\begin{aligned}\text s = \text{ut} + \frac{1}{2}\text{at}^2 \\[1em] -1 = (1)(1) + \frac{1}{2}(-\text a)(1)^2 \\[1em] -1 = 1 - \frac{\text a}{2} \\[1em] \frac{\text a}{2} = 2 \\[1em] \text a = 4\ \text{ms}^{-2}\end{aligned}$$

Substituting this in the expression for the acceleration,

$$\begin{aligned}\frac{\text e\sigma}{2\epsilon_0 \text m} = 4 \\[1em] \sigma = \frac{8\ \text m\epsilon_0}{\text e} \\[1em] \sigma = 8\left[\frac{\text m\epsilon_0}{\text e}\right]\end{aligned}$$

Comparing with $\alpha\left[\frac{\text m\epsilon_0}{\text e}\right]$,

$\alpha = 8$

Hence the value of α is 8.

Given,

• A charge is kept at the central point P of a cylindrical region
• The two edges subtend a half-angle θ at P
• Electric flux through the curved surface is $\Phi$ when θ = 30°, and $\frac{\Phi}{\sqrt{\text n}}$ when θ = 60°

The solid angle subtended by a disc at a point on its axis, the semi-vertex angle of the cone so formed being θ, is

$\omega = 2\pi(1 - \cos\theta)$

The charge q at P lies inside the closed cylinder, so the total flux through the whole cylinder is $\frac{\text q}{\epsilon_0}$, corresponding to the total solid angle of 4π steradian.

The two plane ends of the cylinder together subtend a solid angle

$$\begin{aligned}\omega_{ends} = 2 \times 2\pi(1 - \cos\theta) \\[1em] \omega_{ends} = 4\pi(1 - \cos\theta)\end{aligned}$$

The flux through the two plane ends is therefore

$$\begin{aligned}\Phi_{ends} = \frac{\text q}{4\pi\epsilon_0} \times 4\pi(1 - \cos\theta) \\[1em] \Phi_{ends} = \frac{\text q}{\epsilon_0}(1 - \cos\theta)\end{aligned}$$

Hence the flux through the curved surface is

$$\begin{aligned}\Phi_{curved} = \frac{\text q}{\epsilon_0} - \frac{\text q}{\epsilon_0}(1 - \cos\theta) \\[1em] \Phi_{curved} = \frac{\text q\cos\theta}{\epsilon_0}\end{aligned}$$

When θ = 30° :

$$\begin{aligned}\Phi = \frac{\text q}{\epsilon_0}\cos 30^\circ \\[1em] \Phi = \frac{\text q}{\epsilon_0} \times \frac{\sqrt{3}}{2}\end{aligned}$$

When θ = 60° :

$$\begin{aligned}\Phi' = \frac{\text q}{\epsilon_0}\cos 60^\circ \\[1em] \Phi' = \frac{\text q}{\epsilon_0} \times \frac{1}{2}\end{aligned}$$

Taking the ratio,

$$\begin{aligned}\frac{\Phi'}{\Phi} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\[1em] \frac{\Phi'}{\Phi} = \frac{1}{\sqrt{3}} \\[1em] \Phi' = \frac{\Phi}{\sqrt{3}}\end{aligned}$$

Comparing with $\frac{\Phi}{\sqrt{\text n}}$,

$\text n = 3$

Hence the value of n is 3.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: No two electric lines of force can intersect each other. If two lines of force were to intersect, then at the point of intersection two tangents could be drawn to the two lines of force. This would mean two directions of electric field intensity at the same point, which is not possible.

Reason (R) is also correct: The tangent drawn at any point of an electric line of force gives the direction of the electric field at that point. This is precisely why two intersecting lines would give two directions at one point.

