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.

Before grounding

$ \begin{aligned} & \mathrm{V}_{\text {sphere }}=\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {net }}=\left(\mathrm{V}_{\mathrm{C}}\right)_{\mathrm{q}}+\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {ind }} \\ & \mathrm{V}_{\text {sphere }}=\frac{\mathrm{kq}}{\ell}+0=\frac{9 \times 10^9 \times 10^{-8}}{0.2}=\frac{90}{0.2}=\frac{900}{2}=450 \text { volt } \end{aligned} $

After grounding

$ \begin{aligned} & \frac{\mathrm{kQ}}{\ell}+\frac{\mathrm{kq}_{\mathrm{s}}}{\mathrm{R}}=\mathrm{V}_{\text {sphere }}^{\prime}=0 \\ & \mathrm{q}_{\mathrm{s}}=-\frac{\mathrm{R}}{\ell} \mathrm{q}=-\frac{1}{2} \times 10^{-8}=-5 \times 10^{-9} \\ & \mathrm{q}_{\mathrm{s}}=-5 \times 10^{-9} \text { Coulomb } \end{aligned} $

Charge flower from sphere to ground $=5 \times 10^{-9}$ Coulomb

After grounding is removed

$ \begin{aligned} & \left(\mathrm{V}_{\text {sphere }}\right)_{\text {final }}=\frac{\mathrm{kq}}{\ell^{\prime}}+\frac{\mathrm{kq}_{\mathrm{s}}}{\mathrm{R}} \\ & =\frac{9 \times 10^9 \times 10^2 \times 10^{-8}}{30}-\frac{9 \times 10^9 \times 5 \times 10^{-9} \times 10^2}{10} \\ & =\frac{9 \times 1000}{30}-450=300 \mathrm{volt}-450 \mathrm{volt}=-150 \mathrm{volt} \end{aligned} $

$\begin{aligned}\left(E_4\right)_I= & \frac{\sigma_0}{2 \epsilon_0}[1-1+1-1-1+1]=0 \\ \left(V_{\text {First }}\right)_I & =\frac{-\sigma_0}{2 \epsilon_0}[-1+2-3+4-5] \mathrm{d} \\ & =\frac{-\sigma_0}{2 \epsilon_0}[-3] \mathrm{d}=\frac{\sigma_0 3 \mathrm{~d}}{2 \epsilon_0}\end{aligned}$

$\begin{aligned} & \begin{aligned}\left(V_{\text {Last }}\right)_I & =\frac{-\sigma_0}{2 \epsilon_0}[1-2+3-4+5] \\ & =\frac{\sigma_0}{2 \epsilon_0}[-3 \mathrm{~d}]\end{aligned} \\ & \begin{aligned}\left(V_{\text {First }}-V_{\text {Last }}\right)_I & =\frac{3 \sigma_0 \mathrm{~d}}{\epsilon_0} \\ & =\frac{3 \times 9 \times 10^{-6} \times 1 \times 10^{-6}}{9 \times 10^{-12}}=3 \mathrm{volt}\end{aligned}\end{aligned}$

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.

The electric field inside sphere is zero. So dipole will oscillate when $r>R$.

$\begin{aligned} & \text { For } \mathrm{r}>\mathrm{RE}=\frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2} \\ & \omega=\sqrt{\frac{\mathrm{PE}}{\mathrm{I}}}=\sqrt{\frac{\mathrm{P}_0 \sigma \mathrm{R}^2}{\mathrm{I} \varepsilon_0 \mathrm{r}^2}} \end{aligned}$

when $r=2 R$

$\omega=\sqrt{\frac{\mathrm{P}_0 \sigma}{4 \mathrm{I} \varepsilon_0}}$

when $r=10 R$

$\omega=\sqrt{\frac{\mathrm{P}_0 \sigma}{100 \mathrm{I} \varepsilon_0}}$

When, x = q, the situation is symmetric

$\Rightarrow$ Electric field at O would be zero.

