Electrostatics

To determine the magnitude of the charge, we have the following given information:

Electric field $E = 500 \, \mathrm{N/C}$

Electric potential $V = 30 \, \mathrm{V}$

First, we use the formula for the potential gradient, which relates the electric field and potential through the distance $x$ from the point charge:

$ E = \frac{V}{x} $

Solving for $x$:

$ x = \frac{30}{500} = 0.06 \, \mathrm{m} $

Now, we employ the formula for the electric potential due to a point charge:

$ V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{x} $

Substitute the known values to solve for the magnitude of the charge $q$:

$ 30 = \frac{9 \times 10^9}{0.06} \times q $

Rearranging gives:

$ q = \frac{30 \times 0.06}{9 \times 10^9} = 2 \times 10^{-10} \, \mathrm{C} $

Thus, the magnitude of the charge is:

$ 2 \times 10^{-10} \, \mathrm{C} $

Given,

' 'The charges corresponding to number's of a clock $-q,-2 q \ldots 12 q$.

Electric field is given by

$ E=k q / r^2 \Rightarrow E \propto q $

Let's take a pair of charges ( $-9 q$ and $-3 q$ )

The value of $E$ due to $-9 q>-3 q$

and net magnitude $=-6 q=E_0$

Same goes for all pairs with magnitude of $(-6 q)$.

The angle between each field,

$ \theta=\frac{360^{\circ}}{12}=30^{\circ} $

From the property of vector addition, if two equal vector is separated by $120^{\circ}$, then resultant vector is same as initial vector.

For making $120^{\circ}$, take vector 1 to vector 5 Resultant of vector 1 and $5=E_0=$ Vector 3 direction.

Resultant of vector 2 and $6=E_0=$ Vector 4 direction

Now, vector (1 and 5) and (2 and 6) are eliminated

Now, vector $3=E_0+E_0=2 E_0$

Vector $4=E_0+E_0=2 E_0$

Vector 3 denotes 10 AM

Vector 4 denotes 9 AM

So, the resultant vector of 9 AM and 10 AM is in midway direction.

i.e., $9: 30$.

The electric field $\vec{E}$ is related to the electric potential $V$ by the following relation:

$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$

Given the potential $V = \frac{1}{2}(y^2 - 4x)$, we need to find the partial derivatives with respect to $x$ and $y$.

Partial derivative with respect to $ x $:

$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} \left[\frac{1}{2}(y^2 - 4x)\right] = \frac{1}{2}(-4) = -2$

Partial derivative with respect to $ y $:

$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} \left[\frac{1}{2}(y^2 - 4x)\right] = \frac{1}{2}(2y) = y$

Now, let's plug these into the formula for the electric field:

$\vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}\right) = -(-2\hat{i} + y\hat{j}) = 2\hat{i} - y\hat{j}$

We need to find the electric field at the point $x = 1\,\text{m}$ and $y = 1\,\text{m}$. So, we substitute $y = 1$ into the equation:

$\vec{E} = 2\hat{i} - (1)\hat{j} = (2\hat{i} - \hat{j})\,\text{V/m}$

So, the correct answer is:

Option C: $(2\hat{i} - \hat{j})\,\text{V/m}$

Let two identical spheres have charge q. And distance between them = r

$\therefore$ Force between the spheres $(F) = {{k{q^2}} \over {{r^2}}}$

Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.

So, both sphere A and C have charge $ = {q \over 2}$

Now, C get's in contact with B. So their total charge $\left( {q + {q \over 2}} \right)$ gets divided equally.

So, charge on both B and C is $ = {{q + {q \over 2}} \over 2} = {{3q} \over 4}$

Now, C is place midpoint between A and B.