The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For the two spherical conductors it is given that σ₁ = σ₂. Let r₁ and r₂ be their radii, so that

$$\begin{aligned}\frac{\text q_1}{4\pi \text r_1^2} = \frac{\text q_2}{4\pi \text r_2^2} \\[1em] \frac{\text q_1}{\text q_2} = \frac{\text r_1^2}{\text r_2^2}\end{aligned}$$

Reason (R) is also correct: The surface density is the charge per unit area, and the electric field near the surface of a spherical conductor is $\text E = \text k\frac{\text q}{\text r^2}$. Taking the ratio of the intensities near the two surfaces,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text q_1}{4\pi\epsilon_0 \text r_1^2} \times \frac{4\pi\epsilon_0 \text r_2^2}{\text q_2} \\[1em] \frac{\text E_1}{\text E_2} = \frac{\text q_1}{\text q_2} \times \frac{\text r_2^2}{\text r_1^2}\end{aligned}$$

Substituting $\frac{\text q_1}{\text q_2} = \frac{\text r_1^2}{\text r_2^2}$,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text r_1^2}{\text r_2^2} \times \frac{\text r_2^2}{\text r_1^2} \\[1em] \frac{\text E_1}{\text E_2} = 1 \\[1em] \text E_1 = \text E_2\end{aligned}$$

The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The Coulomb force, or electric force, given by Coulomb's law is valid over nuclear distances (about 10^-15 m) as well as over very large distances. Hence the range of the Coulomb force is infinite.

Reason (R) is also correct: From Coulomb's law,

$\text F = \frac{\text k|\text q_1||\text q_2|}{\text r^2}$

the force does act between two charged particles. This generally holds only for charged objects whose sizes are much smaller than the distance between them.

However, the Reason merely states between what the Coulomb force acts; it says nothing about how far the force extends. The infinite range follows from the fact that the force falls off as $\frac{1}{\text r^2}$ and never becomes exactly zero at any finite distance. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a point charge is rotated in a circle around another charge, the work done is zero.

Reason (R) is also correct: The work done is the dot product of force and displacement,

$\text W = \vec{\text F} \cdot \vec{\text S} = \text{FS}\cos\theta$

In rotating the charge in a circle, the electrostatic force is along the radius while the direction of motion is along the tangent, so the angle between the two at any instant is 90°. Therefore

$$\begin{aligned}\text W = \text{FS}\cos 90^\circ \\[1em] \text W = 0\end{aligned}$$

The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a conducting medium is placed between two charges, the electric force between them becomes zero.

Reason (R) is also correct: The dielectric constant of any material is

$$\begin{aligned}\text K = \frac{\text E_0}{\text E} = \frac{\text F_0}{\text F} \\[1em] \text F = \frac{\text F_0}{\text K}\end{aligned}$$

where F₀ is the force when no material is present between the charges and F is the force after the material is introduced. Thus the reduction in the force is inversely proportional to the dielectric constant.

Since the dielectric constant of a conducting medium is infinity,

$$\begin{aligned}\text F = \frac{\text F_0}{\infty} \\[1em] \text F = 0\end{aligned}$$

The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: In electrostatics, the electric lines of force can never form closed loops. An electric line of force originates from a positive charge and terminates on a negative charge, so it cannot return to its starting point.

Reason (R) is also correct: The electric flux is a measure of the number of lines of force passing through a surface, and by Gauss' theorem the flux through a closed surface enclosing a charge q is $\frac{\text q}{\epsilon_0}$. Hence the number of lines of force originating or terminating on a charge is proportional to the magnitude of the charge.

However, the Reason concerns only how many lines of force a charge has; it says nothing about whether a line of force can close on itself. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

Note: The printed key gives option 3, treating the Reason as false. The Reason is a true statement but does not explain the Assertion, so the correct option is 2.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Charge is quantized. The charge q on a body is always given by

$\text q = \text{ne}$

where n is any positive or negative integer and e is the elementary charge.

