Electrostatics

The given situation is shown below

Since, electric field is directed along the $X$-axis, hence, point $A$ and $C$ lie on equipotential surface

$\therefore$ Potential at point $C=$ Potential at point $A$

$\begin{array}{rlrl} \Rightarrow & V_C =V_A \\ \therefore & V_B-V_A =V_B-V_C \\ & =-E \Delta x=-5(2-0)=-10 \mathrm{~V} \end{array}$

According to given situation, electric field produced in the region of drilled hole at a distance $x$ from its centre,

$E=\frac{Q x}{4 \pi \varepsilon_0 R^3}\quad \text{.... (i)}$

Force on charge $(-q)$ when it is placed at a distance $x$,

$\begin{aligned} & F=-q E \\ & \Rightarrow \quad m a=-q E \\ & a=\frac{-q E}{m}=\frac{-q}{m} \cdot \frac{Q x}{4 \pi \varepsilon_0 R^3} \\ & \Rightarrow \quad a=\frac{-Q q x}{4 \pi \varepsilon_0 m R^3} \\ & \Rightarrow \quad-\omega^2 x=\frac{-Q q x}{4 \pi \varepsilon_0 m R^3} \\ & \therefore \quad \omega^2=\frac{Q q}{4 \pi \varepsilon_0 m R^3} \\ & (2 \pi f)^2=\frac{Q q}{4 \pi \varepsilon_0 m R^3} \\ & \Rightarrow \quad f=\frac{1}{2 \pi}\left(\frac{Q q}{4 \pi \varepsilon_0 m R^3}\right)^{1 / 2} \\ & \end{aligned}$

We know that, electric field $(E)$ and electric potential are related as

$E=\frac{-d V}{d r}$

For the region of constant potential $(V)$

$E=0$

According to Gauss's, law, electric flux through closed Gaussian surface

$\phi=\oint E . d s=\frac{q}{\varepsilon_0}\quad \text{... (i)}$

$\begin{aligned} & \text { if } \quad E=0 \\ & \Rightarrow \quad \oint E . d s=0 \\ \end{aligned}$

Hence, from Eq (i) , $q=0$

Inside a charged hollow metal sphere, electric field $(E)$ is zero but electric potential $(V)$ is not zero and equal to same potential as on its surface. Work done to move a charge particle (either $+v e$ or $-v e$) on a equipotential surface between any two points is always zero.

$\begin{aligned} \text { Since, } W & =q\left(V_1-V_2\right) \\ & =q(V-V) \quad \left(\because V_1=V_2=V\right)\\ & =0 \end{aligned}$

Potential energy of system of two like charges placed at a distance $r$ is given as,

$U=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}$

When both like charges $q_1$ and $q_2$ are brought closer, then $r$ decreases, hence potential energy $U$ increases.

Therefore, statements (A), (B) and (C), all are true.

According to first condition,

$F=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \quad \text{... (i)}$

Electric field produced due to dielectric material,

$\begin{aligned} E & =\frac{9}{4 \pi \varepsilon_0\left(r-\frac{r}{5}+\sqrt{k} \frac{r}{5}\right)^2} \\ & =\frac{q}{4 \pi \varepsilon_0 \frac{r^2}{25}(4+\sqrt{k})^2} \\ & =\frac{q}{4 \pi \varepsilon_0 \frac{r^2}{25}}(4+\sqrt{9})^2 \quad (\because k=9)\\ E & =\frac{q}{4 \pi \varepsilon_0 r^2}\left(\frac{25}{49}\right) \end{aligned}$

Force between the charges,

$\begin{aligned} F^{\prime} & =q E=q \cdot \frac{q}{4 \pi \varepsilon_0 r^2}\left(\frac{25}{49}\right) \\ & =\frac{25}{49} \frac{q^2}{4 \pi \varepsilon_0 r^2} \\ & =\frac{25}{49} F \end{aligned}$

