Electrostatics

Electrostatic field lines originate from positive charges and terminate on negative charges. Unlike magnetic field lines, they do not form closed loops because the electrostatic force is conservative and a charge cannot return to its starting point along a continuous line of force without a change in potential.

Hence, statement (A) is incorrect.

By convention, electric field lines represent the direction of force on a positive test charge. If a charge is positive $(\mathrm{q}>0)$, the field lines point radially outward from the charge.

Hence, the statement B is correct.

Gauss's Law is a fundamental law of electromagnetism that relates electric flux to enclosed charge. It is mathematically derived based on the inverse-square nature of Coulomb's Law ( $\mathrm{F} \propto \frac{1}{\mathrm{r}^2}$ ). So, Gauss-Law is valid only for inverse square force (like gravitational force too).

Hence, the statement C is correct.

The electrostatic field is a conservative field. In a conservative field the work done in moving a charge between two points is independent of the path taken, and the total work done over any closed loop is exactly zero.

Hence, the statement D is Correct.

Coulomb's force is a central force, i.e., it always acts along the line joining the centres of two charges. The torque on the particle $(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}})$ is zero because the central force $(\overrightarrow{\mathrm{F}})$ is anti-parallel to the position vector, so to make the angular momentum $\left(\vec{\tau}=\frac{d \overrightarrow{\mathrm{~L}}}{d \mathrm{t}}\right)$ will be constant. For the constant angular momentum, the motion is in a plane. Therefore, the motion of a particle under Coulomb's force must take in a plane.

Hence, the statement D is correct.

Therefore, statements B, C, D, and E are correct, while statement A is false.

Hence, the correct option is (C).

When 3 nC is at infinity then electric potential due to charges 1 nC and 2 nC is zero. So the initial potential is $\mathrm{V}_{\mathrm{i}}$

The total potential at the third corner due to $\mathrm{q}_1(1 \mathrm{nC})$ and $\mathrm{q}_2(2 \mathrm{nC})$ is the sum of their individual potentials:

$ \mathrm{V}_{\mathrm{f}}=\frac{\mathrm{kq}_1}{\mathrm{r}}+\frac{\mathrm{kq}_2}{\mathrm{r}}=\frac{\mathrm{k}\left(\mathrm{q}_1+\mathrm{q}_2\right)}{\mathrm{r}} $

The work done is given by $\mathrm{W}=\mathrm{q}_3 \times \Delta \mathrm{V}$, where $\Delta \mathrm{V}=$ potential difference between initial and final position.

$ \mathrm{W}=\mathrm{q}_3 \times\left[\frac{\mathrm{k}\left(\mathrm{q}_1+\mathrm{q}_2\right)}{\mathrm{r}}-0\right] $

$\Rightarrow $ $ \mathrm{W}=\frac{\mathrm{k} \times \mathrm{q}_3 \times\left(\mathrm{q}_1+\mathrm{q}_2\right)}{\mathrm{r}} $

Substituting the values,

$ \mathrm{W}=\frac{\left(9 \times 10^9\right) \times\left(3 \times 10^{-9}\right) \times\left(1 \times 10^{-9}+2 \times 10^{-9}\right)}{0.03} $

$\Rightarrow $ $ \mathrm{W}=\frac{(27) \times\left(3 \times 10^{-9}\right)}{0.03}=27 \times 10^{-7} \mathrm{~J} $

As $1 \mathrm{~J}=10^6 \mu \mathrm{~J}$

$ \mathrm{W}=27 \times 10^{-7} \times 10^6 \mu \mathrm{~J} $

$\Rightarrow $ $ W=27 \times 10^{-1} \mu J=2.7 \mu J $

So, the work done in bring a charge of 3 nC from infinity to the third corner of the triangle is $2.7 \mu \mathrm{~J}$.

Hence, the correct option is (C).

The electric field $\vec{E}$ is related to the electric potential $V$ by $E=-\frac{d V}{d r}$.

Since the potential V depends only on the radial distance r , the electric field is radial.

Given the potential function :

$ \mathrm{V}=\mathrm{ar}^3+\mathrm{b} $

Then the electric field is,

$ \mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dr}}=-\left[\frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{ar}^3+\mathrm{b}\right)\right]=-3 \mathrm{ar}^2 $

We need to find the total charge enclosed in a sphere of unit radius ( $\mathrm{R}=1$ ).

The electric flux passing through the surface of this sphere is given by $\Phi_E=\oint \vec{E} \cdot d \vec{A}$

At $R=1$, the electric field magnitude is :

$ E(1)=-3 a(1)^2=-3 a $

The surface area of the sphere is $4 \pi R^2=4 \pi(1)^2=4 \pi$.

Thus, the flux is :

$ \Phi_{\mathrm{E}}=\mathrm{E} \times \text { Area }=(-3 \mathrm{a}) \times(4 \pi)=-12 \pi \mathrm{a} $

According to Gauss's Law,

$ \Phi_{\mathrm{E}}=\frac{\mathrm{Q}_{\text {enclosed }}}{\epsilon_0} \Rightarrow \mathrm{Q}_{\text {enclosed }}=\Phi_{\mathrm{E}} \times \epsilon_0 $

$\Rightarrow $ $Q_{\text {enclosed }}=(-12 \pi a) \times \epsilon_0$

$\Rightarrow $ $\mathrm{Q}_{\text {enclosed }}=-12 \pi \mathrm{a} \epsilon_0=\alpha \times \pi \mathrm{a} \epsilon_0 \Rightarrow \alpha=-12$

The electric potential V due to a charged spherical shell of radius R and charge Q is given by :

$ \mathrm{V}=\left\{\begin{array}{l} \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{R}}, \mathrm{r}<\mathrm{R} \\ \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{R}}, \mathrm{r}=\mathrm{R} \\ \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}, \mathrm{r}>\mathrm{R} \end{array}\right. $

Point A is on the surface of shell A, inside shell B, and inside shell C. The total potential is the sum of potentials due to each shell (Superposition Principle).