$\Rightarrow$ (A) is correct.

When x = $-$q, we can think of x as q + ($-$2q) $\Rightarrow$ Magnitude of electric field at $O = {1 \over {4\pi { \in _0}}}{{(2q)} \over {{{\left( {2 \times {{\sqrt 3 a} \over 2}} \right)}^2}}}$

$ = {1 \over {4\pi { \in _0}}}{{2q} \over {3{a^2}}} = {q \over {6\pi { \in _0}{a^2}}}$

$\Rightarrow$ (B) is correct.

For x = 2q, potential at O is

${V_0} = 6 \times {1 \over {4\pi { \in _0}}} \times {q \over {\sqrt 3 a}} + {1 \over {4\pi { \in _0}}}{q \over {\sqrt 3 a}}$

$ = {{7q} \over {4\sqrt 3 \pi { \in _0}a}}$

$\Rightarrow$ (C) is correct.

For $x = - 3q,\,{V_0} = 2 \times {1 \over {4\pi { \in _0}}} \times {q \over {\sqrt 3 a}} = {q \over {2\sqrt 3 \pi { \in _0}a}}$

$\Rightarrow$ (D) is not correct.

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.

Let Gaussian surface be same as the non-conducting spherical shell. The charge enclosed by this Gaussian surface is the charge on the line segment PQ i.e.,

${q_{enc}} = PQ\lambda = \sqrt 3 R\lambda $,

where we used law of cosines in triangle OPQ to get $PQ = \sqrt {{R^2} + {R^2} - 2(R)(R)\cos 120^\circ } = \sqrt 3 R$. The electric flux through the shell (Gaussian surface) is given by Gauss's law

$\phi = {q_{enc}}/{ \in _0} = \sqrt 3 R\lambda /{ \in _0}$.

The electric field at a point due to the infinitely long line charge is radial i.e., perpendicular to the line PQ. The non-conducting spherical shell does not affect or alter the electric field. Thus, electric field at a point lying on the surface of the shell is perpendicular to the z-axis i.e., its z component is zero. The figure shows electric field on the surface of the shell for a positive line charge. Note that the magnitude of electric field is inversely proportional to the distance of the point from the line charge.

Since charge Q is outside the hemispherical surface, the net flux passing through the curved surface of hemispherical surface and flat surface is zero.

Therefore,

$\phi$_curved + $\phi$_flat = 0 ....... (1)

Hence, option (B) is incorrect.

Now,

${\phi _{flat}} = \int {\overrightarrow E .\,d\overrightarrow A = \int {EdA\cos \theta } } $

and $E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{{(\sqrt {{R^2} + {r^2}} )}^2}}} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {({R^2} + {r^2})}}$

Also,

$\cos \theta = {R \over {\sqrt {{R^2} + {r^2}} }}$

and

$A = 2\pi r \Rightarrow dA = 2\pi rdr$

Therefore,

${\phi _{flat}} = \int\limits_0^R {{1 \over {4\pi {\varepsilon _0}}}{Q \over {{R^2} + {r^2}}}2\pi rdr{R \over {\sqrt {{R^2} + {r^2}} }}} $

${\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\int\limits_0^R {{{rdr} \over {{{({R^2} + {r^2})}^{3/2}}}}} $

Substituting R² + r² = t, we get

$2rdr = dt$

$ \Rightarrow {\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\int\limits_0^R {{{dt} \over 2}{1 \over {{t^{3/2}}}} = {{QR} \over {2{\varepsilon _0}}}\left[ {{1 \over 2}{{{t^{ - 1/2}}} \over { - 1/2}}} \right]_0^R} $

Substituting $t = {R^2} + {r^2}$, we get

${\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\left[ {{1 \over 2}{{{{({R^2} + {r^2})}^{ - 1/2}}} \over { - 1/2}}} \right]_0^R$

$ = {{QR} \over {2{\varepsilon _0}}}\left[ {{{ - 1} \over {\sqrt {{R^2} + {r^2}} }}} \right]_0^R = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt {{R^2} + {R^2}} }} + {1 \over {\sqrt {{R^2}} }}} \right)$

$ = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt {2{R^2}} }} + {1 \over {\sqrt {{R^2}} }}} \right) = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 R}} + {1 \over R}} \right)$

$ = {{QR} \over {2{\varepsilon _0}}}{1 \over R}\left( {{{ - 1} \over {\sqrt 2 }} + 1} \right) = {Q \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 }} + 1} \right)$

Using Eq. (1), we get

${\phi _{curved}} = - {\phi _{flat}} = - {Q \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 }} + 1} \right) = - {Q \over {2{\varepsilon _0}}}\left( {1 - {1 \over {\sqrt 2 }}} \right)$

Hence, option (A) is correct.

The potential at any point on the circumference of the flat surface is ${1 \over {4\pi {\varepsilon _0}}}{Q \over {\sqrt 2 R}}$.

Thus, the circumference of flat surface is equipotential.

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,

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

Let $\rho$ be the charge density of the spherical charge distribution of radius r₁ centred at the origin O. A spherical cavity of radius r₂ centred at P with distance OP = a = r₁ $-$ r₂ is made in the spherical charge distribution.

The sphere with cavity is equivalent to a sphere of uniform charge density $-$$\rho$ and radius r₂ centred at P embedded in the original sphere. Thus, the electric field at a point Q in the cavity is superposition of (i) electric field at Q due to the sphere of charge density $\rho$ and radius r₁ centred at O (say $\overrightarrow E $₁), and (ii) electric field at Q due to the sphere of charge density $-$$\rho$ and radius r₂ centred at P (say $\overrightarrow E $₂). Let $\overrightarrow a $, $\overrightarrow r $, and $\overrightarrow r $ $-$ $\overrightarrow a $ be the vectors as shown in the figure. The electric fields $\overrightarrow E $₁, $\overrightarrow E $₂, and their superposition $\overrightarrow E $₁₂ are given by

${\overrightarrow E _1} = {1 \over {4\pi { \in _0}}}{{{4 \over 3}\pi |\overrightarrow r {|^3}\rho } \over {|\overrightarrow r {|^2}}}\widehat r = {\rho \over {3{ \in _0}}}\overrightarrow r $,

${\overrightarrow E _2} = - {\rho \over {3{ \in _0}}}(\overrightarrow r - \overrightarrow a )$,

${\overrightarrow E _{12}} = {\overrightarrow E _1} + {\overrightarrow E _2} = {\rho \over {3{ \in _0}}}\overrightarrow a $.

Thus, the electric field at a point within the cavity is uniform and its magnitude and direction both depend on $\overrightarrow a $.

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

Case 1 : If q is shifted towards right by x, we get

$F = {F_2} - {F_1} = {{\lambda q} \over {2\pi {\varepsilon _0}}}\left( {{1 \over {{d \over 2} - x}} - {1 \over {{d \over 2} + x}}} \right)$ (towards left)

Case 2 : If $-$q is shifted towards right by x

$F = {F_2} - {F_1} = {{\lambda q} \over {2\pi {\varepsilon _0}}}\left( {{1 \over {{d \over 2} - x}} - {1 \over {{d \over 2} + x}}} \right)$ (towards right)

Thus, +q exhibits SHM while $-$q continues to move towards rightwards.

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

${E_1}(r) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{r^2}}}$

${E_2}(r) = {\lambda \over {2\pi {\varepsilon _0}r}}$

${E_3}(r) = {\sigma \over {2{\varepsilon _0}}}$

At $r = {r_0}$,

${E_1}({r_0}) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}}$

${E_2}({r_0}) = {\lambda \over {2\pi {\varepsilon _0}{r_0}}}$

${E_3}({r_0}) = {\sigma \over {2{\varepsilon _0}}}$

As ${E_1}({r_0}) = {E_2}({r_0}) = {E_3}({r_0})$ (Given)

$\therefore$ ${1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = {\lambda \over {2\pi {\varepsilon _0}{r_0}}} = {\sigma \over {2{\varepsilon _0}}}$

Then

${1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = {\sigma \over {2{\varepsilon _0}}}$

or $Q = \sigma \pi r_0^2$

Hence, option (a) is incorrect.

