Electric potential Questions in English

(A) The work done in bringing a charge $q_0$ from infinity to a point is given by $W = q_0 V$,where $V$ is the electric potential at that point due to the existing charges.
In a regular hexagon of side $d$,the distance of each corner from the centre is $d$.
The electric potential $V$ at the centre due to the six charges is the sum of the potentials due to each individual charge:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{+q}{d} + \frac{-q}{d} + \frac{+q}{d} + \frac{-q}{d} + \frac{+q}{d} + \frac{-q}{d} \right)$
$V = \frac{1}{4 \pi \varepsilon_0 d} (q - q + q - q + q - q)$
$V = 0$
Therefore,the work done $W = q_0 \times 0 = 0$.

(A) For a uniformly charged thin spherical shell of radius $R$ and charge $Q$:
$1$. Inside the shell $(r < R)$,the electric field is zero,which implies that the electric potential is constant and equal to the potential at the surface,i.e.,$V_{\text{inside}} = \frac{kQ}{R}$.
$2$. Outside the shell $(r \ge R)$,the shell behaves like a point charge placed at its center,so the potential follows the inverse relation $V_{\text{outside}} = \frac{kQ}{r}$,where $r$ is the distance from the center.
$3$. Thus,the graph of $V$ versus $r$ is a horizontal line for $r \le R$ and a rectangular hyperbola for $r > R$.

(C) The formula for electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.
Given: $V = 50 \; V$,$Q = 5 \times 10^{-9} \; C$,and $k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \; N m^2 C^{-2}$.
Substituting the values into the formula: $50 = \frac{9 \times 10^9 \times 5 \times 10^{-9}}{r}$.
Simplifying the numerator: $50 = \frac{45}{r}$.
Solving for $r$: $r = \frac{45}{50} = 0.9 \; m$.
Converting meters to centimeters: $r = 0.9 \times 100 \; cm = 90 \; cm$.

(D) The charges on the shells are:
$q_x = \sigma(4\pi a^2)$
$q_y = -\sigma(4\pi b^2)$
$q_z = \sigma(4\pi c^2)$
Since the potential of shell $X$ is equal to the potential of shell $Z$ $(V_x = V_z)$:
$V_x = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_x}{a} + \frac{q_y}{b} + \frac{q_z}{c} \right)$
$V_z = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_x}{c} + \frac{q_y}{c} + \frac{q_z}{c} \right)$
Equating $V_x$ and $V_z$:
$\frac{\sigma 4\pi a^2}{a} - \frac{\sigma 4\pi b^2}{b} + \frac{\sigma 4\pi c^2}{c} = \frac{\sigma 4\pi a^2}{c} - \frac{\sigma 4\pi b^2}{c} + \frac{\sigma 4\pi c^2}{c}$
Simplifying the equation:
$a - b + \frac{c^2}{c} = \frac{a^2 - b^2 + c^2}{c}$
$a - b + c = \frac{a^2 - b^2 + c^2}{c}$
$c(a - b + c) = a^2 - b^2 + c^2$
$c(a - b) + c^2 = a^2 - b^2 + c^2$
$c(a - b) = a^2 - b^2$
$c(a - b) = (a - b)(a + b)$
$c = a + b$
Given $a = 2\,cm$ and $b = 3\,cm$:
$c = 2 + 3 = 5\,cm$.

(D) Let $q$ be the charge and $r$ be the radius of each small drop.
The potential of each small drop is given by $V = \frac{Kq}{r} = 10 \, mV$.
When $n = 64$ drops combine to form a bigger drop of radius $R$ and charge $Q$,the volume remains conserved:
$n \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$64 r^3 = R^3 \implies R = 4r$.
The total charge on the bigger drop is $Q = nq = 64q$.
The potential of the bigger drop is $V' = \frac{KQ}{R} = \frac{K(64q)}{4r} = 16 \times (\frac{Kq}{r})$.
Substituting the value of $V$,we get $V' = 16 \times 10 \, mV = 160 \, mV$.

