Electrostatics

Electric field due to infinitely long wire

$\overrightarrow E = - {{2k\lambda } \over r}\widehat r$

We know, $V = - \int {\overrightarrow E \,.\,\overrightarrow {dr} } $

$ = - \int {{{ - 2k\lambda } \over r}\widehat r\,.\,\overrightarrow {dr} } $

$v = \int {{{2k\lambda } \over r}dr = 2k\lambda \,\ln r + c} $

Net potential due to all wires,

$V = {v_1} + {v_2} + {v_3}$

$ = 2k\lambda \ln \left( {\sqrt {{x^2} + {y^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{y^2} + {z^2}} } \right) + 2k\lambda \ln \left( {\sqrt {{z^2} + {x^2}} } \right) + c$

for equipotential surface, $v = c$

$ \Rightarrow 2k\lambda \left[ {\ln \left( {\sqrt {{x^2} + {y^2}} \,.\,\sqrt {{y^2} + {z^2}} \,.\,\sqrt {{z^2} + {x^2}} } \right)} \right] = c$

$ \Rightarrow \sqrt {({x^2} + {y^2})\,.\,({y^2} + {z^2})\,.\,({z^2} + {x^2})} = c$

$ \Rightarrow ({x^2} + {y^2})({y^2} + {z^2})({z^2} + {x^2}) = c$

where, c = constant.

In $\Delta OMA,{{OM} \over {OA}} = \cos 60^\circ = {1 \over 2}$

$ \Rightarrow OM = {l \over 2}$ (as $OA = l$)

$MN = ON - OM = l - {l \over 2} = {l \over 2}$

Using work $-$ energy theorem,

${W_{all}} = \Delta k$

$ \Rightarrow {W_c} = {k_f} - {k_i}$

$ \Rightarrow qE{l \over 2} = {1 \over 2}m{v^2} - o$ (As ${v_i} = o$ and $w = FS$)

$ \Rightarrow {v^2} = {{qEl} \over m}$

$ \Rightarrow v = \sqrt {{{qEl} \over m}} $

$\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^2} \\ & 9 \times 10^{-3}=\frac{9 \times 10^9 \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^2} \\ & \mathrm{r}^2=\frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4} \\ & \mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm} \end{aligned}$

$\begin{aligned} & \mathrm{U}_1=\frac{4 \mathrm{Kq}_0^2}{\mathrm{a}}+\frac{2 \mathrm{Kq}_0^2}{\sqrt{2} \mathrm{a}}=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4+\sqrt{2}) \\ & \mathrm{U}_2=\frac{\mathrm{Kq}_0^2}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}(4+\sqrt{2})=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4 \sqrt{2}+2) \\ & \mathrm{U}_2-\mathrm{U}_1=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2) \end{aligned}$

$ u = \frac{1}{2}\varepsilon_0 E^2 $

The energy density of an electric field in free space is given by the expression above. For a parallel plate capacitor with plate area $A$ and separation $d$, the volume between the plates is

$ V = Ad. $

Multiplying the energy density by the volume gives the total energy stored:

$ U = u \times V = \frac{1}{2}\varepsilon_0 E^2 \times Ad = \frac{1}{2}\varepsilon_0 E^2 Ad. $

Thus, the potential energy stored in the capacitor is

$ \frac{1}{2}\varepsilon_0 E^2 Ad. $

This corresponds to the option:

$ \text{Option D: } \frac{1}{2}\varepsilon_0 E^2 Ad. $

The work done on an electric dipole in a uniform electric field when it is rotated from an angle $\theta_1$ to an angle $\theta_2$ is given by:

$ W = -pE(\cos \theta_2 - \cos \theta_1) $

where:

$ p = q \cdot d $ is the dipole moment,

$ E $ is the electric field strength,

$ \theta_1 $ and $ \theta_2 $ are the initial and final angles between the dipole moment and the electric field.

Given:

Charge $ q = 4 \, \mu \mathrm{C} = 4 \times 10^{-6} \, \mathrm{C} $

Distance $ d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m} $

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

Initial angle $\theta_1 = 0^\circ$ (aligned with the field, equilibrium position)

Final angle $\theta_2 = 180^\circ$

First, calculate the dipole moment $ p $:

$ p = q \cdot d = 4 \times 10^{-6} \, \mathrm{C} \cdot 0.18 \, \mathrm{m} = 7.2 \times 10^{-7} \, \mathrm{Cm} $

Calculate the work done $ W $:

$ W = -pE(\cos 180^\circ - \cos 0^\circ) $

$ W = -7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot (-1 - 1) $

$ W = 7.2 \times 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot 2 $

$ W = 1.44 \times 10^{-2} \, \mathrm{J} = 14.4 \, \mathrm{mJ} $

Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is 14.4 mJ (Option D).