So, the potential at A ;

• 1. Due to A (Surface, $\mathrm{r}=\mathrm{a}) \Rightarrow \mathrm{V}_{\mathrm{AA}}=\frac{\mathrm{k} \mathrm{q}_1}{\mathrm{a}}$

• 2. Due to B (Inside, $\mathrm{r}<\mathrm{b}) \Rightarrow \mathrm{V}_{\mathrm{AB}}=\frac{\mathrm{kq}_2}{\mathrm{~b}}$

• 3. Due to C (Inside, $\mathrm{r}<\mathrm{c}) \Rightarrow \mathrm{V}_{\mathrm{AC}}=\frac{\mathrm{kq}_3}{\mathrm{c}}$

Hence, total potential at point A is,

$ V_A=\left(\frac{\mathrm{kq}_1}{\mathrm{a}}+\frac{\mathrm{kq}_2}{\mathrm{~b}}+\frac{\mathrm{kq}_3}{\mathrm{c}}\right)=\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_1}{\mathrm{a}}+\frac{\mathrm{q}_2}{\mathrm{~b}}+\frac{\mathrm{q}_3}{\mathrm{c}}\right) $

For the point B which is outside shell A , on the surface of shell B , and inside shell C . So, the potential at point B,

• 1. Due to A (Outside, distance b$) \Rightarrow \mathrm{V}_{\mathrm{BA}}=\frac{\mathrm{k} \mathrm{q}_1}{\mathrm{~b}}$

• 2. Due to B (Surface, $\mathrm{r}=\mathrm{b}) \Rightarrow \mathrm{V}_{\mathrm{BB}}=\frac{\mathrm{kq}_2}{\mathrm{~b}}$

• 3. Due to C (Inside, $\mathrm{r}<\mathrm{c}) \Rightarrow \mathrm{V}_{\mathrm{BC}}=\frac{\mathrm{kq}_3}{\mathrm{c}}$

Hence, total potential at point B is:

$ V_B=\frac{1}{4 \pi \epsilon_0}\left(\frac{q_1}{b}+\frac{q_2}{b}+\frac{q_3}{c}\right) $

$\Rightarrow $ $ V_B=\frac{1}{4 \pi \epsilon_0}\left(\frac{q_1+q_2}{b}+\frac{q_3}{c}\right) $

Point C is outside shell A , outside shell B , and on the surface of shell C . So, potential at point C is,

• 1. Due to A (Outside, distance c$) \Rightarrow \mathrm{V}_{\mathrm{CA}}=\frac{\mathrm{kq}_1}{\mathrm{c}}$

• 2. Due to B (Outside, distance c$) \Rightarrow \mathrm{V}_{\mathrm{CB}}=\frac{\mathrm{kq}_2}{\mathrm{c}}$

• 3. Due to C (Surface, $\mathrm{r}=\mathrm{c}) \Rightarrow \mathrm{V}_{\mathrm{CC}}=\frac{\mathrm{kq}_3}{\mathrm{c}}$

Hence, total potential at point C is :

$ V_C=\frac{1}{4 \pi \epsilon_0}\left(\frac{q_1}{c}+\frac{q_2}{c}+\frac{q_3}{c}\right) $

$\Rightarrow $ $V_C=\frac{1}{4 \pi \epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)$

So, $\mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_1}{\mathrm{a}}+\frac{\mathrm{q}_2}{\mathrm{~b}}+\frac{\mathrm{q}_3}{\mathrm{c}}\right), \mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_1+\mathrm{q}_2}{\mathrm{~b}}+\frac{\mathrm{q}_3}{\mathrm{c}}\right), \mathrm{V}_{\mathrm{C}}=\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_1+\mathrm{q}_2+\mathrm{q}_3}{\mathrm{c}}\right)$.

Hence, the correct option is (D).

For a system of discrete point charges $q_i$ located at positions $\vec{r}_i$, the electric dipole moment $\vec{P}$ with respect to the origin is defined as the vector sum of the product of each charge and its position vector.

$\overrightarrow{\mathrm{P}}=\sum_{\mathrm{i}} \mathrm{q}_{\mathrm{i}} \overrightarrow{\mathrm{r}}_{\mathrm{i}} $

$\Rightarrow $ $\overrightarrow{\mathrm{P}}=\mathrm{q}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{q}_2 \overrightarrow{\mathrm{r}}_2+\mathrm{q}_3 \overrightarrow{\mathrm{r}}_3+\cdots$

For charge $1\left(\mathrm{q}_1\right)$ :

$\mathrm{q}_1=+2 \mathrm{q}$ and it is at coordinates $(0,-3 \mathrm{a})$

So, position vector with respect to origin is, $=\overrightarrow{\mathrm{r}}_1=0 \hat{1}-3 a \hat{\jmath}=-3 a \hat{\jmath}$

So, electric dipole moment is, $\vec{p}_1=2 q(-3 a \hat{\jmath})=-6 a q \hat{\jmath}$

For charge $2\left(\mathrm{q}_2\right)$ :

$\mathrm{q}_2=+3 \mathrm{q}$ and it is at coordinates $(2 \mathrm{a}, 0)$

So, position vector with respect to origin is, $=\overrightarrow{\mathrm{r}}_2=2 a \hat{\mathrm{i}}-0 \hat{\mathrm{j}}=2 a \hat{\mathrm{i}}$

So, electric dipole moment is, $\vec{p}_2=3 q(2 a \hat{\mathrm{i}})=6 a q \hat{\mathrm{i}}$

For charge $3\left(\mathrm{q}_3\right)$ :

$\mathrm{q}_2=-4 \mathrm{q}$ and it is at coordinates $(-2 \mathrm{a}, 0)$

So, position vector with respect to origin is, $=\overrightarrow{\mathrm{r}}_3=-2 \mathrm{a} \hat{\mathrm{i}}-0 \hat{\mathrm{j}}=-2 \mathrm{a} \hat{\mathrm{i}}$

So, electric dipole moment is, $\vec{p}_3=-4 q(-2 a \hat{i})=8 a q \hat{i}$

So, the net electric dipole moment is the sum of the individual moments :

$ \vec{P}=\vec{P}_1+\vec{P}_2+\vec{P}_3 $

$\Rightarrow $ $\overrightarrow{\mathrm{P}}=(-6 q a \hat{\jmath})+(6 q a\hat{\imath})+(8 q a \hat{\imath})$

$\Rightarrow $ $\overrightarrow{\mathrm{P}}=14 q a \hat{\imath}-6 q a \hat{\jmath}$

$\Rightarrow $ $ \overrightarrow{\mathrm{P}}=2 \mathrm{qa}(7 \hat{\imath}-3 \hat{\jmath}) $

Hence, the correct option is (A).

The dipole moment is given by $\mathrm{p}=\mathrm{q} \times 2 \mathrm{l}$.

For Dipole A : $\mathrm{q}_{\mathrm{A}}=2 \times 10^{-6} \mathrm{C}, 2 \mathrm{l}=1 \mathrm{~cm}=0.01 \mathrm{~m}$.

$ \mathrm{p}_{\mathrm{A}}=\left(2 \times 10^{-6}\right)(0.01)=2 \times 10^{-8} \mathrm{C} \cdot \mathrm{~m} $

For Dipole B : $\mathrm{q}_{\mathrm{B}}=4 \times 10^{-6} \mathrm{C}, 2 \mathrm{l}=1 \mathrm{~cm}=0.01 \mathrm{~m}$.

$ p_B=\left(4 \times 10^{-6}\right)(0.01)=4 \times 10^{-8} \mathrm{C} \cdot \mathrm{~m} $

The distance between the centres is 80 cm .

Point P is equi-distant from both, so it is at the midpoint : $\mathrm{r}=40 \mathrm{~cm}=0.4 \mathrm{~m}$.

Point P is on the axial line of dipole A and on the equatorial line of dipole B .

