Electrostatics

Here we are assuming that " $\ell$ " is very large just for the sake of symmetry. Outside cylinder will have zero electric field inside, so the flux generated on the plate will be due to inner cylinder only in sections AB and CD , as section be will be at that place where electric field is zero.

Flux through element will be =

$ \begin{aligned} \mathrm{d} \phi & =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}} \\ \mathrm{~d} \phi & =\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \mathrm{dy} \cdot \ell \cdot \cos \theta .........(1) \end{aligned} $

from figure we can say that

$ \begin{aligned} & \cos \theta=\frac{R}{r} \Rightarrow r=R \sec \theta \\ & \tan \theta=\frac{y}{R} \Rightarrow y=R \tan \theta \\ & \Rightarrow d y=R \sec ^2 \theta d \theta \\ & d \phi=\frac{2 k \lambda}{R \sec \theta} R \cdot \sec ^2 \theta \cdot \ell \cdot \cos \theta d \theta \\ & d \phi=2 k \lambda \ell d q \end{aligned} $

$\begin{aligned} & \int_0^{\phi_{\mathrm{AB}}} \mathrm{d} \phi=2 \mathrm{k} \lambda \ell \int_{\pi / 4}^{\pi / 3} \mathrm{~d} \theta \\ & \phi_{\mathrm{AB}}=2 \mathrm{k} \lambda \ell\left[\frac{\pi}{3}-\frac{\pi}{4}\right] \\ & \phi_{\mathrm{AB}}=2 \mathrm{kQ}\left[\frac{\pi}{12}\right] \\ & \phi_{\mathrm{AB}}=2 \times \frac{1}{4 \pi \epsilon_0 \epsilon_{\mathrm{r}}} \mathrm{Q}\left[\frac{\pi}{12}\right] \\ & \phi_{\mathrm{AB}}=\frac{\mathrm{Q}}{120 \epsilon_0} \\ & \phi_{\text {plate }}=\phi_{\mathrm{AB}}+\phi_{\mathrm{BC}}+\phi_{\mathrm{CD}} \quad\left(\phi_{\mathrm{CD}}=\phi_{\mathrm{AB}}\right) \\ & \phi_{\text {plate }}=\frac{\mathrm{Q}}{60 \epsilon_0}\end{aligned}$

The electric field $\vec{E}$ due to a dipole at a point is given by:

On the axial line (along the dipole axis): $E_{\text {axial }}=\frac{2 k p}{r^3}$ (along the dipole moment )

where, $K=\frac{1}{4 \pi \varepsilon_0}$

On the equatorial line (perpendicular bisector):

$ \text { Equatorial }=\frac{-k p}{r^3}\binom{\text { opposite to }}{\text { the dipole moment }} $

Here, $x$ is on the equatorial line for both dipoles:

$ \vec{E}=\overrightarrow{E_1}+\overrightarrow{E_2} $

$ \text { } \begin{aligned} & So\,\,\, \vec{E}=-\frac{k p}{r^3} \hat{j}-\frac{k p}{r^3} \hat{j} \\ & \Rightarrow \vec{E}=-\frac{2 k p}{r^3} \hat{j}=\frac{-2}{r^3}\left(\frac{1}{4 \pi \varepsilon_0}\right) p{\hat{j}} \\ & \Rightarrow \vec{E}=\frac{-p}{2 \pi \varepsilon_0 r^3} \hat{j} \\ & \text { So }(p) \rightarrow(2) \end{aligned} $

Here, $x$ is on the equatorial line for both dipoles.