$ U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} $

Here,

$ q_1 = 7 \times 10^{-6} \, C $

$ q_2 = -4 \times 10^{-6} \, C $

The separation between the charges is along the x-axis from $ -7 \, \text{cm} $ to $ 7 \, \text{cm} $, so

$ r = 0.14 \, m $

The Coulomb constant is approximated as

$ \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \frac{Nm^2}{C^2} $

Substitute these values into the equation:

$ U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-4 \times 10^{-6})}{0.14} $

Calculate the product of the charges:

$ (7 \times 10^{-6})(-4 \times 10^{-6}) = -28 \times 10^{-12} \, C^2 $

Now, divide by the distance:

$ \frac{-28 \times 10^{-12}}{0.14} = -2 \times 10^{-10} \, C^2/m $

Finally, multiply by the Coulomb constant:

$ U = 9 \times 10^9 \times (-2 \times 10^{-10}) = -1.8 \, J $

Thus, the electrostatic potential energy of the configuration is:

$ -1.8 \, J $

This corresponds to Option D.

$\begin{aligned} & \frac{\mathrm{kq}}{(\mathrm{r}-\mathrm{a})^2}=\frac{\mathrm{kq}}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{kq}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)} \cos \theta \\ & \frac{1}{(\mathrm{r}-\mathrm{a})^2}=\frac{1}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\ & \frac{1}{(\mathrm{r}-\mathrm{a})^2}-\frac{1}{(\mathrm{r}+\mathrm{a})^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\ & \frac{4 \mathrm{ra}}{\left(\mathrm{r}^2-\mathrm{a}^2\right)^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{2 r}{\left(r^2-a^2\right)^2}=\frac{1}{\left(r^2+a^2\right)^{\frac{3}{2}}} \\ & \frac{4 r^2}{\left(r^2-a^2\right)^4}=\frac{1}{\left(r^2+a^2\right)^3} \\ & \Rightarrow 4 r^2\left(r^2+a^2\right)^3=\left(r^2-a^2\right)^4 \\ & 4 r^8\left(1+\frac{a^2}{r^2}\right)^3=r^8\left(1-\frac{a^2}{r^2}\right)^4 \\ & 4\left(1+\frac{a^2}{r^2}\right)^3=\left(1-\frac{a^2}{r^2}\right)^4\end{aligned}$

Exact value cannot be solved in exam for this equation to be true

$ \left|\frac{\mathrm{a}}{\mathrm{r}}\right|>1 \Rightarrow \mathrm{a}>\mathrm{r} $

But point charge $Q$ lies between charges of dipole

1 hence electric field cannot be zero.

There for it should be bonus.

But by solving from mathematical software we are getting $\mathrm{a} / \mathrm{r} \approx 3$.

Let's analyze the given expression step by step.

We are given the electric flux:

$\phi = \alpha \sigma + \beta \lambda$

where:

$\sigma$ is a surface charge density with units of charge per unit area (C/m²).

$\lambda$ is a linear charge density with units of charge per unit length (C/m).

For the equation to be dimensionally consistent, the two terms on the right must have the same units as the flux $\phi$.

For the term $\alpha \sigma$:

Since $\sigma$ has units $\text{C/m}^2$, the constant $\alpha$ must have units that convert $\sigma$ into the same units as the flux.

For the term $\beta \lambda$:

Since $\lambda$ has units $\text{C/m}$, the constant $\beta$ carries its own units to ensure the term matches the flux’s dimensions.

Since both terms add to give the flux, they must share the same overall dimension. Thus, the dimensions of $\alpha \sigma$ and $\beta \lambda$ must be equal:

$[\alpha] \cdot \frac{Q}{L^2} = [\beta] \cdot \frac{Q}{L}$

Here, $Q$ represents the unit of charge and $L$ represents a unit of length. Canceling the common factors, we find:

$\frac{[\alpha]}{[\beta]} = L$

That is, the ratio $\frac{\alpha}{\beta}$ has the dimension of length.