Now, ${\lambda \over {2\pi {\varepsilon _0}{r_0}}} = {\sigma \over {2{\varepsilon _0}}}$ or ${r_0} = {\lambda \over {\pi \sigma }}$

Hence, option (b) is incorrect.

At $r = {r_0}/2$,

${E_1}({r_0}/2) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{{({r_0}/2)}^2}}} = {4 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = 4{E_1}({r_0})$

or, ${E_1}({r_0}) = {1 \over 4}{E_1}({r_0}/2)$

${E_2}({r_0}/2) = {\lambda \over {2\pi {\varepsilon _0}({r_0}/2)}} = {{2\lambda } \over {2\pi {\varepsilon _0}{r_0}}} = 2{E_2}({r_0})$

or, ${E_2}({r_0}) = {1 \over 2}{E_2}({r_0}/2)$

$\because $${E_1}({r_0}) = {E_2}({r_0})$

$\therefore$ ${1 \over 4}{E_1}({r_0}/2) = {1 \over 2}{E_2}({r_0}/2)$

or, ${E_1}({r_0}/2) = 2{E_2}({r_0}/2)$

Hence, option (c) is correct.

${E_3}({r_0}/2) = {\sigma \over {2{\varepsilon _0}}} = {E_3}({r_0})$

$\because$ ${E_2}({r_0}) = {E_3}({r_0})$

$\therefore$ ${1 \over 2}{E_2}({{{r_0}} \over 2}) = {E_3}({{{r_0}} \over 2})$

or, ${E_2}({r_0}/2) = 2{E_3}({r_0}/2)$

Hence, option (d) is incorrect.

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

Here we will use the concept of vectors and concept of electric field.

Let's take a point P in the overlapping region.

From $\Delta OPQ$,

By triangle law of vector addition,

$\overrightarrow {{r_1}} - \overrightarrow {{r_2}} = \overrightarrow d \quad \text{... (1)}$

Net electrostatic field at point P;

${\overrightarrow E _{net}} = {{K{q_1}} \over {r_1^3}}\overrightarrow {{r_1}} = {{K{q_2}} \over {r_2^3}}\overrightarrow {{r_2}} $

$ \Rightarrow {\overrightarrow E _{net}} = {{K\left( {\rho {4 \over 3}\pi r_1^3} \right)\overrightarrow {{r_1}} } \over {r_1^3}} - {{K\left( {\rho {4 \over 3}\pi r_2^3} \right)\overrightarrow {{r_2}} } \over {r_2^3}}$

$ \Rightarrow {\overrightarrow E _{net}} = {4 \over 3}\rho \pi \times {1 \over {4\pi {\varepsilon _0}}}\left( {\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \right)$

$ \Rightarrow {\overrightarrow E _{net}} = {\rho \over {3{\varepsilon _0}}}\overrightarrow d $

Hence, the electrostatic field is constant in magnitude and has same direction.

Therefore, options (C) and (D) are correct.

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 electric field at point $O$ due to the charges at vertices $A$ and $D$ is $4K$ along the direction OD. Similarly, due to the charges at vertices $B$ and $E$, the electric field is $2K$ along the direction OE, and due to the charges at vertices $C$ and $F$, it is $2K$ along the direction OC. Given the uniform geometry of this setup, the resulting electric field is $6K$ along OD.