$ \begin{aligned} &\text { Electric field at } x \text { due to dipole } 1 \text {, }\\ &\vec{E}_1=-\frac{k p}{r^3} \hat{j} \end{aligned} $

Electric field at $x$ due to dipole 2,

$ \vec{E}_2=\frac{k p}{r^3} \hat{j} \quad\left(\begin{array}{l} \text { as dipole moment } \\ \text { is in }-\hat{j} \text { direction }) \end{array}\right. $

so. $\vec{E}=\vec{E}_1+\vec{E}_2$

$ \begin{aligned} & =\frac{-k p}{r^3} \hat{j}+\frac{k p}{r^3} \hat{j} \\ & \Rightarrow \vec{E}=0 \end{aligned} $

So $($ Q $) \longrightarrow(1)$

Here, $x$ is on the equatorial

line for dipole 1 and on the axial line for dipole 2.

So, $\vec{E}_1=\frac{-k p}{r^3} \hat{j}$

and $\vec{E}_2=\frac{2 k p}{r^3} \hat{i}$

$ \begin{aligned}So\,\,\,\,\, \vec{E} & =\vec{E}_1+\vec{E}_2 \\ & =\frac{-k p}{r^3} \hat{j}+\frac{2 k p}{r^3} \hat{i} \\ & =\frac{k p}{r^3}(-\hat{j}+2 \hat{i}) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \vec{E}=\frac{k p}{r^3}(2 \hat{i}-\hat{j}) \\ & \Rightarrow \vec{E}=\frac{p}{4 \pi \varepsilon_0 r^3}(2 \hat{i}-\hat{j}) \end{aligned}\\ &\text { So }(R) \rightarrow(4) \end{aligned} $

Here, $x$ is on the axial line for both dipoles.

so $\vec{E}_1=\frac{2 k p}{r^3} \hat{i}$ and $\overrightarrow{E_2}=\frac{2 k p}{r^3} \hat{i}$

So

$ \begin{aligned} \vec{E} & =\vec{E}_1+\vec{E}_2 \\ & =\frac{2 k p}{r^3} \hat{i}+\frac{2 k p}{r^3} \hat{i} \\ \Rightarrow \vec{E} & =\frac{4 k p}{r^3} \hat{i} \\ \Rightarrow \vec{E} & =4\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{p}{r^3} \hat{i} \\ \Rightarrow \vec{E} & =\frac{p}{\pi \varepsilon_0 r^3} \hat{i} \end{aligned} $

So $(S) \rightarrow(5)$

Hence, option (c) is correct.

Consider a circular hoop of radius $a$ with centre $O$.

Bead $A$ is fixed to the hoop, while bead $B$ can slide freely along its circumference.

Forces on bead $B$:

Electrostatic force due to bead $A$, acting along the line $AB$.

Normal reaction from the hoop, directed radially inward toward $O$.

Because these two forces are the only interactions, bead $B$ reaches equilibrium at the point $C$—the position diametrically opposite $A$—where the Coulomb repulsion is exactly balanced by the hoop’s normal reaction.

Displace bead $B$ slightly from its equilibrium point $C$ by a small angle $\theta$ measured at the centre $O$.

1. Geometry of the configuration

In the isosceles triangle $OAB$ (see figure)

$\angle A=\dfrac{\theta}{2}$

$\angle O = 180^{\circ}-\theta$

The side opposite $\angle O$ is $AB$. Using the cosine rule,

$ AB^{2}=a^{2}+a^{2}-2a^{2}\cos(180^{\circ}-\theta)=4a^{2}\cos^{2}\!\frac{\theta}{2}. $

Hence the separation of the charges is

$ r=AB=2a\cos\frac{\theta}{2}. $

2. Coulomb force on $B$

$ F=\frac{q^{2}}{4\pi\varepsilon_{0}r^{2}} =\frac{q^{2}}{16\pi\varepsilon_{0}a^{2}\cos^{2}\dfrac{\theta}{2}}. $

3. Components along the hoop

The force makes an angle $\dfrac{\theta}{2}$ with the radial line $OB$.

Normal component: $F\cos\dfrac{\theta}{2}$  (balanced by the normal reaction of the hoop).

Tangential component: $F\sin\dfrac{\theta}{2}$  (provides the restoring acceleration).