Field due to A (Axial) :

$ \mathrm{E}_{\mathrm{A}}=\frac{1}{4 \pi \epsilon_0} \frac{2 \mathrm{p}_{\mathrm{A}}}{\mathrm{r}^3}=\frac{\left(9 \times 10^9\right) 2\left(2 \times 10^{-8}\right)}{(0.4)^3}=\frac{360}{0.064}=5625 \mathrm{~N} / \mathrm{C} $

It is along the axis of dipole A (horizontal)

Field due to B (Equatorial) :

$ \mathrm{E}_{\mathrm{B}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{r}^3}=\frac{\left(9 \times 10^9\right) 4 \times 10^{-8}}{(0.4)^3}=\frac{360}{0.064}=5625 \mathrm{~N} / \mathrm{C} $

It acts opposite to the direction of dipole B (vertical).

Since the vectors are perpendicular and equal in magnitude $\left(\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{B}}=\mathrm{E}\right)$ :

$ \mathrm{E}_{\mathrm{net}}=\sqrt{\mathrm{E}_{\mathrm{A}}^2+\mathrm{E}_{\mathrm{B}}^2}=\mathrm{E} \sqrt{2} $

$ \mathrm{E}_{\mathrm{net}}=5625 \sqrt{2} \frac{\mathrm{~N}}{\mathrm{C}}=0.5625 \sqrt{2} \times 10^4=\frac{9}{16} \sqrt{2} \times 10^4 \frac{\mathrm{~N}}{\mathrm{C}} $

So, net electric field at point P is, $\mathrm{E}_{\text {net }}=\frac{9}{16} \sqrt{2} \times 10^4 \mathrm{~N} / \mathrm{C}$.

Hence, the correct option is (B).

The total electrostatic energy $(\mathrm{U})$ of this system consists of two parts :

1. The energy of each individual charge due to the external electric field.

2. The interaction energy between the two charges.

The external electric field is given as $E=\frac{\mathrm{A}}{r^2}$.

The relation between electric field and potential ( V ) is:

$ V(r)=-\int E \cdot d r=-\int \frac{A}{r^2} d r=\frac{A}{r} $

The energy of a charge $q$ at distance $r$ is $U=q V(r)=\frac{q A}{r}$.

For $\mathrm{q}_1=7 \mu \mathrm{C}=7 \times 10^{-6} \mathrm{C}$ :

Position $\mathrm{r}_1=9 \mathrm{~cm}=0.09 \mathrm{~m}$.

$ U_1=\frac{q_1 A}{r_1}=\frac{\left(7 \times 10^{-6}\right)\left(9 \times 10^5\right)}{0.09}=\frac{6.3}{0.09}=70 \mathrm{~J} $

For $\mathrm{q}_2=-2 \mu \mathrm{C}=-2 \times 10^{-6} \mathrm{C}$ :

Position $\mathrm{r}_2=9 \mathrm{~cm}=0.09 \mathrm{~m}$.

$ U_2=\frac{q_2 A}{r_2}=\frac{\left(-2 \times 10^{-6}\right)\left(9 \times 10^5\right)}{0.09}=-\frac{1.8}{0.09}=-20 \mathrm{~J} $

The energy due to the interaction between the two point charges is given by :

$ \mathrm{U}_{\mathrm{int}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{~d}} $

The distance (d) between ( $-9,0,0$ ) and ( $9,0,0$ ) is $18 \mathrm{~cm}=0.18 \mathrm{~m}$.

Using $\mathrm{k}=9 \times 10^9 \frac{\mathrm{~N} \cdot \mathrm{~m}^2}{\mathrm{C}^2}$ :

$ U_{\text {int }}=\frac{\left(9 \times 10^9\right)\left(7 \times 10^{-6}\right)\left(-2 \times 10^{-6}\right)}{0.18} $

$\Rightarrow $ $U_{\mathrm{int}}=-\frac{0.126}{0.18}=-0.7 \mathrm{~J}$

So, total electrostatic energy $\left(\mathrm{U}_{\text {total }}\right)$,

$ \mathrm{U}_{\text {total }}=\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_{\mathrm{int}} $

$\Rightarrow $ $\mathrm{U}_{\text {total }}=70+(-20)+(-0.7)$

$\Rightarrow $ $\mathrm{U}_{\text {total }}=50-0.7=49.3 \mathrm{~J}$

According to Gauss's Law, the total electric flux $\Phi$ through a closed surface is

$ \Phi=\frac{Q_{\text {enclosed }}}{\epsilon_0} $

For the flux contribution from charge at vertex $\mathrm{A}(2 \mathrm{q})$

As the charge 2 q is placed at a corner (vertex) of the cube.

A corner of a cube is a meeting point for 8 identical cubes stacked together

Since the charge is shared equally among 8 cubes, the portion of the flux entering the specific cube is $\frac{1}{8}$ of the total flux produced by the charge.

$ \Phi_{\mathrm{A}}=\frac{1}{8} \times \frac{2 \mathrm{q}}{\epsilon_0}=\frac{2 \mathrm{q}}{8 \epsilon_0}=\frac{\mathrm{q}}{4 \epsilon_0} $

For the flux contribution from charge at the centre of face CDEF (q)

The charge q is placed at the center of the face CDEF and a face of a cube is shared by exactly 2 identical cubes placed side-by-side.

Since the charge is shared equally between 2 cubes, the portion of the flux entering the specific cube is $\frac{1}{2}$ of the total flux produced by the charge.

$ \Phi_{\text {face }}=\frac{1}{2} \times \frac{q}{\epsilon_0}=\frac{q}{2 \epsilon_0} $

The total flux passing through the cube is the sum of the contributions from both charges.

$ \Phi_{\text {total }}=\Phi_{\mathrm{A}}+\Phi_{\text {face }} $

$\Rightarrow $ $\Phi_{\text {total }}=\frac{\mathrm{q}}{4 \epsilon_0}+\frac{\mathrm{q}}{2 \epsilon_0}$

$\Rightarrow $ $ \Phi_{\text {total }}=\frac{q+2 q}{4 \epsilon_0}=\frac{3 q}{4 \epsilon_0} $

Therefore, the total electric flux passing through the cube is $\frac{3 q}{4 \epsilon_0}$.

Hence, the correct option is (B).

Electric potential is a scalar quantity. According to the principle of superposition the total electric potential at a point is the algebraic sum of the individual potentials created by each charge.

The potential $\mathrm{V}_{\mathrm{i}}$ at the center (at distance r ) due to a single point charge +q at a vertex is :

$ \mathrm{V}_{\mathrm{i}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}} $

Since there are 5 identical positive charges situated at the exact same distance r from the center O , the total potential V is simply 5 times the potential of a single charge :

$ V=V_1+V_2+V_3+V_4+V_5 $

$\Rightarrow $ $\mathrm{V}=5\left(\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}}\right)$

$ \mathrm{V}=\frac{\mathrm{5q}}{4 \pi \epsilon_0 \mathrm{r}} $

Electric field is a vector quantity, which means both magnitude and direction must be considered.