(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
From the figure,the distance of point $P$ from the charge $+q$ is $r_1 = 2 \ cm = 2 \times 10^{-2} \ m$.
The distance of point $P$ from the charge $-q$ is $r_2 = 3 \ cm + 3 \ cm + 2 \ cm = 8 \ cm = 8 \times 10^{-2} \ m$.
The potential at point $P$ is given by:
$V = V_{+q} + V_{-q} = \frac{Kq}{r_1} + \frac{K(-q)}{r_2}$
$V = Kq \left( \frac{1}{2 \times 10^{-2}} - \frac{1}{8 \times 10^{-2}} \right)$
$V = Kq \left( \frac{4 - 1}{8 \times 10^{-2}} \right) = Kq \left( \frac{3}{8 \times 10^{-2}} \right)$
$V = Kq \left( \frac{3}{8} \right) \times 10^2 \ V$.
Thus,the potential in units of $10^2 \ V$ is $\left( \frac{3}{8} \right) qK$.

(C) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{kQ}{r}$.
The distance of point $P(\sqrt{3}, \sqrt{3})$ from the origin $(0,0)$ is $r_1 = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = \sqrt{6} \text{ m}$.
The distance of point $Q(\sqrt{6}, 0)$ from the origin $(0,0)$ is $r_2 = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} \text{ m}$.
Since $r_1 = r_2 = \sqrt{6} \text{ m}$,the potential at point $P$ is $V_P = \frac{kQ}{r_1}$ and the potential at point $Q$ is $V_Q = \frac{kQ}{r_2}$.
Therefore,the potential difference $V_P - V_Q = \frac{kQ}{r_1} - \frac{kQ}{r_2} = 0 \text{ V}$.

(B) The potential $V$ at the centre of a charged ring is given by $V = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{R}$.
For a half ring of radius $R$,the total charge $Q = \lambda \times (\pi R)$.
Since every point on the half ring is at the same distance $R$ from the centre,the potential is $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$.
Substituting $Q = \lambda \pi R$,we get $V = \frac{1}{4\pi \epsilon_0} \frac{\lambda \pi R}{R} = \frac{\lambda}{4 \epsilon_0}$.
Given $\lambda = 4 \times 10^{-9} \ C/m$ and $k = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \ N \ m^2/C^2$.
$V = k \lambda \pi = (9 \times 10^9) \times (4 \times 10^{-9}) \times \pi$.
$V = 36 \pi \ V$.
Comparing this with $x \pi \ V$,we find $x = 36$.

(C) For a uniformly charged thin spherical shell of radius $R$ and charge $q$,the electric potential at any point inside the shell (including the center) is constant and equal to the potential at the surface.
The potential at any point $r \leq R$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.
Since point $C$ is at the center $(r = 0)$ and point $P$ is on the surface $(r = R)$,both points are at the same potential.
Therefore,$V_C = V_P = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.
The potential difference between points $C$ and $P$ is $V_C - V_P = 0$.

(A) For a charged conducting cylinder of radius $R$ and linear charge density $\lambda$,the electric field $E$ at a distance $r > R$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.
When a charge is given to the inner cylinder,an electric field exists in the region between the two cylinders,resulting in a potential difference $V = \int_{R_1}^{R_2} E \, dr$.
If a charge is given to the outer cylinder,the electric field inside the cavity of the outer cylinder is zero due to the properties of a hollow conductor (Gauss's Law). Thus,no potential difference exists between the cylinders.
If the same charge density $\lambda$ is given to both,the inner cylinder creates a field,but the outer cylinder does not contribute to the field in the region between them. Therefore,a potential difference will still exist due to the inner cylinder.
Thus,option $A$ is correct because charging the inner cylinder creates a radial electric field between the cylinders,leading to a potential difference.

(A) For a uniformly charged spherical shell,the electrostatic potential inside the shell is constant and equal to the potential on its surface.
Given: Potential on the surface $V_{surface} = 120 \ V$ and radius $R = 10 \ cm$.
$1$. At the centre $(r = 0 \ cm)$: Since $r < R$,the potential is equal to the surface potential,so $V_{centre} = 120 \ V$.
$2$. At a distance $r = 5 \ cm$ from the centre: Since $r < R$,the potential is constant throughout the interior,so $V_{r=5cm} = 120 \ V$.
$3$. At a distance $r = 15 \ cm$ from the centre: Since $r > R$,the shell acts as a point charge. The potential is given by $V = \frac{kQ}{r}$.
We know $V_{surface} = \frac{kQ}{R} = 120 \ V$,so $kQ = 120 \times 10 = 1200 \ V \cdot cm$.
Thus,$V_{r=15cm} = \frac{1200}{15} = 80 \ V$.
Therefore,the potentials are $120 \ V, 120 \ V, 80 \ V$.