Reason (R) is incorrect: The charge on an electron is e = 1.6 × 10^-19 C, which is very much less than 1 C. Hence it is not true that a charge less than 1 C is not possible. Quantization means that the charge must be an integral multiple of e, not that it must exceed 1 C.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a cavity within a conductor, the electric field is zero.

Reason (R) is also correct: The charge given to a conductor resides only at its surface.

Since the charges reside only on the outer surface of the conductor, no charge resides on the wall of the empty cavity. An electric line of force must begin on a positive charge and end on a negative charge, and in electrostatics it cannot form a closed loop; with no charge inside the cavity or on its wall, no line of force can exist within the cavity. Moreover, the wall of the cavity is a part of the conductor, so all its points are at the same potential, whereas a line of force always runs from a higher to a lower potential. Hence the electric field in the cavity is zero.

The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Note: The printed explanation infers E = 0 from zero flux through a Gaussian surface, although the chapter states that zero flux does not imply E = 0; the keyed option 1 is nevertheless correct.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: An electrostatic field line never forms a closed loop. The field originates from a positive charge and terminates on a negative charge, and since isolated positive and negative charges do exist, the field lines do not close on themselves.

Reason (R) is also correct: The electrostatic field is a conservative field, which means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final positions and is independent of the path taken.

Since the work done depends only on the initial and the final positions, the work done by the field in taking a charge round any closed path is zero. If a field line formed a closed loop, a positive charge taken round that loop along the line would have the force on it in the direction of its motion at every point, so the work done would be positive and not zero. Hence the conservative nature of the field rules out closed field lines, and the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Note: The printed key gives option 2. Since a closed field line would make the work done round it non-zero, contrary to the conservative nature of the field, the Reason explains the Assertion and the correct option is 1.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a conductor is given a charge, no excess charge appears inside it; the excess charge can reside only on its surface.

Reason (R) is also correct: The electric field inside a conductor is zero.

Since the electric field inside the conductor is zero, the flux through any Gaussian surface drawn wholly within the material of the conductor is zero (E = 0 implies φ = 0). By Gauss' theorem the charge enclosed by every such surface is then zero, so no excess charge can appear inside the conductor and it resides only on the surface. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Note: The printed key gives option 2. Since the zero field inside the conductor is the very premise from which Gauss' theorem shows that no inner charge appears, the Reason explains the Assertion and the correct option is 1.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: Only a charged particle which is initially at rest moves along the electric field line. If the charged particle is already in motion and its initial velocity is at an angle with the electric field line, the particle may follow a parabolic path, just as a projectile does.

Reason (R) is also incorrect: The electric field lines diverge from a positive charge and converge at a negative charge, and not the other way round as stated.

Therefore, both assertion and reason are false.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: When a charged particle enters an electric field in a perpendicular direction, it describes a path of radius of curvature r given by

$$\begin{aligned}\text{qE} = \frac{\text{mv}^2}{\text r} \\[1em] \text r = \frac{\text{mv}^2}{\text{qE}} \\[1em] \text r = \frac{2(\text{K.E.})}{\text{qE}} \qquad \left(\because \text{K.E.} = \frac{1}{2}\text{mv}^2\right)\end{aligned}$$

so that $\text r \propto \frac{1}{\text q}$. The electron and the proton carry charges of equal magnitude and are given the same kinetic energy, so r is the same for both of them. Hence the path of the electron is not more curved than that of the proton.

Reason (R) is also incorrect: As the above relation shows, r depends upon the kinetic energy, which is the same for both the particles, and not upon their masses. So the smaller mass of the electron does not make its curve larger.

Therefore, both assertion and reason are false.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The acceleration of a charged particle in an electric field is

$\vec{\text a} = \frac{\text q\vec{\text E}}{\text m}$

In a non-uniform field $\vec{\text E} \space$varies from point to point, and so does $\vec{\text a}$. But the acceleration depends only on the field at the location of the particle and on its charge and mass; it does not depend on the velocity of the charged particle.