Now, let’s match this with the given options:

Option A: Displacement (commonly measured as a length)

Option B: Charge (has the unit Coulomb)

Option C: Electric Field (has units of V/m or N/C)

Option D: Area (has units of length squared, $L^2$)

Since $\frac{\alpha}{\beta}$ has the dimensions of a length, it logically corresponds to a displacement.

Thus, the correct answer is:

Option A – displacement.

Electric potential and field at point A:

At point A (on the $x$-axis), the electric field is $E_A = \frac{2kP}{r^3} = E_0$ and the potential is $V_A = \frac{kP}{r^2} = V_0$.

Electric potential and field at point B:

Point B is on the $y$-axis, which is at the same distance from the origin as point A, but in a direction perpendicular to the dipole moment.

The field at point B is given by $E_B = \frac{kP}{(2r)^3} = \frac{E_0}{16}$. This is because the distance from O to B is $2r$, and the formula for the field off the axis (on the equatorial line) is used.

The potential at point B is $V_B = \frac{k \vec{p} \cdot \hat{r}}{(2r)^2} = 0$, since $\vec{p}$ and $\hat{r}$ are perpendicular at B, so their dot product is zero.

$\phi = {{{q_{enc}}} \over {{\varepsilon _0}}}$, ${q_{enc}}$ = enclosed charge

here, we need three other identical cubes to enclose the charge fully.

So, total four cubes enclose the charge

Hence, ${\phi _{cube}} = {\phi \over 4}$

$ = {{{q_{enc}}} \over {4{\varepsilon _0}}} = {{\lambda \left( {{a \over 2}} \right)} \over {4{\varepsilon _0}}} \Rightarrow {\phi _{cube}} = {{\lambda a} \over {8{\varepsilon _0}}}$

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} $

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

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

Effective air distance between charges is,

$ d=(r-t)+t \sqrt{K} $

$r=$ distance between charges

$t=$ thickness of dielectric slab

$K=$ dielectric constant

∴ Force is $F=\frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{(r-t+t \sqrt{K})^2}$

$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q_1 Q_2}{r^2}\left(\frac{r}{r-t+t \sqrt{K}}\right)^2 \\ & =F_{\text {air }}\left(\frac{r}{r-t+t \sqrt{K}}\right)^2 \end{aligned} $

Here, $t=30 \%$ of $r=\frac{30}{100} r$

and $F=\frac{F_{\text {air }}}{2.56}$

$ \begin{aligned} & \text { So, } \frac{1}{2.56}=\left(\frac{r}{r-\frac{30}{100} r+\frac{30}{100} r \times \sqrt{K}}\right)^2 \\ & \Rightarrow \quad \frac{10}{16}=\frac{r}{r-0.3 r+0.3 r \sqrt{K}} \\ & \Rightarrow \quad \frac{10}{16}=\frac{1}{0.7+0.3 \sqrt{K}} \\ & \Rightarrow \quad 7+3 \sqrt{K}=16 \\ & \quad 3 \sqrt{K}=9 \\ & \quad \sqrt{K}=3 \text { or } K=9 . \end{aligned} $

$729 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3$

$ \begin{aligned} \Rightarrow \quad(9 r)^3 & =R^3 \\ R^3 & =9 r \end{aligned} $

Potential on small sphere,

$ \begin{aligned} & \boldsymbol{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ \Rightarrow & 3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ \Rightarrow & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Charge on big sphere, $Q=729 q$

∴ Potential on big sphere,

$ V^{\prime}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R} $

$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{729 q}{9 r}=81 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ & =81 \times 3 \\ & =243 \mathrm{volt} \\ & \text { [From Eq. (i)] } \end{aligned} $

Step 1: Formula for electric field around a long wire
The electric field ($E$) at a distance $x$ from a very long straight wire with charge per unit length ($\lambda$) is given by:

$E = \frac{2 K \lambda}{x}$
Here, $K$ is Coulomb's constant ($K = 9 \times 10^9~\mathrm{N\,m^2\,C^{-2}}$).