(A) The potential at point $C$ due to charges $q_1$ and $q_2$ is:
$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AC} + \frac{q_2}{BC} \right)$
Given $AC = 40\ cm = 0.4\ m$. Since $AB = 30\ cm$ and $AC = 40\ cm$,by Pythagoras theorem,$BC = \sqrt{0.3^2 + 0.4^2} = 0.5\ m$.
So,$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right)$.
The potential at point $D$ due to charges $q_1$ and $q_2$ is:
$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AD} + \frac{q_2}{BD} \right)$
Given $AD = 40\ cm = 0.4\ m$ (radius of the circular path) and $BD = AD - AB = 40\ cm - 30\ cm = 10\ cm = 0.1\ m$.
So,$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right)$.
The change in potential energy $\Delta U$ is given by:
$\Delta U = q_3 (V_D - V_C) = q_3 \left[ \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right) - \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right) \right]$
$\Delta U = \frac{q_3}{4 \pi \epsilon_0} \left( \frac{q_2}{0.1} - \frac{q_2}{0.5} \right) = \frac{q_3}{4 \pi \epsilon_0} (10 q_2 - 2 q_2) = \frac{8 q_2 q_3}{4 \pi \epsilon_0}$.
Comparing this with $\frac{q_3 K}{4 \pi \epsilon_0}$,we get $K = 8 q_2$.

(B) $1$. For a uniformly charged spherical shell of radius $R$ and charge $Q$,the electric field inside the shell is zero $(E = 0)$.
$2$. Since the electric field is zero inside,the potential $V$ is constant throughout the interior and is equal to the potential at the surface,given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
$3$. The work done $W$ in moving a test charge $q$ between two points $1$ and $2$ is given by $W = q(V_2 - V_1)$.
$4$. Since $V_1 = V_2$ at any two points inside the shell,$W = q(0) = 0$.
$5$. Thus,Assertion $A$ is true because the potential is constant,and Reason $R$ is true and provides the correct explanation for why the work done is zero.

(A) The electric potential $V$ at a point due to a spherical shell is given by:
$1$. Inside the shell $(r < \text{radius})$: $V = \frac{KQ}{\text{radius}}$
$2$. Outside the shell $(r > \text{radius})$: $V = \frac{KQ}{r}$
For a point at $r = \frac{3R}{2}$:
- This point is outside the inner shell (radius $R$), so the potential due to the inner shell is $V_1 = \frac{KQ}{3R/2} = \frac{2KQ}{3R}$.
- This point is inside the outer shell (radius $2R$), so the potential due to the outer shell is $V_2 = \frac{K(-Q)}{2R} = -\frac{KQ}{2R}$.
The total potential $V$ is the sum of the potentials due to both shells:
$V = V_1 + V_2 = \frac{2KQ}{3R} - \frac{KQ}{2R}$
$V = \frac{4KQ - 3KQ}{6R} = \frac{KQ}{6R}$

(D) The work done $W$ in moving a charge $q_0$ from point $I$ to $F$ is given by $W = q_0 (V_F - V_I)$.
Here,the charge being moved is $q_0 = -q$.
The potential at infinity is $V_F = 0$.
The distance of each corner from the centre of the cube of side $a$ is $r = \frac{\sqrt{3}}{2} a$.
The potential at the centre $V_I$ due to $8$ charges $q$ is $V_I = 8 \times \frac{1}{4 \pi \epsilon_0} \frac{q}{r} = 8 \times \frac{1}{4 \pi \epsilon_0} \frac{q}{(\sqrt{3} a / 2)} = \frac{4 q}{\sqrt{3} \pi \epsilon_0 a}$.
Therefore,$W = -q (0 - V_I) = q V_I = q \left( \frac{4 q}{\sqrt{3} \pi \epsilon_0 a} \right) = \frac{4 q^2}{\sqrt{3} \pi \epsilon_0 a}$.