Applying Newton’s second law tangentially,

$ ma\frac{d^{2}\theta}{dt^{2}} =-F\sin\frac{\theta}{2} =-\frac{q^{2}}{16\pi\varepsilon_{0}a^{2}\cos^{2}\dfrac{\theta}{2}}\, \sin\frac{\theta}{2}. $

(The minus sign shows the couple is restoring.)

4. Small-angle approximation

$ \sin\frac{\theta}{2}\simeq\frac{\theta}{2},\qquad \cos\frac{\theta}{2}\simeq1. $

Substituting,

$ \frac{d^{2}\theta}{dt^{2}} =-\frac{q^{2}}{32\pi\varepsilon_{0}ma^{3}}\;\theta =-\omega^{2}\theta, $

where

$ \boxed{\displaystyle \omega^{2} =\frac{q^{2}}{32\pi\varepsilon_{0}ma^{3}} }. $

The motion is simple harmonic, with angular frequency $\omega$ given above.

For a point charge Q at the origin, we have

$E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{d^2}}}$

$ \Rightarrow E \propto {1 \over {{d^2}}}$

Thus, the correct mapping is $R \to 1$.

For a small dipole with point charges Q at (0, 0, l) and $-$Q at (0, 0, $-$l), 2l << d, we have

$E = {1 \over {4\pi {\varepsilon _0}}}{{2Q \times 2l} \over {{d^3}}}$

$ \Rightarrow E \propto {1 \over {{d^3}}}$

Thus, the correct mapping is $S \to 2$.

For an infinite line charge coincident with x-axis with uniform linear charge density $\lambda$, we have

$E = {\lambda \over {2\pi {\varepsilon _0}d}}$

$ \Rightarrow E \propto {1 \over d}$

Thus, the correct mapping is $Q \to 3$.

For two infinite wires carrying uniform linear charge density parallel to the x-axis, we have

$E = {\lambda \over {2\pi {\varepsilon _0}(d - l)}} - {\lambda \over {2\pi {\varepsilon _0}(d + l)}} = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{1 \over {d - l}} - {1 \over {d + l}}} \right)$

$ \Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{{d + l - d + l} \over {(d - l)(d + l)}}} \right) = {\lambda \over {2\pi {\varepsilon _0}}}{{(2l)} \over {({d^2} - {l^2})}}$

Since $2l < < d$

$ \Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}{{2l} \over {{d^2}}}$

$ \Rightarrow E \propto {1 \over {{d^2}}}$

Thus, the correct mapping is $R \to 4$.

For an infinite plane charge coincident with the x-y plane with uniform surface charge density, we have

$E = {\sigma \over {2\pi {\varepsilon _0}}}$

Therefore, E is independent of d.

Thus, the correct mapping is $P \to 5$.

Therefore, option (B) is correct.

The distance between the two plates is h. The potential of the bottom plate is V₀ and that of the top plates is $-$V₀. The electric field between the plates is E = 2V₀/h (directed upwards). The radius of each ball is r (<< h). Let m be the mass and C be the capacitance of each ball.

When the ball touches the bottom plate, it gets a positive charge q = CV₀ (we assume that the charge transfer is instantaneous). This positively charged ball experiences an upward force, F = qE = 2CV$_0^2$/h, which accelerates the ball upwards. Since the force is constant, the ball cannot do SHM (for SHM, the force should be proportional to the displacement and directed towards the centre).

When the ball hits the top plate, it transfers the positive charge to the plate and gets negative charge q = $-$CV₀. This negatively charged ball again experience a force F = qE (downward) and starts accelerating downwards. Thus, the ball keeps moving between the bottom and the top plates carrying a charge +q upwards and $-$q downwards.

As the balls keep on oscillating between plates, current will flow even in steady state.