Step 2: Substitute the given values
We are given:

• $E = 7.5 \times 10^4~\mathrm{N/C}$
• $\lambda = 2.5 \times 10^{-7}~\mathrm{C/m}$
• $K = 9 \times 10^9~\mathrm{N\,m^2\,C^{-2}}$

Step 3: Solve for $x$
Multiply the numbers on the right side first:

• $2 \times 9 = 18$
• $18 \times 2.5 = 45$
• So, top becomes: $45 \times 10^9 \times 10^{-7} = 45 \times 10^{2} = 4500$

Step 4: Rearrange to find $x$
Multiply both sides by $x$ and divide both sides by $7.5 \times 10^4$: $x = \frac{4500}{7.5 \times 10^4} = \frac{4500}{75000} = 0.06~\mathrm{m}$

Step 5: Convert meters to centimeters
$0.06~\mathrm{m} = 6~\mathrm{cm}$

Step 1: Displacement equations

The proton and the alpha particle both start from rest, so their displacement after some time is given by:

$s = \frac{1}{2} a t^2.$

Since they travel the same distance:

$\frac{1}{2} a_\alpha t_\alpha^2 = \frac{1}{2} a_p t_p^2$

This means:

$a_\alpha t_\alpha^2 = a_p t_p^2$

Step 2: Acceleration in an electric field

The acceleration of a charged particle in an electric field $E$ is:

$a = \frac{qE}{m}$

- For a proton, charge $= e$, mass $= m_p$

- For an alpha particle, charge $= 2e$, mass $= 4m_p$

So: - $a_p = \frac{eE}{m_p}$ - $a_\alpha = \frac{2eE}{4m_p} = \frac{eE}{2m_p}$

Step 3: Substitute accelerations into equation

Plug the accelerations into the earlier equation:

$\left(\frac{e E}{2 m_p}\right) t_\alpha^2 = \left(\frac{e E}{m_p}\right) t_p^2$

Step 4: Solve for time ratio

Divide both sides by $\frac{eE}{m_p}$:

$\frac{t_\alpha^2}{2} = t_p^2$

This gives:

$t_\alpha^2 = 2 t_p^2$

Taking square roots:

$t_\alpha = \sqrt{2} t_p$

So,

$\frac{t_p}{t_\alpha} = \frac{1}{\sqrt{2}}$

or

$t_p : t_\alpha = 1 : \sqrt{2}$

$ \begin{aligned} &\text { Net force on the charge at origin }=0\\ &\begin{aligned} & \Rightarrow \frac{1}{4 \pi \varepsilon_0}\left[\begin{array}{l} \frac{2 \times 10^{-6} Q}{\left(1 \times 10^{-2}\right)^2}+\frac{2 \times 10^{-6} \times 4 \times 10^{-6}}{\left(2 \times 10^{-2}\right)^2} \\ +\frac{2 \times 10^{-6} \times 12 \times 10^{-6}}{\left(4 \times 10^{-2}\right)^2} \end{array}\right]=0 \\ & \Rightarrow \frac{Q}{10^{-4}}+\frac{4 \times 10^{-6}}{4 \times 10^{-4}}+\frac{12 \times 10^{-6}}{16 \times 10^{-4}}=0 \\ & \Rightarrow Q+10^{-6}+\frac{3}{4} \times 10^{-6}=0 \\ & \Rightarrow Q+\frac{7}{4} \times 10^{-6}=0 \\ & \Rightarrow Q=-\frac{7}{4} \times 10^{-6} \mathrm{C}=-\frac{7}{4} \mu \mathrm{~F}=-1.75 \mu \mathrm{~F} \end{aligned} \end{aligned} $

According to work energy theorem,

$ \begin{aligned} & W=q \Delta V \\ & \mathrm{KE}_f-\mathrm{KE}_i=q \Delta V \\ & \frac{1}{2} m v^2-0=2 \times 10^{-6} \times 160 \\ & \Rightarrow \frac{1}{2} \times 10^{-5} v^2=3.2 \times 10^{-4} \\ & \Rightarrow v^2=64 \Rightarrow v=8 \mathrm{~m} / \mathrm{s} \end{aligned} $

Acceleration of electron in electric field is

$ a_e=\frac{F_e}{m}=-\frac{e E}{m} $

Acceleration of positron in electric field.