The net electric field at the centre is the vector sum of the electric fields produced by each individual charge :

$ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2+\overrightarrow{\mathrm{E}}_3+\overrightarrow{\mathrm{E}}_4+\overrightarrow{\mathrm{E}}_5 $

$\Rightarrow $ $\overrightarrow{\mathrm{E}}_{\mathrm{net}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_1+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_2+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_3+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_4+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_5$

$\Rightarrow $ $\overrightarrow{\mathrm{E}}_1=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\left(\hat{r}_1+\hat{r}_2+\hat{r}_3+\hat{r}_4+\hat{r}_5\right)$

$\Rightarrow $ $ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\left(\frac{\overrightarrow{\mathrm{r}}_1}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_2}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_3}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_4}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_5}{\mathrm{r}}\right)=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^3}\left(\overrightarrow{\mathrm{r}}_1+\overrightarrow{\mathrm{r}}_2+\overrightarrow{\mathrm{r}}_3+\overrightarrow{\mathrm{r}}_4+\overrightarrow{\mathrm{r}}_5\right) $

Using closed polygon rule, $\vec{r}_1+\vec{r}_2+\vec{r}_3+\vec{r}_4+\vec{r}_5=\overrightarrow{0}$

Therefore, $\overrightarrow{\mathrm{E}}_{\text {net }}=0$

So, at the centre of regular polygon,

The electric potential is $V=\frac{5 q}{4 \pi \epsilon_0 r}$ and the electric field is $\vec{E}=0$.

Hence, the correct option is (B).

When small liquid bubbles (or drops) coalesce to form a larger one, two fundamental quantities are conserved: the total charge and the total volume.

Let the radius of each small initial bubble be r and the charge on each be q .

The electric potential ( $\mathrm{V}_1$ ) on the surface of one such small bubble is given by the formula for the potential of a charged sphere:

$ \mathrm{V}_1=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}} $

Three such identical bubbles coalesce to form one large bubble.

Volume of 3 small bubbles $=$ Volume of 1 large bubble

$ 3 \times\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi R^3 $

$\Rightarrow $ $ 3 r^3=R^3 \Rightarrow R=3^{1 / 3} r $

Using conservation of charge, the total charge ( $Q$ ) on the new bigger bubble is simply the algebraic sum of the charges of the three initial bubbles.

$ \mathrm{Q}=\mathrm{q}+\mathrm{q}+\mathrm{q}=3 \mathrm{q} $

The electric potential ( $\mathrm{V}_2$ ) on the surface of the resultant bigger bubble is :

$ V_2=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R} $

$\Rightarrow V_2=\frac{1}{4 \pi \epsilon_0} \frac{3 q}{3^{\frac{1}{3}} r}=3^{\frac{2}{3}}\left(\frac{1}{4 \pi \epsilon_0} \frac{q}{r}\right)$

$ \Rightarrow \mathrm{V}_2=3^{\frac{2}{3}} \mathrm{~V}_1 $

So, the ratio of the potential on one initial bubble ( $\mathrm{V}_1$ ) to that on the resultant bigger bubble ( $\mathrm{V}_2$ ),

$ \frac{V_1}{V_2}=\frac{V_1}{3^{\frac{2}{3}} V_1}=\frac{1}{3^{\frac{2}{3}}} $

So, the ratio is $1: 3^{2 / 3}$. Hence, the correct option is (A).

Electric potential difference between two points is defined as the negative of the work done by the electric field in moving a unit positive charge from the initial point to the final point.

The infinitesimal work done $d W$ by an electric field $\vec{E}$ on a unit charge over a tiny displacement $d \vec{r}$ is $\vec{E} \cdot d \vec{r}$.

Therefore, the infinitesimal change in potential, dV, is :

$ \mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $

To find the total potential difference between an initial position $\overrightarrow{\mathrm{r}}_1$ and a final position $\overrightarrow{\mathrm{r}}_2$, we integrate both sides :

$ \int\limits_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{dV}=-\int\limits_{\overrightarrow{\mathrm{r}}_1}^{\overrightarrow{\mathrm{r}}_2} \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $

$\Rightarrow $ $ \mathrm{V}_2-\mathrm{V}_1=-\int\limits_{\overrightarrow{\mathrm{r}}_1}^{\overrightarrow{\mathrm{r}}_2} \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $

The electric field is given as: $\overrightarrow{\mathrm{E}}=\mathrm{Ax} \hat{\imath}+\mathrm{By} \hat{\jmath}$

The general position displacement vector is: $\mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{dxi}+\mathrm{dy} \hat{\mathrm{j}}+\mathrm{dz} \hat{\mathrm{k}}$

The initial point (Point 1) is the origin: $\left(x_1, y_1\right)=(0,0)$. Let the potential here be $V_0$.

The final point (Point 2) is: $\left(x_2, y_2\right)=(10,20)$. The potential here is $\mathrm{V}=500 \mathrm{~V}$.

$ V_2-V_1=-\int\limits_{(0,0)}^{(10,20)}(A x \hat{\imath}+B y \hat{\jmath}) \cdot(d x \hat{\imath}+d y \hat{\jmath}+d z \hat{k}) $

$\Rightarrow $ $\mathrm{V}_2-\mathrm{V}_1=-\int\limits_{(0,0)}^{(10,20)} \mathrm{Axdx}+\mathrm{Bydy}$

$\Rightarrow $ $V_2-V_1=-\left[\int\limits_0^{10} A x d x+\int\limits_0^{20} B y d y\right]$

$\Rightarrow $ $\mathrm{V}_2-\mathrm{V}_1=-\left[\left[\mathrm{A}\left(\frac{\mathrm{x}^2}{2}\right)\right]_0^{10}+\left[\mathrm{B}\left(\frac{\mathrm{y}^2}{2}\right)\right]_0^{20}\right]$

$\Rightarrow $ $\mathrm{V}_2-\mathrm{V}_1=-\left[\mathrm{A}\left(\frac{10^2}{2}-\frac{0^2}{2}\right)+\mathrm{B}\left(\frac{20^2}{2}-\frac{0^2}{2}\right)\right]$

$\Rightarrow $ $\mathrm{V}_2-\mathrm{V}_1=-\left[\mathrm{A}\left(\frac{100}{2}\right)+\mathrm{B}\left(\frac{400}{2}\right)\right]$

$\Rightarrow $$ V_2-V_1=-[50 A+200 B] $

We are given : $\mathrm{A}=10, \mathrm{~B}=5, \mathrm{~V}_2=500 \mathrm{~V}$

$ 500-V_0=-[50(10)+200(5)] $

$\Rightarrow $ $500-V_0=-[500+1000]$

$\Rightarrow $ $500-V_0=-1500$

$\Rightarrow $ $\mathrm{V}_0=500+1500$

$\Rightarrow $ $\mathrm{V}_0=2000 \mathrm{~V}$

When the bob is at its new equilibrium position, it is stationary. This means the net force acting on it is zero.