Repulsion force between A and C,

${F_{AC}} = {{k\left( {{q \over 2}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 3{q^2}} \over {8{r^2}}}$

Repulsion force between B and C,

${F_{BC}} = {{k\left( {{{3q} \over 4}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 9{q^2}} \over {16{r^2}}}$

$\therefore$ Net force on C,

${F_{net}} = {F_{BC}} - {F_{AC}}$

$ = {{4k \times 9{q^2}} \over {16{r^2}}} - {{4k \times 3{q^2}} \over {8{r^2}}}$

$ = {{k{q^2}} \over {{r^2}}}\left( {{9 \over 4} - {3 \over 2}} \right)$

$ = {{k{q^2}} \over {{r^2}}} \times {3 \over 4}$

$ = {3 \over 4}F$ [as $F = {{k{q^2}} \over {{r^2}}}$]

$\left( {4\pi {r^2}} \right){E_\rho } = {{{Q_{in}}} \over {{\varepsilon _0}}}$

$ = {{\int_0^r {{\rho _0}\left( {{3 \over 4} - {r \over 4}} \right)4\pi {r^2}dr} } \over {{\varepsilon _0}}}$

$ = {{{\rho _0}\pi 4} \over {{\varepsilon _0}}}\left( {{{{r^3}} \over 4} - {{{r^4}} \over {4R}}} \right)$

${E_\rho } = {{{\rho _0}} \over {4{\varepsilon _0}}}\left( {r - {{{r^2}} \over R}} \right)$

$ = {{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {1 - {r \over R}} \right)$

Since ${\overrightarrow E _{net}} = \overrightarrow 0 $ in the bulk of a conductor

$\Rightarrow$ Potential would be constant.

$\Rightarrow$ Statement I is correct

Since a conductor's surface is equipotential, $\overrightarrow E $ just outside is perpendicular to the surface.

$E = {{8\,m} \over e}$ V/m

$l = 1$ m

${v_x} = 2$ m/s

${a_y} = - 8$ m/s²

$t = {l \over {{v_x}}} = {1 \over 2}s$

$ \Rightarrow |{v_y}| = 4$ m/s

$\Rightarrow$ angle of deviation = $\theta$

$\tan \theta = {{{v_y}} \over {{v_x}}}$

$\theta = {\tan ^{ - 1}}\left( {{4 \over 2}} \right) = {\tan ^{ - 1}}(2)$

so $F = {{kq(4 - q) \times {{10}^{ - 12}}} \over {{r^2}}}$

so F_max will be at q = 2 $\mu$C

(x << a) ($\alpha$ is acceleration)

${F_{net}} = - \left( {{{k{q_0}Q} \over {{{(a - x)}^2}}} - {{kQ{q_0}} \over {{{(a + x)}^2}}}} \right)$

$m\alpha = - {{k{q_0}Q} \over {{a^4}}}4ax$

$ \Rightarrow \alpha = - {{4k{q_0}Q} \over {m{a^3}}}x$

So, $T = 2\pi \sqrt {{{4\pi {\varepsilon _0}m{a^3}} \over {4{q_0}Q}}} $

or $T = \sqrt {{{4{\pi ^3}{\varepsilon _0}m{a^3}} \over {{q_0}Q}}} $

After connection

${\sigma _1}{R_1} = {\sigma _2}{R_2}$

Now $E = {\sigma \over {{\varepsilon _0}}}$

$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}} = {2 \over 1}$

Force experienced by the charge q

$F = {{kQqx} \over {{{\left[ {{{\left( {{d \over 2}} \right)}^2} + {x^2}} \right]}^{{3 \over 2}}}}}$

For maximum Coulomb's force for x

${{dF} \over {dx}} = 0$

On solving $x = {d \over {2\sqrt 2 }}$

$\overrightarrow E = -{{dV} \over {dx}}\widehat i$

$\overrightarrow E = - 6x\widehat i$

So, $\overrightarrow E $ at (1, 0, 3) is

$\overrightarrow E = - 6\widehat i$ V/m

${v^2} - {u^2} = 2as$

$ \Rightarrow {0^2} - {200^2} = 2\left( {{{ - qE} \over m}} \right)(S)$

$ \Rightarrow - {200^2} = 2\left[ {{{ - 40 \times {{10}^{ - 6}} \times {{10}^5}} \over {100 \times {{10}^{ - 6}}}}} \right][S]$