$ a_p=\frac{F_p}{m}=\frac{e E}{m} $

Displacement of electron in electric field

$ y_e=0 \times t+\frac{1}{2} a_e t^2=-\frac{1}{2} \frac{e E}{m} t^2 $

Displacement of positron in electric field

$ y_p=\frac{1}{2} a_p t^2=\frac{1}{2} \frac{e E}{m} t^2 $

The distance of separation between electron and positron

$ \begin{aligned} & =\left|y_p-y_e\right| \\ & =\left|\frac{\frac{1}{2} e E t^2}{m}-\left(-\frac{1}{2} \frac{e E t^2}{m}\right)\right| \\ & =\left|\frac{e E t^2}{m}\right|=\frac{e E t^2}{m} \end{aligned} $

The potential at points $B$ and $C$ due to the charges at $O$ and $A$

Potential at $B$

$ V_B=\frac{q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 2 R} $

Potential at $C$

$ V_c=\frac{q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 \sqrt{2} R} $

Work done

$ \begin{aligned} W & =q\left(V_C-V_B\right) \\ & =q\left[\begin{array}{l} \left(\frac{q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 \sqrt{2} R}\right) \\ -\left(\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{4 \pi \varepsilon_0 2 R}\right) \end{array}\right] \\ & =\frac{q^2}{4 \pi \varepsilon_0 R}\left[\frac{1}{\sqrt{2}}-\frac{1}{2}\right] \\ & =\frac{q^2}{4 \pi \varepsilon_0 R}\left[\frac{\sqrt{2}-1}{2}\right] \end{aligned} $

The electromagnetic force between two protons is much stronger than the gravitational force between them.

Let $F_e$ be the electromagnetic force, and $F_g$ be the gravitational force.

According to the problem: $ F_e = 10^n F_g $

We write the formulas for both forces:

Electromagnetic force: $ F_e = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} $

Gravitational force: $ F_g = \frac{G m^2}{r^2} $

Replace $F_e$ and $F_g$ in the equation: $ \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = 10^n \frac{G m^2}{r^2} $

Since $r^2$ is on both sides, it cancels out:

$ \frac{1}{4\pi\varepsilon_0} e^2 = 10^n G m^2 $

Solve for $10^n$:

$ 10^n = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{G m^2} $

Now, use the constants:

$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$

Charge of proton, $e = 1.6 \times 10^{-19}$

Gravitational constant, $G = 6.67 \times 10^{-11}$

Mass of proton, $m = 1.67 \times 10^{-27}$

Plug in the values:

$ 10^n = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2} $

Calculate the numerator: $9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 9 \times 10^9 \times 2.56 \times 10^{-38} = 2.304 \times 10^{-28}$

Calculate the denominator: $6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2 = 6.67 \times 10^{-11} \times 2.79 \times 10^{-54} = 1.86 \times 10^{-64}$

Divide the values: $ \frac{2.304 \times 10^{-28}}{1.86 \times 10^{-64}} = 1.23 \times 10^{36} $

This means $10^n = 1.23 \times 10^{36}$, so $n = 36$.

Total charge on spherical shell.

$ Q=\sigma A=\sigma\left(4 \pi R^2\right)=4 \pi R^2 \sigma $

According to Gauss's law total electric flux through a closed surface.

$ \phi=\frac{Q}{\varepsilon_0}=\frac{4 \pi R^2 \sigma}{\varepsilon_0} $

$\therefore$ Flux through one face

$ \phi_1=\frac{\phi}{6}=\frac{4 \pi R^2 \sigma}{6 \varepsilon_0}=\frac{2 \pi R^2 \sigma}{3 \varepsilon_0} $

Given data:

• Dipole moment, $ p = 2 \times 10^{-10}\ \mathrm{C \, m} $

• Electric field, $ E = 10^4\ \mathrm{N \, C^{-1}} $

• Angle between $ \vec{p} $ and $ \vec{E} $, $ \theta = 30^\circ $

Formula for torque:

$ \tau = p E \sin \theta $

Substitute values:

$ \tau = (2 \times 10^{-10})(10^4) \sin 30^\circ $

$ = 2 \times 10^{-6} \times \frac{1}{2} $

$ = 1 \times 10^{-6}\ \mathrm{N \, m} $

✅ Final Answer:

$ \boxed{10^{-6}\ \mathrm{N \, m}} $

Correct option: A

$ \begin{aligned} &\begin{aligned} O A & =\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 \\ O B & =2 \mathrm{~m} \\ \therefore \quad V_A & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{O A} \\ & =9 \times 10^9 \times \frac{10^{-3} \times 10^{-6}}{2}=4.5 \mathrm{~V} \end{aligned}\\ &\text { Similarly, } V_B=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{O B}\\ &\begin{aligned} & =9 \times 10^9 \times \frac{10^{-3} \times 10^{-6}}{2} \\ & =4.5 \mathrm{~V} \\ \therefore \quad V_A-V_B & =4.5-4.5=0 \mathrm{~V} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_A q_B}{r^2} \\ = & 9 \times 10^9 \times \frac{1.6 \times 10^{-19} \times 3.2 \times 10^{-19}}{9 \times 10^{-4}} \\ = & 512 \times 10^{-25} \mathrm{~N} \end{aligned} $

$ \begin{aligned} & \text { Diagonal } A C=\sqrt{A D^2+D C^2} \\ & =\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 \mathrm{~m} \\ & \therefore O A=O B=O C=O D=\frac{A C}{2}=\frac{2}{2}=1 \mathrm{~m} \end{aligned} $

Electrostatic potential of the centre $(O)$

$ \begin{aligned} V= & V_A+V_B+V_C+V_D \\ = & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{12 \times 10^{-9}}{1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(-20 \times 10^{-9}\right)}{1} +\frac{1}{4 \pi \varepsilon_0} \frac{32 \times 10^{-9}}{1}+\frac{1}{4 \pi \varepsilon_0} \frac{\left(-15 \times 10^{-9}\right)}{1} \\ = & \frac{10^{-9}}{4 \pi \varepsilon_0}[12-20+32-15] \\ = & 9 \times 10^9 \times 10^{-9}(9)=81 \text { volt } \end{aligned} $

$ \begin{aligned} &\text { Initially, }\\ &F=\frac{k\left|q_1 q_2\right|}{r^2} \end{aligned} $

$ \begin{aligned} & =\frac{k \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{r^2} \\ & =\frac{32 \times 10^{-12} k}{r^2} \end{aligned} $

After connecting by conducting wires,

$ q=q_1+q_2=8 \mu C+(-4 \mu C)=4 \mu C $

∴ Charge on each sphere

$ \begin{aligned} q^{\prime} & =\frac{q}{2}=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C} \\ \therefore \quad F^{\prime} & =k \cdot \frac{q^2}{r^2}=\frac{k \times\left(2 \times 10^{-6}\right)^2}{r^2} \\ & =\frac{4 \times 10^{-12} k}{r^2}=\frac{1}{8} \times \frac{32 \times 10^{-12} k}{r^2} \\ & =\frac{F}{8} \end{aligned} $

$ \begin{aligned} & \text { } V=20|r|=20 \sqrt{x^2+y^2+z^2} \\ & \begin{aligned} & \therefore E=-\left(\frac{\partial v}{\partial x} \hat{\mathbf{i}}+\frac{\partial v}{\partial y} \hat{\mathbf{j}}+\frac{\partial v}{\partial z} \hat{\mathbf{k}}\right) \\ &=-\left[\frac{20 x}{\sqrt{x^2+y^2+z^2}} \hat{\mathbf{i}}+\frac{20 y}{\sqrt{x^2+y^2+z^2}} \hat{\mathbf{j}}\right. \left.+\frac{20 z}{\sqrt{x^2+y^2+z^2}} \hat{\mathbf{k}}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{-20(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{\sqrt{x^2+y^2+z^2}} \\ & \text { At point }(4 m, 3 m,-5 m) \\ & \qquad \begin{aligned} E & =\frac{-20(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})}{\sqrt{4^2+3^2+(-5)^2}} \\ & =\frac{-20}{5 \sqrt{2}}(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \\ & =-\sqrt{2}(8 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}) \end{aligned} \end{aligned} $

Given, $q_1+q_2=25 \mu \mathrm{C}$

Also, $F=\frac{k q_1 q_2}{r^2}$

$ \begin{aligned} & \Rightarrow \quad 0.6=\frac{9 \times 10^9\left(q_1 q_2\right)}{(1.5)^2} \\ & \Rightarrow \quad q_1 q_2=\frac{0.6 \times(1.5)^2}{9 \times 10^9} \\ & \Rightarrow q_1 q_2=1.5 \times 10^{-10} \end{aligned} $

Using $\left(q_1-q_2\right)^2=\left(q_1+q_2\right)^2-4 q_1 q_2$

we get

$ \begin{aligned} & \left(q_1-q_2\right)^2=\left(25 \times 10^{-6}\right)^2-4 \times 1.5 \times 10^{-10} \\ & \Rightarrow\left(q_1-q_2\right)^2=625 \times 10^{-12}-6 \times 10^{-10} \\ & =(625-600) \times 10^{-12} \\ & =25 \times 10^{-12} \end{aligned} $

or $\quad q_1-q_2=5 \times 10^{-6} \mathrm{C}=5 \mu \mathrm{C}$

Given:

• Total atoms in the solid = $6 \times 10^{24}$

• $0.005\%$ of atoms lose one electron each.