The free body diagram of bob is,

The gravitational force acting vertically downwards. Its magnitude is $\mathrm{F}_{\mathrm{g}}=\mathrm{mg}$.

Because the bob has a charge $q$ and is in a horizontal electric field $\vec{E}$, it experiences a horizontal force. Its magnitude is $\mathrm{F}_{\mathrm{e}}=\mathrm{qE}$. Let's assume it points to the left.

The force exerted by the string on the bob, acting along the length of the string towards the pivot point with magnitude T.

Let the string make an angle $\theta$ with the vertical when it reaches this new equilibrium position.

On resolving the forces into vertical and horizontal components.

Along vertical direction ( y - axis):

The upward component of the tension must balance the downward weight of the bob.

Therefore: $\mathrm{T} \cos \theta=\mathrm{mg} \ldots$ (i)

Along horizontal direction ( $\mathrm{x}-$ axis): The horizontal component of the tension must balance the outward pull of the electric force.

Therefore : $\mathrm{T} \sin \theta=\mathrm{qE} \ldots$ (ii)

Squaring and adding both the equations,

$ \mathrm{T}^2 \sin ^2 \theta+\mathrm{T}^2 \cos ^2 \theta=\mathrm{q}^2 \mathrm{E}^2+\mathrm{m}^2 \mathrm{~g}^2 $

$\Rightarrow $ $\mathrm{T}^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=\mathrm{q}^2 \mathrm{E}^2+\mathrm{m}^2 \mathrm{~g}^2$

$\Rightarrow $ $\mathrm{T}^2=\mathrm{q}^2 \mathrm{E}^2+\mathrm{m}^2 \mathrm{~g}^2$

$\Rightarrow $ $ \mathrm{T}=\sqrt{\mathrm{m}^2 \mathrm{~g}^2+\mathrm{q}^2 \mathrm{E}^2} $

Therefore, the magnitude of the tension is $T=\sqrt{m^2 g^2+q^2 E^2}$.

Hence, the correct option is (B).

The electric field $(\overrightarrow{\mathrm{E}})$ created by a point charge q at a distance r is given by Coulomb's Law:

$ |\overrightarrow{\mathrm{E}}|=\frac{1}{4 \pi \epsilon_0} \frac{|\mathrm{q}|}{\mathrm{r}^2} $

If the charge is positive $(+q)$, the electric field points away from the charge. If the charge is negative $(-q)$, the electric field points towards the charge.

When multiple charges are present, the net electric field at any point is simply the vector sum of the individual electric fields created by each charge.

Let the centre of the circle be the origin $(0,0)$. All six charges are at a distance R from the center.

Let the magnitude of the electric field produced by a single charge of magnitude $Q$ at the center as $E_0$ :

$ \mathrm{E}_0=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{R}^2} $

The electric field due to charge +Q charge at $90^{\circ}$ is downward with magnitude $\mathrm{E}_0$ and that for the +Q at $270^{\circ}$ it is upward with magnitude $\mathrm{E}_0$. These two vectors are equal and opposite, so they cancel each other out.

The electric field due to charge +Q charge at $30^{\circ}$ is with magnitude $\mathrm{E}_0$ and opposite to that for the +Q at $210^{\circ}$ with magnitude $\mathrm{E}_0$. These two vectors are equal and opposite, so they cancel each other out.

$ \left|\overrightarrow{\mathrm{E}}_{n e t}\right|=\mathrm{E}_0+\mathrm{E}_0=2 \mathrm{E}_0 $

The direction of this net vector is $150^{\circ}$ or $30^{\circ}$ with negative x -axis.

So, the net electric field in vector form is,

$\overrightarrow{\mathrm{E}}_{\mathrm{net}}=\left(2 \mathrm{E}_0\right)\left(\cos 150^{\circ} \hat{\imath}+\sin 150^{\circ} \hat{\jmath}\right)$

$\Rightarrow $ $\overrightarrow{\mathrm{E}}_{\mathrm{net}}=\left(2 \mathrm{E}_0\right)\left(-\frac{\sqrt{3}}{2} \hat{\imath}+\frac{1}{2} \hat{\jmath}\right)$

$\Rightarrow $ $ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=-\mathrm{E}_0(\sqrt{3} \hat{\imath}-\hat{\jmath}) $

Putting the expression for $\mathrm{E}_0$,

$ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=-\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{R}^2}(\sqrt{3} \hat{\imath}-\hat{\jmath}) $

Therefore, the net electric field at the centre of circle is $-\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2}(\sqrt{3} \hat{\imath}-\hat{\jmath})$.

Hence, the correct option is (C).

At the start, the spheres are moving towards each other.

Since there are two spheres of mass $m$ moving with speed $u$, so the initial kinetic energy is,

$ \mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{mu}^2+\frac{1}{2} \mathrm{mu}^2=\mathrm{mu}^2 $

Initial electrostatic potential energy is,

$ \mathrm{U}_{\mathrm{ei}}=\frac{\mathrm{kQ}^2}{4 \mathrm{R}} $

Initial gravitational potential energy is :

$ U_{g i}=-\frac{G \mathrm{~m}^2}{4 \mathrm{R}} $

When the spheres just touch, their centres are separated by a distance equal to the sum of their radii $(R+R=2 R)$. To find the minimum $u$, we assume they come to a momentary rest at this point.

Total final kinetic energy, $\mathrm{K}_{\mathrm{f}}=0$

Final electrostatic potential energy is,

$ \mathrm{U}_{\mathrm{ef}}=\frac{\mathrm{k} \mathrm{Q}^2}{2 \mathrm{R}} $

Final gravitational potential energy is :

$ \mathrm{U}_{\mathrm{gf}}=-\frac{\mathrm{Gm}^2}{2 \mathrm{R}} $

The total energy remains constant :

$ K_i+U_{e i}+U_{g i}=K_f+U_{e f}+U_{g f} $

$\Rightarrow $ $\mathrm{mu}^2+\frac{\mathrm{kQ}^2}{4 \mathrm{R}}-\frac{\mathrm{Gm}^2}{4 \mathrm{R}}=0+\frac{\mathrm{kQ}^2}{2 \mathrm{R}}-\frac{\mathrm{Gm}^2}{2 \mathrm{R}}$

$\Rightarrow $ $\mathrm{mu}^2=\left(\frac{\mathrm{kQ}^2}{2 \mathrm{R}}-\frac{\mathrm{kQ}^2}{4 \mathrm{R}}\right)-\left(\frac{\mathrm{Gm}^2}{2 \mathrm{R}}-\frac{\mathrm{Gm}^2}{4 \mathrm{R}}\right)$

$\Rightarrow $ $\mathrm{mu}^2=\frac{\mathrm{kQ}^2}{4 \mathrm{R}}-\frac{\mathrm{Gm}^2}{4 \mathrm{R}}$