Let a ball attains charge q at bottom plate. Then

${{Kq} \over r} = {V_0} \Rightarrow q = {{{V_0}r} \over K}$

Electric field inside the chamber, $E = [{V_0} - ( - {V_0})]/h$

$ = 2{V_0}/h$

$\therefore$ Acceleration of each ball, $a = {{qE} \over m} = {{2V_0^2r} \over {mhK}}$

$\therefore$ Time taken by a ball to reach the other plate

$t = \sqrt {{{2h} \over a}} = {h \over {{V_0}}}\sqrt {{{mK} \over r}} $

$\therefore$ Average current due to n such balls,

${I_{av}} = {{nq} \over t} = {{n{V_0}r} \over K} \times {{{V_0}} \over h}\sqrt {{r \over {mK}}} \Rightarrow {I_{av}} \propto V_0^2$

Electric field due to uniformly charged dielectric solid sphere of radius R at a point P is given by

$E = \left\{ {\matrix{ {k\,.\,{Q \over {{R^3}}}\,.\,r,} & {r \le R} \cr {k\,.\,{Q \over {{r^2}}},} & {r > R} \cr } } \right.$

In first two cases, P is outside the sphere. Therefore,

${E_1} = {1 \over {4\pi {E_0}}}.\,{Q \over {{R^2}}}$ ..... (1)

${E_2} = {1 \over {4\pi {E_0}}}.\,{{2Q} \over {{R^2}}}$ .... (2)

In Sphere 3, point P located is inside. Therefore,

${E_3} = \left( {{\rho \over {3{E_0}}}} \right)r$

$\rho = {{4Q} \over {(4/3)\pi {{(2R)}^3}}}$

$ = {{12Q} \over {4\pi \,.\,8{R^3}}} = {{3Q} \over {8\pi }}$

${E_3} = \left( {{{3Q} \over {\pi {R^3}}}} \right)R$

${E_3} = {1 \over {4\pi {E_0}}}\,.\,{Q \over {2{R^2}}}$ ..... (3)

${E_2} > {E_1} > {E_3}$

(P) Component of forces along x-axis will vanish. Net force along positive y-axis.

(Q) Component of forces along y-axis will vanish. Net force along positive x-axis

(R) Component of forces along x-axis will vanish. Net force along negative y-axis.

(S) Component of forces along y-axis will vanish. Net force along negative x-axis

The electric field at $Q$

$ \vec{E}_Q=\frac{1}{4 \pi \epsilon_0}\left[\frac{\frac{4}{3} \pi R^3 \rho_1}{(2 R)^2} \hat{\imath}-\frac{\frac{4}{3} \pi R^3 \rho_2}{R^2} \hat{\imath}\right]=\overrightarrow{0} $

gives $\rho_1 / \rho_2=4$

The electric field at $P$,

$ \vec{E}_P=-\frac{1}{4 \pi \epsilon_0}\left[\frac{\frac{4}{3} \pi R^3 \rho_1}{(2 R)^2} \hat{\imath}+\frac{\frac{4}{3} \pi(2 R)^3 \rho_2}{(5 R)^2} \hat{\imath}\right]=\overrightarrow{0}, $

gives $\rho_1 / \rho_2=-32 / 25$.

Given that the proton moves at a $45^{\circ}$ angle to the vertical, we can derive the electric field (E) as follows :

$ qE = mg $ or $ E = \frac{mg}{q} $

Since $ E = \frac{X}{d} $, we can substitute to find $ X $ :

$ \frac{X}{d} = \frac{mg}{q} $ or $ X = \frac{mgd}{q} $

Where :

• $ m = 1.67 \times 10^{-27} \, \mathrm{kg} $


• $ g = 10 \, \mathrm{m/s^2} $


• $ d = 1 \, \mathrm{cm} = 1 \times 10^{-2} \, \mathrm{m} $


• $ q = 1.6 \times 10^{-19} \, \mathrm{C} $

Substituting these values, we get :

$ \begin{aligned} X &= \frac{1.67 \times 10^{-27} \times 10 \times 1 \times 10^{-2}}{1.6 \times 10^{-19}} \, \mathrm{V} \\\\ &= \frac{1.67}{1.6} \times 10^{-9} \, \mathrm{V} \\\\ &= 1 \times 10^{-9} \, \mathrm{V} \end{aligned} $