(D) The potential at point $P$ inside the shell is the sum of the potential due to the charge $Q$ at the center and the potential due to the charge $q$ on the shell.
For a point inside a conducting shell,the potential due to the shell's charge $q$ is constant and equal to the potential at its surface,which is $V_{shell} = \frac{kq}{R}$.
The potential due to the point charge $Q$ at distance $r = R/2$ is $V_{point} = \frac{kQ}{R/2} = \frac{2kQ}{R}$.
Therefore,the total potential at point $P$ is $V_P = V_{point} + V_{shell} = \frac{2kQ}{R} + \frac{kq}{R}$.
Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $V_P = \frac{2Q}{4 \pi \varepsilon_0 R} + \frac{q}{4 \pi \varepsilon_0 R}$.

(C) The potential $V$ at the center of the square due to the four charges $Q$ at the corners is given by $V = 4 \times \frac{kQ}{r}$,where $r$ is the distance from the corner to the center. For a square of side $a$,the diagonal is $a\sqrt{2}$,so $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Thus,$V = 4 \times \frac{kQ}{a/\sqrt{2}} = \frac{4\sqrt{2}kQ}{a}$.
The potential energy $U_i$ of the charge $-Q$ at the center is $U_i = (-Q)V = -\frac{4\sqrt{2}kQ^2}{a}$.
The potential energy $U_f$ at infinity is $0$.
The work done $W$ is $W = U_f - U_i = 0 - (-\frac{4\sqrt{2}kQ^2}{a}) = \frac{4\sqrt{2}kQ^2}{a}$.
Substituting $k = \frac{1}{4\pi\varepsilon_0}$,we get $W = \frac{4\sqrt{2}Q^2}{4\pi\varepsilon_0 a} = \frac{\sqrt{2}Q^2}{\pi\varepsilon_0 a}$.

(D) Let the charges on the spheres of radii $R$ and $r$ be $Q_1$ and $Q_2$ respectively,such that $Q_1 + Q_2 = Q$.
Given that the surface charge densities are equal,$\sigma = \frac{Q_1}{4\pi R^2} = \frac{Q_2}{4\pi r^2}$.
This implies $\frac{Q_1}{R^2} = \frac{Q_2}{r^2} = k'$,where $k'$ is a constant.
Thus,$Q_1 = k' R^2$ and $Q_2 = k' r^2$.
Since $Q_1 + Q_2 = Q$,we have $k'(R^2 + r^2) = Q$,so $k' = \frac{Q}{R^2 + r^2}$.
The potential at the common centre is $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q_1}{R} + \frac{Q_2}{r} \right)$.
Substituting the values of $Q_1$ and $Q_2$:
$V = \frac{1}{4\pi\varepsilon_0} \left( \frac{k' R^2}{R} + \frac{k' r^2}{r} \right) = \frac{k'}{4\pi\varepsilon_0} (R + r)$.
Substituting $k' = \frac{Q}{R^2 + r^2}$:
$V = \frac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)}$.

(B) The kinetic energy gained by a charged particle accelerated through a potential difference '$V$' is given by $K = qV$.
Since both particles start from rest,the kinetic energy is equal to the work done by the electric field: $\frac{1}{2}mv^2 = qV$.
For particle '$A$': $\frac{1}{2}mv_A^2 = qV \implies v_A = \sqrt{\frac{2qV}{m}}$.
For particle '$B$': $\frac{1}{2}mv_B^2 = (4q)V \implies v_B = \sqrt{\frac{8qV}{m}}$.
Taking the ratio of their speeds $(v_A : v_B)$:
$\frac{v_A}{v_B} = \frac{\sqrt{2qV/m}}{\sqrt{8qV/m}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio of their speeds is $1: 2$.