$\Rightarrow $ $\mathrm{mu}^2=\frac{\mathrm{kQ}^2}{4 \mathrm{R}}\left(1-\frac{\mathrm{Gm}^2}{\mathrm{kQ}^2}\right)$

$\Rightarrow $ $\mathrm{u}^2=\frac{\mathrm{kQ}^2}{4 \mathrm{mR}}\left(1-\frac{\mathrm{Gm}^2}{\mathrm{kQ}^2}\right)$

$\Rightarrow $ $u=\sqrt{\frac{k Q^2}{4 m R}\left(1-\frac{G m^2}{k Q^2}\right)}$

The work done ( W ) by an external agent in moving a charge q from point A to point B in the field of a source charge $Q$ is equal to the change in electrostatic potential energy :

$ \mathrm{W}=\mathrm{U}_{\mathrm{B}}-\mathrm{U}_{\mathrm{A}}=\mathrm{q}\left(\mathrm{~V}_{\mathrm{B}} -\mathrm{V}_{\mathrm{A}} \right) $

Where the electrostatic potential V at a distance r from a source charge Q is :

$ V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r} $

The source charge at origin is, $\mathrm{Q}=10^{-8} \mathrm{C}$

The test charge is $\mathrm{q}=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}$

The distance $r$ of a point $(x, y, z)$ from the origin is $r=\sqrt{x^2+y^2+z^2}$

For Point A (4, 4, 2):

$ r_A=\sqrt{4^2+4^2+2^2}=\sqrt{16+16+4}=\sqrt{36}=6 \mathrm{~m} $

For Point B (2, 2, 1):

$ r_B=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3 \mathrm{~m} $

So, the potential at $\mathrm{A}\left(\mathrm{V}_{\mathrm{A}}\right)$ is,

$ V_A=\frac{k Q}{r_A}=\frac{\left(9 \times 10^9\right) 10^{-8}}{6}=\frac{90}{6}=15 V $

And the potential at $\mathrm{B}\left(\mathrm{V}_{\mathrm{B}}\right)$ is,

$ V_B=\frac{k Q}{r_B}=\frac{\left(9 \times 10^9\right) 10^{-8}}{3}=\frac{90}{3}=30 V $

So, the work done by external agent is,

$ \mathrm{W}=\mathrm{q}\left(\mathrm{~V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right) $

$\Rightarrow $ $\mathrm{W}=\left(2 \times 10^{-6}\right)(30-15)$

$\Rightarrow $ $ \mathrm{W}=30 \times 10^{-6} \mathrm{~J} $

Therefore, the work done to be $30 \times 10^{-6} \mathrm{~J}$.

Hence, the correct option is (A).

When two conducting spheres are in contact, they achieve the same electric potential. Consider two metal spheres with radii $ R $ and $ 3R $ that have the same surface charge density $\sigma$.

For a conductive sphere, the potential $ V $ of a sphere can be expressed as $ V = \frac{\sigma r}{\varepsilon_0} $, where $ r $ is the radius of the sphere and $\varepsilon_0$ is the permittivity of free space.

When the two spheres are brought into contact and then separated, they equalize their potentials:

$ V_1 = V_2 $

Thus, the equations become:

$ \sigma_1 r_1 = \sigma_2 r_2 $

Substituting for the radii:

$ \sigma_1 R = \sigma_2 (3R) $

From this, we can solve for the ratio of the surface charge densities:

$ \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3 $

Therefore, the ratio of the surface charge densities on the smaller sphere to the larger sphere after contact is 3.

$\textbf{Assertion A:}$

For a uniformly charged spherical shell, Gauss's law tells us that the electric field inside the shell is zero.

Since the electric field $\vec{E}$ is zero everywhere inside, the potential $V$ must be constant.

Work done in moving a charge from one point to another in an electric field is given by the difference in potential energy, which is $q(V_B - V_A)$.

If the potential difference $V_B - V_A$ is zero (because the potential is constant), then the work done is zero, regardless of the path taken.

Therefore, Assertion A is true.

$\textbf{Reason R:}$

It states that the electrostatic potential inside a uniformly charged spherical shell is constant and equal to that on its surface.

This is correct because the electric field inside is zero, ensuring that the potential remains uniform inside.

Thus, Reason R is also true.

$\textbf{Relationship between A and R:}$

The work done on a test charge moving inside the shell being zero is a direct consequence of the fact that the potential is constant inside.

Hence, Reason R correctly explains why the work done (Assertion A) is zero.

Given this analysis, the correct answer is:

Option B

"Both A and R are true and R is the correct explanation of A."

Therefore, the relationship between charge density and ROC at different points can be expressed as:

$ \sigma \propto \frac{1}{\mathrm{ROC}} $

From this, the relationship between the charge densities at the specified points becomes clear:

$ (\mathrm{ROC})_1 < (\mathrm{ROC})_3 < (\mathrm{ROC})_2 = (\mathrm{ROC})_4 $

This implies:

$ \sigma_1 > \sigma_3 > \sigma_2 = \sigma_4 $

Thus, the charge density is greatest at the point with the smallest ROC and decreases as the ROC increases, keeping in mind that $\sigma_2$ and $\sigma_4$ have equal ROC, hence equal charge densities.

To find the net flux through a Gaussian cube when an infinitely long wire with a uniform linear charge density $ \lambda = 2 \text{ nC/m} $ passes through the cube's diagonal corners, we use Gauss's Law. The relevant expression is:

$ \phi = \frac{q_{\text{enc}}}{\varepsilon_0} $

Here, $ q_{\text{enc}} $ is the charge enclosed by the Gaussian surface, and $ \varepsilon_0 $ is the permittivity of free space.

Given:

The wire passes through two opposite corners of the cube.

The side length of the cube, $ a = \sqrt{3} \text{ cm} = \sqrt{3} \times 10^{-2} \text{ m} $.

The length of the wire inside the cube is equal to its diagonal, which is $ \sqrt{3} \times a $.

Thus, the charge enclosed $ q_{\text{enc}} $ is:

$ q_{\text{enc}} = \lambda \times \text{(length of wire inside the cube)} = \lambda \times \sqrt{3} \times a $

Substitute the given values:

$ \lambda = 2 \times 10^{-9} \text{ C/m}, \quad a = \sqrt{3} \times 10^{-2} \text{ m} $

Calculating:

$ q_{\text{enc}} = 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} $

Now apply Gauss's Law for net flux:

$ \phi = \frac{\lambda \cdot \sqrt{3} \cdot a}{\varepsilon_0} $

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

$ \phi = 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} \times 36 \pi \times 10^9 $

$ \phi = 2.16 \pi \text{ Nm}^2\text{C}^{-1} $

Thus, the net flux through the Gaussian cube is $ 2.16 \pi \text{ Nm}^2\text{C}^{-1} $.