Consider a thin spherical shell of radius $ R $ with its center at the origin, carrying a uniform positive surface charge density. To understand how the electric field $ \left| \overrightarrow{E} \left( r \right) \right| $ and the electric potential $ V(r) $ vary with the distance $ r $ from the center, refer to the following details:

Electric Field due to a Uniformly Charged Thin Spherical Shell

Inside the Shell

For $ r < R $, the electric field is given by:

$ E_{\text{inside}} = 0 \quad [ r < R ] $

On the Surface of the Shell

For $ r = R $, the electric field is:

$ E_{\text{surface}} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2} $

Outside the Shell

For $ r > R $, the electric field is:

$ E_{\text{outside}} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \quad [ r > R ] $

Below is a graphical representation of the electric field $ E $ variation with the distance $ r $ from the center:

Electric Potential due to a Uniformly Charged Thin Spherical Shell

Inside the Shell

For $ r < R $, the electric potential is:

$ V_{\text{inside}} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} $

On the Surface of the Shell

For $ r = R $, the electric potential is:

$ V_{\text{surface}} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R} $

Outside the Shell

For $ r > R $, the electric potential is:

$ V_{\text{outside}} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} $

Below is a graphical representation of the electric potential $ V $ variation with the distance $ r $ from the center:

The flux through an area $\overrightarrow S $ due to an electric field $\overrightarrow E $ is given by

$\phi = \oint {\overrightarrow E \,.\,d\overrightarrow S } $

$ = \overrightarrow E \,.\,\oint {d\overrightarrow S = \overrightarrow E \,.\,\overrightarrow S } $ ($\because$ $\overrightarrow E $ is a constant.) ...... (1)

The area of the shaded region is the cross product of vectors representing the two sides i.e.,

$\overrightarrow S = (a\widehat j) \times (a\widehat i + a\widehat k) = {a^2}(\widehat i - \widehat k)$ ...... (2)

Use equations (1) and (2) to get

$\phi = ({E_0}\widehat i)\,.\,{a^2}(\widehat i - \widehat k) = {E_0}{a^2}$.

The magnetic fields and the induced electric fields form the closed loops.

The forces acting on the oil drop are its weight, buoyant force, and electrostatic force. The buoyant force on the oil drop is very small as compared to other two forces. Thus, the weight of the spherical oil drop is balanced by the electrostatic force

$qE = mg$

$qE = {4 \over 3}\pi {r^3}\rho g$ ..... (i)

where, r = radius of oil drop, $\rho$ = density of oil, q = charge on the oil drop

IN the absence of electric field,

IN equilibrium,

Viscous force on the drop = Weight of the drop

$6\pi \eta r{v_T} = mg$

$6\pi \eta r{v_T} = {4 \over 3}\pi {r^3}\rho g$ ...... (ii)

where, v_T = terminal velocity, $\eta $ = coefficient of viscosity of air

or ${r^2} = {{18 \times \eta \times {v_T}} \over {4 \times \rho \times g}}$

Substituting the given values, we get

${r^2} = {{18 \times 1.8 \times {{10}^{ - 5}} \times 2 \times {{10}^{ - 3}}} \over {4 \times 900 \times 9.8}}$ or $r = {3 \over 7} \times {10^{ - 5}}$ m

From equation (i), we get $q = {{4\pi {r^3}\rho g} \over {3E}}$

Substituting the given values, we get

$q = {{4 \times \pi \times {{\left( {{3 \over 7} \times {{10}^{ - 5}}} \right)}^3} \times 900 \times 9.8} \over {3 \times \left( {{{81\pi } \over 7} \times {{10}^5}} \right)}}$

$ = {{4 \times \pi \times 7 \times 3 \times 3 \times 3 \times {{10}^{ - 15}} \times 900 \times 9.8} \over {3 \times 81\pi \times {{10}^5} \times 7 \times 7 \times 7}} = 8 \times {10^{ - 19}}C$