$ \Rightarrow S = {4 \over {2 \times 4}}$ m = 0.5 m

Electric field at P will be

$E = {{kq} \over {{{(d/2)}^2}}} \times 2 = {{8kq} \over {{d^2}}}$

So, ${{8 \times 9 \times {{10}^9} \times 8 \times {{10}^{ - 6}}} \over {{d^2}}} = 6.4 \times {10^4}$

So, $d = 3$ m

As one moves closer to a positive charge (isolated) the density of electric field line increases and so does the electric field intensity

$\Rightarrow$ Statement I is true

As opposite poles of an electric dipole would experience equal and opposite forces so net force on a dipole in a uniform electric field will be zero

$\Rightarrow$ Statement II is true

$\left| {{E_0}} \right| = {{kq/2} \over {{a^2}}}\sqrt 2 + {{kq} \over {{{\left( {a\sqrt 2 } \right)}^2}}}$

$ = {{kq} \over {\sqrt 2 {a^2}}} + {{kq} \over {2{a^2}}}$

$ = {{kq} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }} + {1 \over 2}} \right),\,k = {1 \over {4\pi {\varepsilon _0}}}$

Flux passing through flat surface = Flux passing through curved surface.

So, $\phi = {q \over {2{\varepsilon _0}}}$

F_net on charge 3, ${F_1} = {{\sqrt 3 k{q^2}} \over {{1^2}}}$

Force between any 2 charges

${F_2} = {{k{q^2}} \over {{1^2}}}$

So, ${{{F_1}} \over {{F_2}}} = \sqrt 3 $

Non-polar bonds do not have any net dipole moment and are generally formed in compound where there is presence of symmetry.

When non polar material placed in electric field, due to redistribution of charges dipole is formed.

So, (R) is incorrect.

For an infinite charged plane

$E = {\sigma \over {2{\varepsilon _0}}}$ for any value of l

$ \Rightarrow {E_1} = {E_2} = {\sigma \over {2{\varepsilon _0}}}$

According to given information:

${{k{Q^2}} \over {{L^2}}}$ = $\mu$mg

Putting the values, we get

L = 12 cm

${{m{v^2}} \over r} = {{2k\rho \times \pi {R^2}q} \over r}$

$ \Rightarrow {1 \over 2}m{v^2} = {{\rho {R^2}q} \over {4{\varepsilon _0}}}$

Since the droplet is at rest

$\Rightarrow$ Net force = 0

$\Rightarrow$ mg = qE

$\Rightarrow$ $q = {{mg} \over E}$ = 2 $\times$ 10^$-$9 C

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.

The given situation is shown in the figure below.

Since, block remains at rest on inclined plane, then

$ \begin{array}{rlrl} & q E =m g \sin 60^{\circ} \\ \Rightarrow & E =\frac{m g \sin 60^{\circ}}{q} \end{array} $

Here, $q=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}$

$ \begin{aligned} & m=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg} \\ \Rightarrow \quad g & =10 \mathrm{~m} / \mathrm{s}^2 \\ \therefore \quad E & =\frac{5 \times 10^{-3} \times 10}{5 \times 10^{-6}} \times \frac{\sqrt{3}}{2} \\ & =\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C} \end{aligned} $

Given, ratio of radii of two metal sphere,

$ \begin{aligned} & & R_1: R_2=4: 7 \\ \Rightarrow & & \frac{R_1}{R_2}=\frac{4}{7} \end{aligned} $

If $q_1$ and $q_2$ be the charges on both the spheres after being contact with charge of $8.8 \times 10^{-7} \mathrm{C}$. then $q_1+q_2=8.8 \times 10^{-7} \mathrm{C}$

$ \begin{aligned} & =0.88 \mu \mathrm{C} \\ & \Rightarrow \quad q_1+q_2=0.88 \mu \mathrm{C} \end{aligned} $