Electric dipole moment,

$p=5 \times 10^{-7} \mathrm{C}$

Angle, $\theta=60^{\circ}$

Electric field, $E=2 \times 10^4 \mathrm{~N} / \mathrm{C}$

Torque,

$\begin{aligned} \tau & =p E \sin \theta \\ & =5 \times 10^{-7} \times 2 \times 10^4 \times \sin 60^{\circ} \\ & =10^{-2} \times \frac{\sqrt{3}}{2}=8.66 \times 10^{-3} \mathrm{~N}-\mathrm{m} \end{aligned}$

$\begin{aligned} & \text { Given, } q_1=10 \mu \mathrm{C}=10^{-5} \mathrm{C} \\ & q_2=12 \mu \mathrm{C}=1.2 \times 10^{-5} \mathrm{C} \\ & r_1=10 \mathrm{~cm}=10^{-1} \mathrm{~m} \end{aligned}$

We have to find work done in bringing them

Closer when, $r_2=6 \mathrm{~cm}$

$=6 \times 10^{-2} \mathrm{~m}$

Work done, $W=$ change in potential energy in two configuration of charges

$\begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_2}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_1} \\ & =\frac{1}{4 \pi \varepsilon_0} q_1 q_2\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \\ & =9 \times 10^9 \times 10^{-5} \times 1.2 \times 10^{-5}\left(\frac{1}{6 \times 10^{-2}}-\frac{1}{10^{-1}}\right) \\ & =1.08\left(\frac{100}{6}-10\right)=\left(\frac{40}{6}\right) \times 1.08=7.2 \mathrm{~J} \end{aligned}$

Which is nearest to 8.1 J hence, option (a) is correct.

Electric potential at centre R = 0

V_centre = 1.5 V_surface

where,

${V_0} = {{kQ} \over R}$

${V_{inside}} = {{kQ} \over {2R}}\left( {3 - {{{r^2}} \over {{R^2}}}} \right)$

${V_{outside}} = {{kQ} \over r} \propto {1 \over r}$

Net charge on electric dipole = + q $-$ q = 0

Hence, according to Gauss's law,

Electric flux, $\phi = {{{q_{net}}} \over {{\varepsilon _0}}} = {0 \over {{\varepsilon _0}}} = 0$

Electric field due to electric dipole is non-zero and varies at point to point.

Hence, statement I is true.

Electric field due to charged solid sphere at a distance r from centre.

Hence, statement II is false.

Hence, option (c) is the correct.

Given, charges at point A and B are $q_A,q_B=10\mu$C and $-10\mu$C

Separation between A and B = 10 cm = 10 $\times$ 10$^{-2}$ m

Let $O$ is mid-point of $A B$.

$\begin{aligned} \therefore \quad O A & =5 \times 10^{-2} \mathrm{~m} \Rightarrow O P=12 \times 10^{-2} \mathrm{~m} \\ \text { and } A P & =\sqrt{12^2+5^2} \\ & =\sqrt{169}=13 \times 10^{-2} \mathrm{~m} \end{aligned}$

As we know that, Electric field $(E)=\frac{k q}{r^2}$

where, $k$ is Coulomb's constant $=9 \times 10^9 \mathrm{C}^2 \mathrm{~m}^{-2} \mathrm{~N}^{-1}, q$ is charge and $r$ is separation between $A$ and $P$.

$E_A=\frac{9 \times 10^9 \times 10 \times 10^{-6}}{\left(13 \times 10^{-2}\right)^2}=5.3 \times 10^6 \mathrm{~N} / \mathrm{C}$

Similarly, electric field due to charge at $B, E_B=5.3 \times 10^6 \mathrm{~N} / \mathrm{C}$

$\therefore$ Net Electric field $\left(E_{\text {net }}\right)=2 E \sin \theta$

$\begin{aligned} & =2 \times 5.3 \times 10^6 \times \frac{5}{13}=\frac{53}{13} \times 10^6 \\ & =4.07 \times 10^6 \mathrm{~N} / \mathrm{C} \\ & =4.1 \times 10^6 \mathrm{~N} / \mathrm{C} \end{aligned}$

As we know that, charge is a conserved quantity i.e. charge of any isolated system remains same.

Hence, net charge of the system doesn’t change.

As we know that, on equipotential surface potential remains same everywhere and electric field lines are always perpendicular to equipotential surface.