(A) The potential at any point $P$ due to a system of charges is given by $V = \sum \frac{1}{4 \pi \varepsilon_0} \frac{q_i}{r_i}$.
In the isosceles triangle $ABC$,charges $+q$ are placed at $A, B,$ and $C$. $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$.
By symmetry,the distance of $D$ from $A$ is $a$,from $B$ is $a$,and from $C$ is $\sqrt{AC^2 + AD^2 - 2(AC)(AD) \cos A} = \sqrt{(2a)^2 + a^2 - 2(2a)(a) \cos A}$.
Similarly,the distance of $E$ from $A$ is $a$,from $C$ is $a$,and from $B$ is $\sqrt{AB^2 + AE^2 - 2(AB)(AE) \cos A} = \sqrt{(2a)^2 + a^2 - 2(2a)(a) \cos A}$.
Since the configuration is symmetric with respect to the angle $A$,the potential at $D$ $(V_D)$ and the potential at $E$ $(V_E)$ are equal.
$V_D = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AD} + \frac{q}{BD} + \frac{q}{CD} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{a} + \frac{q}{a} + \frac{q}{CD} \right)$.
$V_E = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AE} + \frac{q}{CE} + \frac{q}{BE} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{a} + \frac{q}{a} + \frac{q}{BE} \right)$.
Since $CD = BE$ by symmetry,$V_D = V_E$.
The work done $W = Q(V_E - V_D) = Q(0) = 0$.

(A) For a conducting sphere,the charge $Q$ resides entirely on its outer surface.
Inside a conducting sphere,the electric field $E$ is zero everywhere because there is no enclosed charge $(q_{enclosed} = 0)$ according to Gauss's Law.
Since the electric field is zero inside the sphere,the potential $V$ is constant throughout the interior and is equal to the potential at the surface.
The potential at the surface (and thus at the centre) is given by $V = \frac{Q}{4 \pi \epsilon_0 R}$.
Therefore,the electric field at the centre is $0$ and the electric potential at the centre is $\frac{Q}{4 \pi \epsilon_0 R}$.

(D) The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by the formula:
$W = q(V_B - V_A)$
Here,the potential at point $A$ is $V_A = -5 \ V$ and the potential at point $B$ is $V_B = V$.
Substituting these values into the formula:
$W = q(V - (-5))$
$W = q(V + 5)$
Now,solve for $V$:
$\frac{W}{q} = V + 5$
$V = \frac{W}{q} - 5$
Therefore,the correct option is $D$.

(A) The electrical potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$.
For an equilateral triangle with side length $a$,the distance between any two charges is $a$.
Given charges are $q_1 = Q$,$q_2 = 2Q$,and $q_3 = q$.
The total potential energy is $U = \frac{k}{a} (Q \cdot 2Q + 2Q \cdot q + q \cdot Q)$.
Setting $U = 0$ for the system to have zero potential energy:
$2Q^2 + 2Qq + qQ = 0$.
$2Q^2 + 3Qq = 0$.
$Q(2Q + 3q) = 0$.
Since $Q \neq 0$,we have $2Q + 3q = 0$.
Therefore,$q = -\frac{2}{3} Q$.

(A) The surface charge densities are $\sigma_A = +\sigma$,$\sigma_B = -\sigma$,and $\sigma_C = +\sigma$. The charges on the shells are $Q_A = 4\pi a^2 \sigma$,$Q_B = 4\pi b^2 (-\sigma) = -4\pi b^2 \sigma$,and $Q_C = 4\pi c^2 \sigma$.
The potential at the surface of shell $A$ is the sum of potentials due to all three shells: $V_A = V_{A,A} + V_{A,B} + V_{A,C}$.
Since $A$ is inside $B$ and $C$,the potential due to $B$ and $C$ at the surface of $A$ is equal to the potential at their respective surfaces: $V_A = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_A}{a} + \frac{Q_B}{b} + \frac{Q_C}{c} \right)$.
Substituting the values: $V_A = \frac{1}{4\pi\epsilon_0} \left( \frac{4\pi a^2 \sigma}{a} + \frac{-4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c} \right)$.
$V_A = \frac{1}{4\pi\epsilon_0} (4\pi a \sigma - 4\pi b \sigma + 4\pi c \sigma)$.
$V_A = \frac{\sigma}{\epsilon_0} (a - b + c)$.