Reason (R) is also correct: Charge is an invariant quantity, that is, the amount of charge on a particle does not depend on the frame of reference.

The invariance of charge is a separate property and is not the reason why the acceleration is independent of velocity; that follows from the relation $\vec{\text a} = \frac{\text q\vec{\text E}}{\text m}$ itself. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: A negative charge placed in an electric field moves along the direction opposite to that of the electric field, and not along the direction of the field.

Reason (R) is also incorrect: The force on a charge q in a field $\vec{\text E} \space$is $\vec{\text F} = \text q\vec{\text E}$. For a negative charge this becomes

$\vec{\text F} = -\text q\vec{\text E}$

so the force acts in the direction opposite to the electric field, and not in the direction of the electric field.

Therefore, both assertion and reason are false.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: Gauss' theorem is valid for all vector fields obeying the inverse square law. It is based on the inverse square dependence of the electric field on distance, so if the inverse square law were not obeyed, Gauss' law would show a diversion, that is, it would no longer hold.

Reason (R) is also correct: Charge is conserved, and Gauss' theorem is consistent with the conservation of charge.

However, the validity of Gauss' theorem follows from the inverse square dependence of the field on distance, and not from the conservation of charge. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Gauss' law cannot conveniently be used to calculate the electric field near an electric dipole.

Reason (R) is also correct: Gauss' law is applicable for any closed surface, but it is most useful where the charge distribution has spherical or cylindrical symmetry, or is distributed uniformly over a plane. An electric dipole is a system of two equal and opposite point charges separated by a very small and finite distance, and such a charge distribution is not symmetrical.

Since the field of a dipole is neither constant in magnitude nor everywhere normal to any convenient closed surface, E cannot be taken out of the integral $\oint \vec{\text E} \cdot \text d\vec{\text A}$. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: By Gauss' theorem, the electric flux through the cube is $\phi = \frac{\text q}{\epsilon_0}$. A cube has six faces of equal area, and with the point charge at the centre they are all symmetrically situated with respect to it. Hence the flux through each face is

$\frac{1}{6}\phi = \frac{1}{6}\left(\frac{\text q}{\epsilon_0}\right)$

Reason (R) is also correct: Gauss' law is very helpful in calculating the electric field in those cases where the electric field is symmetrical around the source producing it.

However, the Reason concerns the calculation of the electric field from Gauss' law, whereas the Assertion follows from the symmetry of the six faces of the cube about the central charge. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The flux emerging out is 8 × 10³ V-m and that entering is 2 × 10³ V-m. Taking the outgoing flux as positive and the incoming flux as negative, the net flux through the closed surface is

$$\begin{aligned}\phi = 8 \times 10^3 - 2 \times 10^3 \\[1em] \phi = 6 \times 10^3\ \text{V-m}\end{aligned}$$

Reason (R) is also correct: Gauss' theorem in electrostatics may indeed be applied to verify the result. According to it,

$$\begin{aligned}\phi = \frac{\text q}{\epsilon_0} \\[1em] \text q = \epsilon_0 \phi\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text q = (8.85 \times 10^{-12})(6 \times 10^3) \\[1em] = 53.10 \times 10^{-9}\ \text C \\[1em] \text q = 0.053\ \mu\text C\end{aligned}$$

which is the value stated in the Assertion. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When an electric dipole is enclosed by a closed surface, the net flux coming out of the surface is zero, whatever the value of the dipole moment.

Reason (R) is also correct: An electric dipole consists of two equal and opposite charges separated by a small distance.

Since the two charges are equal and opposite, the net charge enclosed by the surface is zero. By Gauss' theorem the net flux is $\frac{1}{\epsilon_0}$ times the charge enclosed,

$$\begin{aligned}\Phi_\text E = \frac{0}{\epsilon_0} \\[1em] \Phi_\text E = 0\end{aligned}$$

The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.