Since, $E=-dV/dx$

= $-$ (Potential gradient)

$\therefore$ Angle between potential and potential gradient = $\frac{\pi}{2}$ rad

Given,

charges $q_1, q_2$ and $q_3$ are $-14 \mathrm{~nC}, 78.85 \mathrm{~nC}$ and $-56 \mathrm{~nC}$.

As we know that,

$\text { flux, } \phi=\frac{Q_{\text {enclosed }}}{\varepsilon_0}$

where, $\varepsilon_0$ is free space permittivity

$\begin{aligned} & =8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ \therefore \quad \phi_{\text {net }} & =\frac{(-14+78.85-56) \times 10^{-9}}{8.854 \times 10^{-12}} \\ & =10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1} \end{aligned}$

Given, $q_1=q_2=8 \propto \mathrm{C}=8 \times 10^{-6} \mathrm{C}$ and side of triangle, $x=0.2 \mathrm{~m}$

Since, electric potential, $V=\frac{k q}{r}$

$\therefore$ Potential at third point, $V_3=\frac{k q_1}{r}+\frac{k q_2}{r}$

$\Rightarrow V_3=\frac{9 \times 10^9 \times 8 \times 10^{-6}}{0.2} \times 2=7.2 \times 10^5 \mathrm{~V}$

As we know that according to Gauss's law,

$\oint \mathbf{E} \cdot d s=\frac{q_{\text {enc. }}}{\varepsilon_0}$

Hence, Gauss's law is used to determine electric field due to symmetric charge distribution inside the Gaussian surface.

Given, charge on outer sphere = q

Inner sphere is grounded.

Since, charge flows till potential (V) becomes same.

and $V=\frac{kq}{r}$

where, $r$ is radius of sphere $\left(r_2 > r_1\right)$.

$\begin{aligned} \frac{k q^{\prime}}{r_1} & =\frac{-k q}{r_2} \\ \Rightarrow \quad q^{\prime} & =-q\left(\frac{r_1}{r_2}\right)\end{aligned}$

As we know that,

Electric field lines emerge from the positive charge and converge to negative charge. Tangent drawn at any point on the field line gives the direction of electric field at that point.

Electric field lines do not intersect each other. If they do, so there will be two directions of force at a single point which is not possible.

Hence, statement (d) is incorrect.

Consider the equilibrium of charge Q at corner A

${F_B} = {F_D} = {1 \over {4\pi {\varepsilon _0}}}{{{Q^2}} \over {{a^2}}} = {{k{Q^2}} \over {{a^2}}}$ [$\because$ ${1 \over {4\pi {\varepsilon _0}}} = k$]

${F_C} = {{k{Q^2}} \over {{{(\sqrt 2 a)}^2}}}$ [$\because$ $AC = \sqrt 2 a$]

$ = {{k{Q^2}} \over {2{a^2}}}$

${F_O} = {{k\,Qq} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}} = {{k\,2Qq} \over {{a^2}}}$ [$\because$ $OC = {a \over {\sqrt 2 }}$]

For equilibrium of charge Q at corner A, the net force on this charge along AC must be zero.

$\therefore$ ${F_O} = {F_C} + {F_B}\cos 45^\circ + {F_D}\cos 45^\circ $

$ = {F_C} + 2{F_B}\cos 45^\circ $ [$\because$ ${F_B} = {F_D}$]

$ = {F_C} + 2{F_B} \times {1 \over {\sqrt 2 }}$ [$\because$ $\cos 45^\circ = {1 \over {\sqrt 2 }}$]

$ \Rightarrow {{k\,2Qq} \over {{a^2}}} = {{k\,{Q^2}} \over {2{a^2}}} + {{2 \times k{Q^2}} \over {{a^2}}} \times {1 \over {\sqrt 2 }}$

$ \Rightarrow 2Qq = {{{Q^2}} \over 2} + {{{Q^2}} \over {\sqrt 2 }} \times 2$

$ \Rightarrow 2q = {Q \over 2} + Q\sqrt 2 $

$ \Rightarrow q = {Q \over 4}(1 + 2\sqrt 2 )$