Electrostatic pressure is

$\left( {{{{\sigma ^2}} \over {2{\varepsilon _0}}}} \right)$ N/m²

The force is

$F = {{{\sigma ^2}} \over {2{\varepsilon _0}}} \times \pi {R^2}$

Therefore, $F \propto {{{\sigma ^2}{R^2}} \over {{\varepsilon _0}}}$

According to Gauss's law, we have

$\mathrm{Flux=\frac{Charge~enclosed}{\varepsilon_0}}$

The enclosed charges are half of the charges in the disc, point charge at $(a/4,-a/4,0)$, and charge in the rod from point $a/4$ to $a/2$, that is, 8 C/4. Therefore,

Flux $ = {{3C - 7C + (8C/4)} \over {{\varepsilon _0}}} = - {{2C} \over {{\varepsilon _0}}}$

Surface charge density is

$\rho = {Q \over {4\pi {R^2}}}$

where R is the radius of the sphere. For cancelling the field inside the cavity due to the shell of radius R, the shell with radius 2R induces a charge $-$Q in its inner surface and hence the total charge on the outer surface is $Q_1+Q_2$. Similarly, the charge on the outermost shell is

${Q_1} + {Q_2} + {Q_3}$

It is given that the surface charge densities $\rho_1,\rho_2$ and $\rho_3$ are equal, that is,

${{{Q_1}} \over {4\pi {R^2}}} = {{{Q_1} + {Q_2}} \over {4\pi {{(2R)}^2}}} = {{{Q_1} + {Q_2} + {Q_3}} \over {4\pi {{(3R)}^2}}}$

$ \Rightarrow {Q_1} = {{{Q_1} + {Q_2}} \over 4} = {{{Q_1} + {Q_2} + {Q_3}} \over 9}$

That is,

${Q_1} = {{{Q_1}} \over 4} + {{{Q_2}} \over 4}$

$ \Rightarrow {Q_2} = 3{Q_1}$

Therefore,

${Q_3} = 5{Q_1}$

Hence, the ratio of the charges given to the shells ${Q_1}:{Q_2}:{Q_3}$ is 1 : 3 : 5.

Case (P) : $E=0,V=0, B=0,\mu=0$.

Case (Q) : $E\ne0,V=0, B=0,\mu=0$.

Case (R) : $E=0,V\ne0, B\ne0,\mu\ne0$.

Case (S) : $E=0,V\ne0, B\ne0,\mu\ne0$.

Case (T) : $E\ne0,V=0, B=0,\mu=0$.

Electric field at point O is vector sum of electric field due to all three point charges individually.

$\overrightarrow E = \overrightarrow {{E_A}} + \overrightarrow {{E_B}} + \overrightarrow {{E_C}} $ and $\overrightarrow {{E_A}} = - \overrightarrow {{E_B}} $

So, $\overrightarrow E = \overrightarrow {{E_C}} = {{2q/3} \over {4\pi {\varepsilon _0}.{R^2}}}$ along $\overrightarrow {OC} $

$ = {q \over {6\pi \varepsilon _0^2.{R^2}}}$ along $\overrightarrow {OC} $

$ = {{ + q'} \over {6\pi \varepsilon _0^2.{R^2}}}\widehat i$

$ = {{ - q'} \over {6\pi \varepsilon _0.{R^2}}}\widehat j$

From geometry, we can find that $\angle$ABC = 30$^\circ$ and $\angle$ACB = 90$^\circ$

So, AB = 2R, AC = R, BC = $\sqrt3$R

Total potential energy of the system is,

$U = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{q/3 \times q/3} \over {2R}} - {{q/3 \times 2q/3} \over R} - {{q/3 \times 2q/3} \over {\sqrt 3 R}}} \right] \ne 0$