If $V$ be the common potential on both the spheres after connection, then

$ \begin{array}{lll} & & q_1=C_1 V \\ \text { and } & & q_2=C_2 V \\ \therefore & & \frac{q_1}{q_2}=\frac{C_1 V}{C_2 V}=\frac{C_1}{C_2}=\frac{4 \pi \varepsilon_0 R_1}{4 \pi \varepsilon_0 R_2} \\ \Rightarrow & & \frac{q_1}{q_2}=\frac{R_1}{R_2}=\frac{4}{7} \\ \Rightarrow & & q_1=\frac{4}{7} q_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we get

$ \begin{array}{ll} & \frac{4}{7} q_2+q_2=0.88 \\ \Rightarrow \quad & q_2=\frac{7}{11} \times 0.88=0.56 \mu \mathrm{C} \\ \therefore \quad & q_1=0.88-q_1 \\ & =0.88-0.56=0.32 \mu \mathrm{C} \end{array} $

i.e charge on smaller sphere,

$ q_1=0.32 \times 10^{-6} \mathrm{C} $

∴ Potential due to smaller sphere at $r=60 \mathrm{~m}$

$ \begin{aligned} V & =9 \times 10^9 \frac{q_1}{r} \\ & =9 \times 10^9 \times \frac{0.32 \times 10^{-6}}{60}=48 \mathrm{~V} \end{aligned} $

For the given situations, arrangement of charges are given as

The magnitude of force on charge, $q_3=5 \mu \mathrm{C}$

placed at mid-point $O$ is given as

$ \begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{(A O)^2}+\frac{1}{4 \pi \varepsilon_0} \frac{q_3 q_2}{(O B)^2} \\ & =9 \times 10^9 \times \frac{10 \times 10^{-6} \times 5 \times 10^{-6}}{\left(5 \times 10^{-2}\right)^2} \\ & -9 \times 10^9 \times \frac{5 \times 10^{-6} \times 10 \times 10^{-6}}{\left(5 \times 10^{-2}\right)^2} \end{aligned} $

$ \begin{aligned} & =\frac{45 \times 10^{-2}}{25 \times 10^{-4}}+\frac{45 \times 10^{-2}}{25 \times 10^{-4}} \\ & =\frac{9}{5} \times(10)^2+\frac{9}{5} \times(10)^2 \\ & =\frac{18}{5} \times(10)^2 \\ & =3.6 \times 10^2 \mathrm{~N} \\ & =360 \mathrm{~N} \end{aligned} $

Capacitance of sphere-1

$ C_1=4 \pi \varepsilon_0 R $

Capacitance of sphere-2

$ \begin{aligned} & C_2=4 \pi \varepsilon_0(3 R) \\ & C_2=12 \pi \varepsilon_0 R \end{aligned} $

We know after connection, charge flows from sphere-1 to sphere-2 till both acquired same potential. If $q_1{ }^{\prime}$ and $q_2{ }^{\prime}$ be charges on sphere-1 and sphere-2 respectively after connection, then

$ \frac{q_1^{\prime}}{q_2^{\prime}}=\frac{C_1 V}{C_2 V} $

(where, $V=$ common potential)

$ \Rightarrow \quad \frac{q_1^{\prime}}{q_2^{\prime}}=\frac{C_1}{C_2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The ratio of charge densities of sphere-1 and sphere-2 after connection is given as

$ \frac{\sigma_1}{\sigma_2}=\frac{\frac{q_1^{\prime}}{4 \pi R^2}}{\frac{q_2^{\prime}}{4 \pi(3 R)^2}}=\frac{q_1^{\prime}}{q_2^{\prime}} \times 9=\frac{C_1}{C_2} \times 9 $

[from Eq. (i)]

$ \therefore \quad \frac{\sigma_1}{\sigma_2}=9 \frac{C_1}{C_2}=9 \cdot \frac{4 \pi \varepsilon_0 R}{4 \pi \varepsilon_0(3 R)}=3 $

Charge placed on the centre of cube, $q=6 \mu \mathrm{C}=6 \times 10^{-6} \mathrm{C}$

According to Gauss's law, total electric flux passing through cube, $\phi=q / \varepsilon_0$ Since, there are 6 faces in cube. Hence, electric flux passing through each face,