The potential at point $O$ is calculated as :

$ V_{\mathrm{O}}=\sum \frac{1}{4 \pi \epsilon_0} \frac{q_i}{L} = \frac{1}{4 \pi \epsilon_0 L} \sum q_i = 0 $

For any point on the line PR, we observe that there are pairs of equal and opposite charges equidistant from these points, making the potential at any point on PR zero. If we consider points on OS, the potential is positive, while for points on OT, the potential is negative. The potential at points on the line ST, at a distance $x$ from $O$ (with $x$ considered positive towards the right), can be shown to be :

$ \begin{aligned} V(x) = \frac{q}{4 \pi \epsilon_0} \left[ \frac{2}{\sqrt{L^2 + x^2 + xL}} - \frac{2}{\sqrt{L^2 + x^2 - xL}} - \frac{4x}{L^2 - x^2} \right]. \end{aligned} $

The positions of all charges are symmetric about the planes $x = +\frac{a}{2}$ and $x = -\frac{a}{2}$. Therefore, the net electric flux crossing the plane $x = +\frac{a}{2}$ is equal to the net electric flux crossing the plane $x = -\frac{a}{2}$.

Similarly, the net electric flux crossing the plane $y = +\frac{a}{2}$ is equal to the net electric flux crossing the plane $y = -\frac{a}{2}$.

According to Gauss's law, the net electric flux crossing the entire region is given by:

$ \phi = \frac{q_{\text{inside}}}{\varepsilon_0} = \frac{3q - q - q}{\varepsilon_0} = \frac{q}{\varepsilon_0} $

The charges are symmetrically placed about the planes $z = +\frac{a}{2}$ and $x = +\frac{a}{2}$. Thus, the net electric flux crossing the plane $z = +\frac{a}{2}$ is equal to the net electric flux crossing the plane $x = +\frac{a}{2}$.

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.

When connected by a wire, charges on A and B are redistributed until potential on both becomes equal. After the charge redistribution, A and B are not influenced by each other because they are far apart. Thus, $E_A^{inside} = 0$ as field inside a conducting shell is zero.

So choice (a) is correct.

Let Q_A and Q_B are the charges on metal shell A and metal sphere B after they are connected by a wire. Since their electric potentials will be equal,

V_A = V_B

$ \Rightarrow {{{Q_A}} \over {4\pi {\varepsilon _0}{R_A}}} = {{{Q_B}} \over {4\pi {\varepsilon _0}{R_B}}} \Rightarrow {{{Q_A}} \over {{Q_B}}} = {{{R_A}} \over {{R_B}}}$

Since ${R_B} < {R_A}$, ${Q_A} > {Q_B}$. So choice (b) is correct.

Now, ${\sigma _A} = {{{Q_A}} \over {4\pi R_A^2}}$ and ${\sigma _B} = {{{Q_B}} \over {4\pi R_B^2}}$

$\therefore$ ${{{\sigma _A}} \over {{\sigma _B}}} = {{{Q_A}} \over {{Q_B}}} \times {\left( {{{{R_B}} \over {{R_A}}}} \right)^2} = {{{R_A}} \over {{R_B}}} \times {\left( {{{{R_B}} \over {{R_A}}}} \right)^2} = {{{R_B}} \over {{R_A}}}$

Hence, choice (c) is also correct.

Electric fields on the surface of shell and sphere are

${E_A} = {{{\sigma _A}} \over {{\varepsilon _0}}}$

and ${E_B} = {{{\sigma _B}} \over {{\varepsilon _0}}}$

$\therefore$ ${{{E_A}} \over {{E_B}}} = {{{\sigma _A}} \over {{\sigma _B}}} < 1$, i.e. ${E_A} < {E_B}$

So, choice (d) is also correct. All the four choices are correct.

Gauss's law is valid only if Coulomb's law holds, i.e. if $E \propto {r^{ - 2}}$. Hence, choice (a) is wrong. Gauss's law cannot be used to calculate a non-uniform field distribution around an electric dipole. So choice (b) is also wrong.

Choice (c) is correct because the directions of electric fields are opposite at a point between two similar charges.

Work done ${W_{A \to B}} = q({V_B} - {V_A}) = ({V_B} - {V_A})$ ($\because$ $q = + \,1\,C$)

Hence, choice (d) is correct.

So the correct choices are (c) and (d).