Step 1: Find how many atoms lose an electron

$ \text{Number of affected atoms} = (0.005\% ) \times (6 \times 10^{24}) $

$ 0.005\% = \frac{0.005}{100} = 5\times10^{-5} $

$ \Rightarrow \text{Number of atoms losing electrons} = 6 \times 10^{24} \times 5\times10^{-5} = 3 \times 10^{20} $

Step 2: Each atom loses one electron

Each lost electron corresponds to a positive charge gained by the solid equal to the charge of that electron ($e = 1.6\times10^{-19}\ \text{C}$).

Step 3: Total charge gained

$ Q = (3\times10^{20}) \times (1.6\times10^{-19}) $

$ Q = 4.8\times10^{1} = 48\ \text{C} $

✅ Final Answer:

$ \boxed{+48\ \text{C}} $

Correct Option: B (+48 C)

Step 1: Write the Formula for Energy Stored

The energy stored in a capacitor is given by the formula: $U = \frac{q^2}{2C}$ where $U$ is energy, $q$ is charge, and $C$ is capacitance.

Step 2: Rearranging to Find Capacitance

We can solve for $C$ as follows: $C = \frac{q^2}{2U}$

Here, $q = 12 \times 10^{-6}~\text{C}$ and $U = 6~\text{J}$. Substitute these values into the formula:

$C = \frac{(12 \times 10^{-6})^2}{2 \times 6}$

Step 3: Calculate Capacitance

Simplify the above equation: $C = 12 \times 10^{-12}~\text{F}$

Step 4: Capacitance of a Spherical Conductor

The capacitance of a spherical conductor is given by: $C = 4\pi\varepsilon_0 r$

Step 5: Find the Radius

Rearrange to find the radius $r$: $r = \frac{C}{4\pi\varepsilon_0}$

We can also use $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9~\text{N m}^2/\text{C}^2$ so:

$r = 9 \times 10^9 \times 12 \times 10^{-12}$

Step 6: Solve for the Radius

Multiply the values: $r = 108 \times 10^{-3} = 0.108~\text{m} = 10.8~\text{cm}$

Force on charge particle in electric field,

$ F=q E=2 \times 20=40 \mathrm{~N} $

According to work energy theorem, Kinetic energy $=W$

$ =F \cdot s=40 \times 0.2=8 \mathrm{~J} $

$\sigma_1=\sigma_2=\sigma$

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

$\therefore$ For the smaller sphere,

$ E_1=\frac{\sigma_1}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0} $

For the larger sphere,

$ E_2=\frac{\sigma_2}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0} $

Since, $\quad \sigma_1=\sigma_2=\sigma$

Thus, $\quad E_1=E_2=E$

As we move along a field line we are moving in direction of reducing potential.

So,  $ V_C>V_B>V_A $

Potential at point $C$, due to charges $+q$ and $-q$.

$ V_C=-\frac{k q}{d}+\frac{k q}{d}=0 $

and potential of point $D$

$ V_0=\frac{-k q}{d}+\frac{k q}{3 d}=\frac{-2 k q}{3 d} $

So, potential difference between $C$ and $D$

$ V_D-V_C=\frac{-2 k q}{3 d} $

So, work done in moving charge $Q$ from $C$ to $D$

$ \begin{aligned} & W=Q\left(V_D-V_C\right) \\ & W=\frac{-2 k Q q}{3 d} \end{aligned} $

$ \begin{aligned}Now,\,\, & K=\frac{1}{4 \pi \varepsilon_0} \Rightarrow W=\frac{-2 Q q}{3\left(4 \pi \varepsilon_0\right) d} \\ & W=\frac{-Q q}{6 \pi \varepsilon_0 d} \end{aligned} $