$ \begin{aligned} \phi_1 & =\frac{\phi}{6} \\ & =\frac{q}{6 \varepsilon_0}=\frac{q}{\frac{3}{2 \pi} \times 4 \pi \varepsilon_0} \\ & =\frac{2 \pi}{3} \times q \cdot \frac{1}{4 \pi \varepsilon_0} \\ & =\frac{2 \pi q}{3} \times 9 \times 10^9 \\ & =\frac{2 \pi}{3} \times 6 \times 10^{-6} \times 9 \times 10^9 \\ & =36 \pi \times 10^3 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C} \end{aligned} $

The given situation is shown below.

Here, $x=\sqrt{3} R$

Now, total potential at the centre of the Ring-1.

$V_1=$ potential due to Ring-1 + Potential due to Ring-2.

$ =\frac{q}{4 \pi \varepsilon_0 R}+\frac{(-q)}{4 \pi \varepsilon_0 \sqrt{\left(R^2+x^2\right)}} $

Similarly, total potential at the centre of the Ring-2.

$ V_2=\frac{-q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 \sqrt{R^2+x^2}} $

Hence, Potential difference,

$ \begin{aligned} & V_1-V_2=2\left(\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{4 \pi \varepsilon_0 \sqrt{R^2+x^2}}\right) \\ &=2\left(\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{4 \pi \varepsilon_0 \sqrt{R^2+(\sqrt{3} R)^2}}\right) \\ & {[\because x=\sqrt{3} R] } \\ &=2\left[\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{8 \pi \varepsilon_0 R}\right] \\ &=2\left[\frac{q}{8 \pi \varepsilon_0 R}\right]=\frac{q}{4 \pi \varepsilon_0 R} \end{aligned} $

Given, mass of charges,

$ \begin{aligned} m_1 & =m_2=1 \mathrm{~g}=10^{-3} \mathrm{~kg} \\ r & =1 \mathrm{~m} \end{aligned} $

∴ Gravitational force between them ,

$ \begin{aligned} F_G & =\frac{G m_1 m_2}{r^2} \\ & =\frac{6.67 \times 10^{-11} \times 10^{-3} \times 10^{-3}}{1^2} \\ & {\left[\therefore G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\right] } \\ & =6.67 \times 10^{-17} \mathrm{~N} \end{aligned} $

Again, charge on $m_1$ and $m_2$,

$ q_1=q_2=1 \text { femto coulomb }=10^{-15} \mathrm{C} $

∴ Electrostatic force between the both charges,

$ \begin{aligned} F_e & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \\ & =9 \times 10^9 \times \frac{10^{-15} \times 10^{-15}}{1^2} \\ & =9 \times 10^{-21} \mathrm{~N} \end{aligned} $

∴ Clearly $F_G \gg F_e$

Hence, gravitational force is dominant force.

The given arrangement of system of charges are shown as

$ \text { In } \triangle A B C, \sin 30^{\circ}=\frac{B C}{A C} $

$ \Rightarrow \quad \frac{1}{2}=\frac{B C}{10} \Rightarrow B C=5 \mathrm{~cm} $

Similarly, $A B=A C \sin 60^{\circ}=10 \times \frac{\sqrt{3}}{2}$

$ =5 \sqrt{3} \mathrm{~cm} $

The given charges form two electric dipole of dipole moment $p_1$ and $p_2$ as shown below

Since, $p_1=A B \times q=5 \sqrt{3} q(\mathrm{C}-\mathrm{m})$

and $p_2=B C \times q=5 q \mathrm{C}-\mathrm{m}$

The magnitude of dipole moment of the combination

$ \begin{aligned} p & =\sqrt{p_1^2+p_2^2} \\ & =\sqrt{(5 \sqrt{3} q)^2+(5 q)^2} \\ \Rightarrow \quad p & =\sqrt{100 q^2}=10 q \end{aligned} $

The given situation is shown below

Acceleration of the charge particle in electric field,

$ a_y=\frac{F}{m}=\frac{q E}{m} $

Time taken by the charged particle to cross the field, $t=x / v$.