Given:

Potential difference $ V = 5 \, \text{V} $

Separation between the plates $ d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} $

The electric field $ E $ is calculated as:

$ E = \frac{V}{d} = \frac{5}{0.5 \times 10^{-3}} = 10^4 \, \text{V/m} $

The torque $ \tau $ experienced by the dipole in the electric field is given by:

$ \tau = PE \sin \theta $

Where:

$ P $ is the dipole moment,

$ E $ is the electric field,

$ \theta = 30^\circ $ is the angle of rotation.

The dipole moment $ P $ is calculated by:

$ P = q \times a = (2 \times 10^{-6} \, \text{C}) \times (0.5 \times 10^{-6} \, \text{m}) = 1 \times 10^{-12} \, \text{C} \cdot \text{m} $

Substituting the values into the torque equation:

$ \tau = (1 \times 10^{-12} \, \text{C} \cdot \text{m}) \times 10^4 \, \text{V/m} \times \sin 30^\circ $

Since $\sin 30^\circ = 0.5$, the torque simplifies to:

$ \tau = 1 \times 10^{-12} \times 10^4 \times 0.5 = 5 \times 10^{-9} \, \text{N} \cdot \text{m} $

Thus, the torque is $ 5 \times 10^{-9} \, \text{N} \cdot \text{m} $.

The assertion suggests that the outer body of an aircraft is made of metal to protect passengers from lightning strikes. The reason provided is that the electric field inside a cavity enclosed by a conductor is zero.

This reasoning is based on the property of metallic conductors, where the electric field within a conductor's cavity is indeed zero. This phenomenon occurs because the charges on a conductor redistribute themselves to cancel any external electric fields within the conductor. Thus, if lightning were to strike an aircraft, the metal body would direct the electric charges along its surface, effectively shielding the interior where passengers are located.

Therefore, the assertion and the reason are both correct, and the reason correctly explains why the metal body of an aircraft shields passengers from lightning. This aligns with the principles of electromagnetic shielding. The key point is that the conductive nature of the aircraft’s metal body ensures no electric field permeates the interior, keeping passengers safe.

We can use Gauss’s law to analyze the dimensions. Gauss’s law states that

$ \Phi_E = \frac{q}{\epsilon_0}, $

where

$\Phi_E$ is the electric flux,

$q$ is the electric charge,

$\epsilon_0$ is the permittivity of free space.

To find the dimensions of $\epsilon_0 \frac{d\Phi_E}{dt}$, follow these steps:

Differentiate Gauss’s law with respect to time:

$ \frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0}\frac{dq}{dt}. $

Multiply the equation by $\epsilon_0$:

$ \epsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}. $

Recognize that the time rate of change of charge, $\frac{dq}{dt}$, is by definition the electric current.

Thus, the dimensions of $\epsilon_0 \frac{d\Phi_E}{dt}$ are those of electric current.

The correct answer is: Option C, electric current.

To determine the value of $ K $ for the given electrical potential scenario, we start by calculating the potential at points C and D.

Potential at Point C:

The potential at point C, $ V_C $, is given by the sum of the potentials due to charges $ q_1 $ and $ q_2 $:

$ V_C = \frac{kq_1}{0.4} + \frac{kq_2}{0.5} $

Potential at Point D:

Similarly, the potential at point D, $ V_D $, is:

$ V_D = \frac{kq_1}{0.4} + \frac{kq_2}{0.1} $

Change in Potential Energy:

The change in potential energy $ \Delta U $ when $ q_3 $ moves from C to D is:

$ \Delta U = (V_D - V_C) \times q_3 $

Substituting the expressions for $ V_D $ and $ V_C $, we get:

$ \Delta U = \left(\frac{kq_2}{0.1} - \frac{kq_2}{0.5}\right) \times q_3 $

Simplifying the Expression:

By simplifying the expression, we find:

$ \Delta U = \left(\frac{5kq_2}{0.5}\right) \times q_3 - \left(\frac{kq_2}{0.5}\right) \times q_3 $

$ \Delta U = \frac{4kq_2}{0.5} \times q_3 = 8kq_2 q_3 $

$ \Delta U = \frac{8q_2 q_3}{4\pi \varepsilon_0} $

Thus, the value of $ K $ is $ 8q_2 $.

In this scenario, two infinite identical charged sheets and a charged spherical body with charge density ' $\rho$ ' are positioned as shown in the figure. We aim to determine the relationship between the electrical fields at points A, B, C, and D.

When considering the electric fields:

Electric Field Due to Charged Sheets:

The infinite charged sheets produce a uniform electric field. The electric field due to a single infinite sheet of charge is constant on either side of the sheet and is directed perpendicularly away from the sheet.

Electric Field Due to a Spherical Charge:

Outside the charged sphere, the electric field behaves as if the entire charge were concentrated at the center of the sphere. The electric field decreases with the square of the distance from the center, following the inverse square law.

Given these principles:

At Points C and D: These are located in different regions regarding the charged sheets and spherical body. Due to the inverse square nature of the electric field from the spherical charge and the uniform field of the sheets, the field strength will vary, resulting in $\vec{E}_C \neq \vec{E}_D$.

At Points A and B: Both points are influenced by the uniform field of the charged sheets. However, if the distance or position concerning the sphere also affects these points, this could lead to differences in field strength, thus $\vec{E}_A > \vec{E}_B$.

This analysis leads us to conclude that the electric field relations are $\mathrm{E}_{\mathrm{C}} \neq \mathrm{E}_{\mathrm{D}}$ and $\mathrm{E}_{\mathrm{A}} > \mathrm{E}_{\mathrm{B}}$.

$\begin{aligned} & \mathrm{E}=\frac{\sigma}{2 \varepsilon_0} \\ & \tau=\mathrm{PE} \sin \theta \\ & =\left[\left(2 \times 10^{-6}\right)\left(\frac{2}{10}\right)\right]\left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right]\left(\frac{1}{2}\right) \\ & =\frac{10}{8.85}=1.12 \mathrm{Nm} \end{aligned}$

The potential inside a uniformly charged spherical shell is equal to the potential on its surface. This means:

$ V_{\text{in}} = V_{\text{surface}} = \frac{kQ}{R} = 120 \, \text{V} $

This maintains the potential at both the center of the shell and any point inside it up to the surface. Therefore, the potential at the center of the shell and at any point inside the shell (like at a distance $ r = 5 \, \text{cm} $) is also $ 120 \, \text{V} $.

For a point outside the shell, at a distance $ r = 15 \, \text{cm} $, the potential is given by the formula:

$ V = \frac{kQ}{r} = \frac{120 \times 10}{15} = 80 \, \text{V} $

Assertion (A): The net dipole moment of a polar linear isotropic dielectric substance is not zero, even when there is no external electric field.

Reason (R): In the absence of an external electric field, the different permanent dipoles in a polar dielectric substance are oriented in random directions.

