Electrostatics

$(a,b)$ We know, ${V_0} = {{Kq} \over R} = Vsurface$

Now, ${V_i} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {R^2}} \right)\,\,\,\,\,$ [For $r < R$]

At the center of sphere $r=0.$ Here

$V = {3 \over 2}{V_0}$

Now, ${5 \over 4}{{Kq} \over R} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {r^2}} \right)$

${R_2} = {R \over {\sqrt 2 }}$

${3 \over 4}{{Kq} \over R} = {{Kq} \over {{R^3}}}$

${1 \over 4}{{Kq} \over R} = {{Kq} \over {R{}_4}}$

${R_4} = 4R$

Also, ${R_1} = 0$ and ${R_2} < \left( {{R_4} - R{}_3} \right)$

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} $

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 $r=10 R$

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

$ \mathrm{V}(z) =\frac{\sigma}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right) $

$ \mathrm{U}(z)_{\mathrm{net}} =\frac{\sigma q}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+\mathrm{cz} $

$ =c\left[\frac{\sigma q}{2 c \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+z\right]$

When, $z=0$ and $\beta=1 / 4$

$ \mathrm{U}_{(z) \text { net }}=c(4 \mathrm{R})=4 R c $

When, $z=z_0=\frac{25}{7} \mathrm{R}$ and; $\mathrm{B}=\frac{1}{4}$

$ U_{(z) \text { net }}=C\left[4 \times \frac{26 R}{7}-3 \times \frac{25 R}{7}\right]=\frac{29 R c}{7} $

When, $z=z_0=\frac{3 R}{7}$ and $\beta=\frac{1}{4}$

$ \mathrm{U}_{(z) \text { net }}=3 R c $

When, $Z=\frac{R}{\sqrt{3}}$ and $\beta=\frac{1}{4}$

$ \mathrm{U}_{(z) n \mathrm{t}}=2.887 \mathrm{Rc} $

As per option (A), particle reaches at origin with the $\mathrm{KE}$

$ \frac{d(U)_z}{d z}=0 \text { at } z=\frac{3 R}{\sqrt{7}} $

So, at $\beta = \frac{1}{4}$ and $z=\frac{3 R}{\sqrt{7}}$

$ \mathrm{U}_{z(\text { net })}=\sqrt{7} R c=2.645 R c $

Similarly, for option (B),

$\mathrm{U}_{(z) \text { net }}$ at $z=\frac{\mathrm{R}}{\sqrt{3}}$ the kinetic energy at origin

will become negative and mearly equal to $3 R c$

Also, for option (C)

$\mathrm{U}_{(z) \text { net }}$ at $\mathrm{Z}=\frac{\mathrm{R}}{\sqrt{3}}>\mathrm{U}_{(z) \text { net }}$ at $z=\frac{3 \mathrm{R}}{\sqrt{7}}$

Particle will return to $z_0$

As per option $D,(\beta>1 $ and $z_0>0) $ $U_{z(\text { net })}$ increases continuously with $\mathrm{Z}$. So, particle reaches the origin.

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.

In air,



T cos$\theta$ = mg ..... (i)

T sin$\theta$ = qE ..... (ii)

qE = mg tan$\theta$ ..... (iii)

In liquid,



T' cos$\theta$ = mg $-$ U

T' cos$\theta$ = mg - $\rho $vg ..... (iv)

T' sin$\theta$ = ${{qE} \over K}$ ...... (v)

Dividing Eq. (v) by Eq. (ii), we get

${{T'} \over T} = {1 \over K} \Rightarrow T' = {T \over K} = {T \over {21}}$ ..... (vi)

Dividing Eq. (iv) by Eq. (v), we get

${{qE} \over K} = $ (mg - $\rho $vg)tan $\theta $

$qE = $ K(mg - $\rho $vg)tan $\theta $ ..... (vii)

From Eqs. (vii) and (iii), we get

mg tan$\theta $ = K(mg - $\rho $vg)tan $\theta $

$ \Rightarrow $ m = K(m - $\rho $v)

$ \Rightarrow $ dv = K(dv - $\rho $v)

$ \Rightarrow $ d = Kd - K$\rho $

$ \Rightarrow $ d = 21d - 21$\rho $ [ $ \because $ K = 21]