Upward component of velocity of charge particle emerging, from field region,

$ \begin{aligned} v_y & =u_y t+\frac{1}{2} a_y t^2 \\ & =0 \times t+\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{x}{v}\right)^2 \\ v_y & =\frac{1}{2} \frac{q E x^2}{m v^2} \end{aligned} $

This equation is in the form of $y=k x^2$, which represents the equation of parabola.

Given, $\sigma=8.85 \times 10^{-6} \mathrm{C} / \mathrm{m}^2$

Kinetic energy of electron,

$ K=8 \times 10^{-17} \mathrm{~J} $

We know that, electric field due to metal plate is given by

$ E=\frac{\sigma}{\varepsilon_0} $

∴ Force on electron in this electric field,

$ F=-e E=-\frac{e \sigma}{\varepsilon_0} $

∴ Acceleration on electron,

$ a=\frac{F}{m}=-\frac{e \sigma}{m \varepsilon_0} $

Let $u$ be the initial velocity of electron, then

$ v^2=u^2+2 a s $

but $v=0$

$ \begin{aligned} &\begin{aligned} & \therefore \quad 0=u^2+2\left(\frac{-e \sigma}{m \varepsilon_0}\right) s \\ & \Rightarrow \quad s=\frac{u^2 \varepsilon_0 m}{2 e \sigma} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { but } \quad K=\frac{1}{2} m u^2 \\ & \Rightarrow \quad u^2=\frac{2 K}{m} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned}\\ &\text { ∴ From Eqs. (i) and (ii), we get }\\ &s=\frac{2 K \varepsilon_0 m}{2 m e \sigma}=\frac{K \varepsilon_0}{e \sigma} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad s & =\frac{8 \times 10^{-17} \times 8.85 \times 10^{-12}}{1.6 \times 10^{-19} \times 8.85 \times 10^{-6}} \\ & =5 \times 10^{-4} \mathrm{~m}=0.5 \mathrm{~mm} \end{aligned} $

The given situation is shown below.

When electron reaches at point $B$, then change in its potential energy is equal to gain in kinetic energy.

i.e. $\Delta U=K$

$\Rightarrow \mathrm{PE}$ at point $B-\mathrm{PE}$ at point $A=\frac{1}{2} m_e v^2$

$ \begin{aligned} & \Rightarrow \quad \frac{1}{4 \pi \varepsilon_0} \frac{q_1 \cdot q}{r_2^2}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q}{r_1^2}=\frac{1}{2} m_e v^2 \\ & \Rightarrow \quad \frac{1}{4 \pi \varepsilon_0}\left[\frac{e q}{r_2^2}-\frac{e q}{r_1^2}\right]=\frac{1}{2} m_e v^2 \\ & \Rightarrow \quad 9 \times 10^9\left[\frac{e q}{2^2}-\frac{e q}{4^2}\right]=\frac{1}{2} m_e v^2 \\ & \Rightarrow \quad 9 \times 10^9\left(\frac{3 e q}{16}\right)=\frac{1}{2} m_e v^2 \\ & \Rightarrow v=\sqrt{\frac{3 e q \times 9 \times 10^9}{8 m_e}} \\ & \Rightarrow v=\sqrt{\frac{3 \times 1.6 \times 10^{-19} \times 2 \times 10^{-8} \times 9 \times 10^9}{8 \times 9 \times 10^{-31}}} \\ & =3.46 \times 10^6 \end{aligned} $

Which is nearest to $4 \times 10^6 \mathrm{~m} / \mathrm{s}$.

The given situation is shown below

From the above figure total charge enclosed by the sphere of radius $2.5 \mathrm{~cm}=5 \mathrm{q}$

$\therefore$ According to gauss law, total electric flux passing through spherical surface,

$\begin{aligned} \phi & =\frac{\text { total charge enclosed }}{\varepsilon_0} \\ & =\frac{5 q}{\varepsilon_0} \end{aligned}$

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.