$ \begin{aligned} &\text { Electric flux }\\ &\begin{aligned} \phi & =\mathbf{E} \cdot \mathbf{A}=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}) \\ & =9\left(\mathrm{Nm}^2 / \mathrm{C}\right) \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \text { } & \frac{1}{2} m v^2=q V \\ \Rightarrow \quad v & =\sqrt{\frac{2 q V}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 180}{9 \times 10^{-31}}} \\ & =8 \times 10^6 \mathrm{~m} / \mathrm{s}=8000 \times 10^3 \mathrm{~m} / \mathrm{s} \\ & =8000 \mathrm{~km} / \mathrm{s} \end{aligned} $

Initial potential energy of the system of

$ U_i=3 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} $

Final potential energy of the system

$ U_f=3 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{\frac{L}{2}}=6 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} $

$\therefore$ Required work done

$ \begin{aligned} W & =U_f-U_i \\ & =6 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L}-3 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{3 q^2}{L} \end{aligned} $

$27 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3$

$ \Rightarrow \quad R=3 r $

Potential on small drop

$ V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}=9 \times 10^9 \times \frac{10^{-12}}{10^{-3}}=9 \mathrm{volt} $

Potential on big drop

$ =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{27 q}{R}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{27 q}{3 r} $

$ \begin{aligned} & =9 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ & =9 \times 9=81 \mathrm{~V} \end{aligned} $

$F_0=98 \mathrm{~N}$

$ \begin{aligned} & \text { Using, } \frac{d}{k}=\frac{d_1}{k_1}+\frac{d_2}{k_2} \\ & \Rightarrow \quad \frac{9}{k}=\frac{6}{4}+\frac{3}{9} \\ & \Rightarrow \quad k=\frac{54}{11} \end{aligned} $

$\therefore$ New force, $F^{\prime}=\frac{F_0}{k}$

$ =\frac{98}{\frac{54}{11}}=98 \times \frac{11}{54}=19.96 \mathrm{~N} $

Which is closest to 18 N .

$U_{\text {initial }}=U_{12}+U_{13}+U_{23}$

$ =\frac{k q^2}{2 a}-\frac{k q^2}{a}-\frac{k q^2}{a}=\frac{k q^2}{2 a}-\frac{2 k q^2}{a}=\frac{-3 k q^2}{2 a} $

After exchanging the final charge with one of the negative charge, there are two - $q$ charges separated by a distance $a$ and one $+q$ charge separated by a distance ' $a$ ' from one of the $-q$ charges.

$ \begin{aligned} & \therefore U_{\text {final }}=U_{12}^{\prime}+U_{13}^{\prime}+U_{23}^{\prime} \\ & =\frac{k q^2}{a}-\frac{k q^2}{2 a}-\frac{k q^2}{a}=\frac{-k q^2}{2 a} \\ & \begin{aligned} \therefore \Delta U & =U_{\text {final }}-U_{\text {initial }} \\ & =-\frac{k q^2}{2 a}-\left(\frac{-3 k q^2}{2 a}\right)=\frac{2 k q^2}{2 a} \\ & =\frac{k q^2}{a}=\frac{q^2}{4 \pi \varepsilon_0 a} \quad\left[\because k=\frac{1}{4 \pi \varepsilon_0}\right] \end{aligned} \end{aligned} $

$\begin{aligned} \phi & =\frac{q_{\mathrm{en}}}{\epsilon_0} \\ & =\frac{(5+1-2) q}{\epsilon_0} \\ & =\frac{4 q}{\epsilon_0} \end{aligned}$

When two conducting spheres of radii $a$ and $b$ are connected by a conducting wire, they come to the same potential because conductors in contact share charges until their potentials become equal. The potential of a charged sphere is given by $V = \frac{kQ}{R}$, where $V$ is the potential, $k$ is Coulomb's constant, $Q$ is the charge on the sphere, and $R$ is the radius of the sphere.

For the two spheres at the same potential, we have:

$\frac{kQ_a}{a} = \frac{kQ_b}{b}$

Where:

• $Q_a$ and $Q_b$ are the charges on the spheres with radii $a$ and $b$, respectively.
• $k$ cancels out from both sides as it is a constant.

From the above equation, to find the ratio of charges $\frac{Q_a}{Q_b}$, we rearrange it as follows:

$\frac{Q_a}{Q_b} = \frac{a}{b}$

Therefore, the ratio of the charges of the two spheres respectively is $\frac{a}{b}$.

So, the correct answer is Option C: $\frac{a}{b}$.