Explanation:

Assertion (A): The statement is incorrect. In a polar linear isotropic dielectric substance, the net dipole moment is zero when an external electric field is absent. This is because the dipoles within the material are randomly oriented and cancel each other out.

Reason (R): The statement is correct. In the absence of an external electric field, the dipoles in a polar dielectric substance remain inherently polarized but are oriented randomly. This randomness in orientation leads to a cancellation of the net dipole moment.

Therefore, while the reason is correct, it does not support the validity of the assertion as stated.

Step 1: Formula for Potential Difference
The potential difference ($\Delta V$) between two points in a uniform electric field is found by multiplying the electric field ($E$) by the distance ($\Delta d$) between the points: $\Delta V = E \times \Delta d$

Step 2: Find the Electric Field
The total potential difference between the two plates is $V$, and they are 10 cm apart. So, the electric field between the plates is: $E = \frac{V}{10}$ Here, distance is in centimeters.

Step 3: Calculate Potential Difference Between A and B
Points A and B are 4 cm apart. So, the potential difference between A and B will be: $V_{AB} = E \times 4 = \frac{V}{10} \times 4 = \frac{4V}{10} = \frac{2V}{5}$

To determine the position along the positive $ z $-axis where the electric field due to a uniformly charged circular loop is at its maximum, we follow these steps:

Expression for Electric Field ($ E $):

The electric field $ E $ at a point along the $ z $-axis for a charged circular loop can be expressed as:

$ E = \frac{KQr}{(x^2 + R^2)^{3/2}} $

Here, $ K $ is the Coulomb's constant, $ Q $ is the total charge, $ r $ is a constant involving charge distribution, $ x $ is the distance along the $ z $-axis, and $ R $ is the radius of the loop.

Maximizing the Electric Field:

To find where $ E $ is maximum, we take the derivative of $ E $ with respect to $ x $ and set it to zero:

$ \frac{dE}{dx} = 0 $

Solve for $ x $:

Solving the equation from the derivative, we find:

$ x = \frac{R}{\sqrt{2}} $

Substitute Given Radius:

Given the radius $ R = a\sqrt{2} $, substituting into the expression for $ x $:

$ x = \frac{\sqrt{2}a}{\sqrt{2}} = a $

Thus, the position along the positive $ z $-axis where the electric field is maximum is at $ x = a $.

Final charge distribution will be

$\therefore \mathrm{F}_{\mathrm{net}}=\frac{3 \sigma}{2 \epsilon_0} \mathrm{q}$

Let $E_p=0$

$\begin{aligned} & \therefore \frac{k q}{x^2}=\frac{k 9 q}{(d-x)^2} \\ & \Rightarrow \frac{d-x}{x}=3 \Rightarrow x=\frac{d}{4} \end{aligned}$

$\therefore$ co-ordinate of P is $\left(\frac{\mathrm{d}}{4}, 0,0\right)$

Step 1: Balancing Forces on the Bob

At equilibrium, the electric force ($qE$) pulling the bob toward the sheet is balanced by the weight of the bob ($mg$) pulling it down.

Step 2: Formula for Electric Field Near Sheet

The electric field ($E$) created by a large sheet of charge is given by $E = \dfrac{\sigma}{2\varepsilon_0}$.

Step 3: Replace $E$ in the Force Equation

The equation from earlier becomes: $q \left[\frac{\sigma}{2 \varepsilon_0}\right] = mg$

Step 4: Solve for $\sigma$

We solve for $\sigma$ by rearranging: $\sigma = \frac{2\varepsilon_0 mg}{q}$

Step 5: Substitute the Values

Put in the given values: $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m, $m = 100$ mg $= 100 \times 10^{-6}$ kg, $g = 10$ m/s$^2$, $q = 10 \times 10^{-6}$ C.

So, $\sigma = \frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}}$

Step 6: Calculate

We do the multiplication and division: $\sigma = 17.7 \times 10^{-10}~\mathrm{C}/\mathrm{m}^2$

Step 7: Convert to nC/m$^2$

$1~\mathrm{nC} = 10^{-9}~\mathrm{C}$, so $\sigma = 1.77~\mathrm{nC}/\mathrm{m}^2$

Given an electric flux $\phi = -2 \times 10^4 \, \text{Nm}^2\text{C}^{-1}$ through a spherical Gaussian surface with a radius of 8.0 cm (r = 0.08 m), we are to find the value of the point charge $Q$. According to Gauss's Law:

$ \phi = \frac{Q}{\varepsilon_0} $

Where $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$.

Rearranging the equation to solve for $Q$, we have:

$ Q = \phi \varepsilon_0 $

Substituting the given values:

$ Q = -2 \times 10^4 \times 8.85 \times 10^{-12} $

Calculating this yields:

$ Q = -17.7 \times 10^{-8} \, \text{C} $

Thus, the value of the point charge is $Q = -17.7 \times 10^{-8} \, \text{C}$.

We know, electric field due to uniformly charged infinite plane sheet,

$E = {{{\sigma _0}} \over {2{\varepsilon _0}}}$

perpendicularly directed away from the sheet for positive charge

So, $\overrightarrow E = {{{\sigma _0}} \over {2{\varepsilon _0}}} \uparrow \theta $

note : E is constant

Let length of dipole = d

so dipole moment, $\overrightarrow p = qd \uparrow $

We know,

Torque, $\overrightarrow \tau = \overrightarrow p \times \overrightarrow E $

$\overrightarrow \tau = qd\widehat i \times {{{\sigma _0}} \over {2{\varepsilon _0}}}\widehat i$

$\widehat \tau = O$ (as $\widehat i \times \widehat i = 0$)

force, on charge $ - q,\overrightarrow {{F_1}} = - qE\widehat i$

on charge $ + q,\overrightarrow {{F_2}} = qE\widehat i$

So net force on dipole, $\overrightarrow F = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} $

$ = - qE\widehat i + qE\widehat i$

$\overrightarrow F = O$

We know, potential energy of dipole

$u(O) = - pE\cos \theta $

here, $O = 0^\circ $ (given)

So, ${u_{\min }} = - pE$

Hence, option B is correct.

$\begin{aligned} & \mathrm{I} \omega 2 \theta=\mathrm{q} \ell \mathrm{E}_0 \theta \\ & 2 \mathrm{~m}\left(\frac{\ell}{2}\right)^2 \omega^2=\mathrm{q} \ell \mathrm{E}_0 \\ & \omega^2=\frac{2 \mathrm{qE}_0}{\mathrm{~m} \ell} \\ & \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{~m} \ell}{2 \mathrm{qE}_0}} \end{aligned}$

(A) $\rightarrow 0$ (III)

(B) $\rightarrow \frac{\sigma}{2 \varepsilon_0}$

(C) $\rightarrow \frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2}$ (No row matching)

$(\mathrm{D}) \rightarrow \frac{\sigma}{\varepsilon_0}(\mathrm{I})$

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}}}$

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

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: $\frac{a}{b}$.