$ \Rightarrow $ 20d = 21$\rho $

$ \Rightarrow $ 20d = 21 $ \times $ 800

$ \Rightarrow $ d = 840 kg/m³

$a = {{qE} \over m}{10^{10}} \times 400\sqrt 3 \,m/{s^2}$

or $a = 4\sqrt 3 \times {10^{12}}\,m/{s^2}$

$ \therefore $ $R = {{{u^2}\sin 2\theta } \over a}$

$ \Rightarrow \sin 2\theta = {{Ra} \over {{u^2}}}$

or $\sin 2\theta = {{5 \times 4\sqrt 3 \times {{10}^{12}}} \over {4 \times 10 \times {{10}^{12}}}}$

$ \Rightarrow \sin 2\theta = {{\sqrt 3 } \over 2}$

$ \Rightarrow 2\theta = 60^\circ ,\,120^\circ $

$ \Rightarrow \theta = 30^\circ $ or $\,60^\circ $ for same range

$ \therefore $ $T = {{2u\sin \theta } \over a}$

$ = {{2 \times 2\sqrt {10} \times {{10}^6}} \over {2 \times 4\sqrt 3 \times {{10}^{12}}}}$

$ \Rightarrow T = \sqrt {{{10} \over {12}}} \times {10^{ - 6}} \Rightarrow T = \sqrt {{5 \over 6}} \times {10^{ - 6}}$

R >> Dipole size.

Circle is equipotential.

So, E_net should be perpendicular to surface hence,

${{k{p_0}} \over {{r^3}}} = {E_0}$

$ \Rightarrow $ $R = {\left( {{{k{p_0}} \over {{E_0}}}} \right)^{1/3}}$

At point B, net electric field will be zero.

E_B = 0

${({E_A})_{net}} = {{2k{p_0}} \over {{R^3}}} + {E_0} = 3{E_0}$

Electric field at point A, ${E_A} = {3 \over {\sqrt 2 }}{E_0}[\widehat i + \widehat j]$

${({E_B})_{net}}$ = 0

(a) h > 2R and r > R



$\phi = {Q \over {{\varepsilon _0}}}$, clearly from Gauss' Law

(b) suppose $h = {{8R} \over 5}$ and $r = {{3R} \over 5}$



$ \therefore $ $\phi $ = 0

so, for $h < {{8R} \over 5}$ then $\phi $ = 0.

(c) For h = 3R and r = ${{4R} \over 5}$



Shaded charge = $2\pi (1 - \cos \,53^\circ ) \times {Q \over {4\pi }} = {Q \over 5}$

$ \therefore $ ${q_{enclosed}} = {{2Q} \over 5}$

$ \therefore $ $\phi = {{2Q} \over {5{\varepsilon _0}}}$

$ \therefore $ for h > 2R and $r = {{4R} \over 5}$

$ \therefore $ $\phi = {{2Q} \over {5{\varepsilon _0}}}$

(d) like option c for h = 2R and $r = {{3R} \over 5}$

${q_{enclosed}} = 2 \times 2\pi (1 - \cos \,37^\circ ){Q \over {4\pi }} = {Q \over 5}$

$ \therefore $ Electric flux, $\phi = {Q \over {5{\varepsilon _0}}}$

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.

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.

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 $x = +\frac{a}{2}$.

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).

Number of electric field lines originating from Q₁ is more than terminating at Q₂.

$\therefore$ $|{Q_1}| > |{Q_2}|$

Here, Q₁ is positive while Q₂ is negative.

The electric field at a distance x towards the right of Q₂ is given by

$|\overrightarrow E | = {1 \over {4\pi { \in _0}}}{{{Q_1}} \over {{{(d + x)}^2}}} - {1 \over {4\pi { \in _0}}}{{{Q_2}} \over {{x^2}}}$,

where d is the separation between Q₁ and Q₂. Since ${Q_1} > {Q_2}$, $|\overrightarrow E |$ becomes zero for some finite x.

(A) Since, potential is same, so electrostatic energy stored is zero.

(B)

$ \text { (C) There is no charge in air, } \therefore r=\mathrm{R}_0 $

(D) From $r=\mathrm{R}_0$ to $\infty$.

We can assume point charge at centre equivalent to $q$.

$ \text { So, } \overrightarrow{\mathrm{E}} \propto \frac{1}{r^2} \text { so, it is discontinuous } $