$F = {{q/3 \times q/3} \over {4\pi {\varepsilon _0} \times {{(\sqrt 3 R)}^2}}} = {{{q^2}} \over {54\pi {\varepsilon _0}{R^2}}}$

Potential at point O is,

$V = {{q/3 \times q/3 - 2q/3} \over {4\pi {\varepsilon _0}R}} = 0$

Time constant ($\tau$) = RC$_{eq}$

Where R = Resistance

C$_{eq}$ = equivalent capacitance

And, $v = {{ - dx} \over {dt}} \Rightarrow dx = - vdt$

$ \Rightarrow \int_{d/3}^x {dx = - v\int_0^t {dt} } $

$ \Rightarrow [x]_{d/3}^x = - v[t]_0^t$

$x - {d \over 3} = - vt$

$x = {d \over 3} - vt$ ..... (i)

And also, ${1 \over {{C_{eq}}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}} = {1 \over {{{{\varepsilon _0}A} \over {(d - x)}}}} + {1 \over {{{{\varepsilon _0}Ak} \over x}}}$

${1 \over {{C_{eq}}}} = {{d - x} \over {{\varepsilon _0}A}} + {x \over {{\varepsilon _0}Ak}}$

$ = {1 \over {{\varepsilon _0}A}}\left[ {d - x + {x \over k}} \right]$

$ = {1 \over {{\varepsilon _0}A}}\left[ {{{(d - x)k + x} \over k}} \right]$

${1 \over {{C_{eq}}}} = {1 \over {{\varepsilon _0}Ak}}[dk - xk + x]$

$\therefore$ ${C_{eq}} = {{{\varepsilon _0}A} \over {kd + x(1 - k)}}$ .... (ii)

Given, $A = 1,k = 2,x = {d \over 3} - vt$

Hence, ${C_{eq}} = {{{\varepsilon _0} \times 1} \over {2d + \left[ {{d \over 3} - vt} \right](1 - 2)}}$

$ = {{6{\varepsilon _0}} \over {5d + 3vt}}$

Therefore, $\tau = R{C_{eq}} = {{6R{\varepsilon _0}} \over {5d + 3vt}}$

Statement 1 is True.

The earth is used as a reference with zero potential. The reasons are :

(i) Due to large size of the earth, its potential does not change even if a small amount of charge is given to the earth or taken away from it. Also, all conductors on the earth which are not given any external charge, are very nearly at the potential V_e. The choice of reference is quite arbitrary and is largely governed by convenience.

(ii) The earth is a good conductor. It is easily accessible for electrical circuits geographically distributed over it and hence it is used as a common reference with zero potentil.

Statement 2 is true.

V = ${Q \over {4\pi {\varepsilon _0}R}}$

Therefore, statement 2 is correct explanation for statement 1.

At $r = R$,

By applying Gauss law,

$\phi = {{{q_{enclosed}}} \over {{\varepsilon _0}}}$

$ \Rightarrow \oint {E.ds = {{{q_{enclosed}}} \over {{\varepsilon _0}}}} $

$ \Rightarrow E\oint {ds = {{{q_{enclosed}}} \over {{\varepsilon _0}}}} $

$ \Rightarrow E.4\pi {R^2} = {{Ze} \over {{\varepsilon _0}}}$

$ \Rightarrow E = {1 \over {4\pi {\varepsilon _0}}}{{Ze} \over {{R^2}}}$

i.e., $E\,\mu {1 \over {{R^2}}}$

Here, electric field is independent of 'a'.

For a = 0, the graph is as shown, the equation for the graph line is

The charge in the dotted element shown in figure is

$dq = \sigma \times 4\pi {r^2}dr$

$\therefore$ $dq = \left( {d - {d \over R}r} \right)4\pi {r^2}dr$

$ \Rightarrow Ze = \int_0^R {4\pi d{r^2}dr - \int_0^R {{{4\pi d} \over R}{r^3}dr} } $

$Ze = 4\pi d{{{R^3}} \over 3} - {{4\pi d} \over R}{{{R^4}} \over 4}$

$\therefore$ ${{Ze} \over {4\pi d{R^3}}} = {1 \over 3} - {1 \over 4} = {1 \over {12}}$

$\therefore$ $d = {{3Ze} \over {\pi {R^3}}}$

The electric field inside a uniformly charged sphere is proportional to r. The volume charge density inside a uniformly charged sphere is constant and hence a = R.

When cavity is not present:

Electric field inside the solid sphere

$ \overrightarrow{\mathrm{E}_{1}}=\frac{\rho}{3 \varepsilon_{0}} \overrightarrow{r_{1}} $ ..... (i)

Solid sphere after making cavity inside solid sphere,

$ \overrightarrow{\mathrm{E}_{2}}=\frac{\rho}{3 \varepsilon_{0}} \overrightarrow{r_{2}} $ ...... (ii)

${\overrightarrow E _{net}} = \overrightarrow {{E_1}} - \overrightarrow {{E_2}} = {\rho \over {3{\varepsilon _0}}}\overrightarrow {{r_1}} - {\rho \over {3{\varepsilon _0}}}\overrightarrow {{r_2}} $

$=\frac{\rho}{3 \varepsilon_{0}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) $ ...... (iii)

From eq (iii), we get

$\therefore \quad \mathrm{E}_{\mathrm{net}}=\frac{\rho}{3 \varepsilon_{0}}(\vec{r})$

Hence, electric field inside the emptied space (cavity) is non-zero and uniform.

Consider that,

$\mathrm{P}=(-a, 0,0)$ and $\mathrm{Q}=(0, a, 0)$ points lie on the equatorial plane of dipole.

The charges makes an electric dipole.

Therefore, potential at $\mathrm{P}=$ Potential at $\mathrm{Q}=0$

Work done $(\mathrm{W})=q\left(\mathrm{~V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}\right)=0$

Let the radii of the inner and outer cylinders be $r_1$ and $r_2$, respectively. The problem has cylindrical symmetry and Gauss's law can be applied to find the electric field $E$.

In the case (A), surface charge density $\sigma$ is given to the inner cylinder. Take a cylinder of radius $r\left(r_1<\right. r

$ E(2 \pi r l)=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{\sigma\left(2 \pi r_1 l\right)}{\epsilon_0}, \quad \text { i.e., } \quad E=\frac{\sigma r_1}{\epsilon_0 r} . $

Potential difference between the two cylinders is given by

$ \Delta V=-\int_{r_1}^{r_2} E \mathrm{~d} r=-\frac{\sigma r_1}{\epsilon_0} \ln \left(\frac{r_2}{r_1}\right) $

In the case (B), charge density $\sigma$ is given to the outer cylinder. Apply Gauss's law to get $E=0$ (because $q_{\mathrm{enc}}=0$ ) and hence $\Delta V=0$.

In the case (C), uniform linear charge density $\lambda$ is kept along the common axis of cylinders. Apply Gauss's law as in case (A) to get the electric field

$ E(2 \pi r l)=\lambda l / \epsilon_0, \quad \text { i.e., } \quad E=\lambda /\left(2 \pi \epsilon_0 r\right) . $

Potential difference between the two cylinders is

$ \Delta V=-\int_{r_1}^{r_2} E \mathrm{~d} r=-\frac{\lambda}{2 \pi \epsilon_0} \ln \left(\frac{r_2}{r_1}\right) $

In the case (D) charge density $\sigma$ is given to both the cylinders. Using the argument similar to the case (A), we get, $\Delta V=-\frac{\sigma r_1}{\epsilon_0} \ln \left(\frac{r_2}{r_1}\right)$.

Net charge on sphere = $-20+20$ = 0

When we brought +q charge near neutral conducting sphere induction no produced.

So, net charge on sphere should be zero.

When +q is placed option (d).

Outside the conducting sphere, rearrange of charge taken place on sphere.