(B) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
Point $A$ is at the midpoint of the side with charges $+2q$ and $+2q$. The distance of $A$ from each of these charges is $L$.
The distance of $A$ from each of the charges $-2q$ at the opposite corners can be found using the Pythagorean theorem. The horizontal distance is $2L$ and the vertical distance is $L$, so the distance $r = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$.
The total potential at $A$ is the sum of potentials due to all four charges:
$V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{2q}{L} + \frac{2q}{L} + \frac{-2q}{L\sqrt{5}} + \frac{-2q}{L\sqrt{5}} \right]$
$V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{4q}{L} - \frac{4q}{L\sqrt{5}} \right]$
$V_A = \frac{4q}{4\pi\epsilon_0 L} \left[ 1 - \frac{1}{\sqrt{5}} \right] = \frac{q}{\pi\epsilon_0 L} \left[ 1 - \frac{1}{\sqrt{5}} \right]$.

(B) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side $a = 6 \ cm = 0.06 \ m$.
Since there are $6$ vertices,each with a charge $q = 2 \ \mu C = 2 \times 10^{-6} \ C$,the total potential $V$ at the centre is the sum of the potentials due to each charge.
$V = 6 \times \left( \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{a} \right)$
Substituting the values:
$V = 6 \times 9 \times 10^9 \times \frac{2 \times 10^{-6}}{0.06}$
$V = 54 \times 10^9 \times \frac{2 \times 10^{-6}}{6 \times 10^{-2}}$
$V = 54 \times 10^9 \times \frac{1}{3} \times 10^{-4}$
$V = 18 \times 10^5 \ V = 1.8 \times 10^6 \ V$.

(A) Let the point where the net potential is zero be at a distance $x$ from charge $A$ $(2 \mu C)$ along the line joining $A$ and $B$.
The potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.
For the net potential to be zero,the sum of potentials from both charges must be zero:
$V_A + V_B = 0$
$\frac{k(2 \times 10^{-6})}{x} + \frac{k(-3 \times 10^{-6})}{1 - x} = 0$
$\frac{2}{x} = \frac{3}{1 - x}$
$2(1 - x) = 3x$
$2 - 2x = 3x$
$2 = 5x$
$x = \frac{2}{5} = 0.4 \ m$.
Thus,the distance from $A$ is $0.4 \ m$.

(D) The potential $V$ of a conducting sphere of radius $r$ carrying a charge $q$ is given by the formula $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
Given that both spheres have the same charge,let $q_1 = q_2 = q$.
The potential of the first sphere is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_1}$.
The potential of the second sphere is $V_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$.
The ratio of their potentials is $\frac{V_1}{V_2} = \frac{\frac{1}{4\pi\epsilon_0} \frac{q}{r_1}}{\frac{1}{4\pi\epsilon_0} \frac{q}{r_2}} = \frac{r_2}{r_1}$.
Therefore,the correct option is $D$.

(B) The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by the formula: $W = q(V_B - V_A)$.
Given:
Work done $W = 90 \ J$
Charge $q = 3 \ C$
Potential at $A$,$V_A = -10 \ V$
Potential at $B$,$V_B = V_1$
Substituting the values in the formula:
$90 = 3 \times (V_1 - (-10))$
$90 = 3 \times (V_1 + 10)$
Divide both sides by $3$:
$30 = V_1 + 10$
$V_1 = 30 - 10$
$V_1 = 20 \ V$.

(B) The charges are placed on the circumference of a circle of diameter $2d$. The radius of the circle is $r = \text{Diameter} / 2 = (2d) / 2 = d$.
Since the charges form a square inscribed in the circle,each charge is at a distance $r = d$ from the center.
The electric potential $V$ at the center due to a single point charge $q$ at distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$.
Since there are four identical charges,the total potential $V_{total}$ at the center is the sum of the potentials due to each charge:
$V_{total} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{d} + \frac{q}{d} + \frac{q}{d} + \frac{q}{d} \right)$
$V_{total} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{4q}{d} \right) = \frac{q}{\pi \varepsilon_0 d}$.
Thus,the potential at the center is proportional to $q/d$.

(C) The potential at the centre $O$ of the hexagon due to a charge $q$ at each vertex is the sum of the potentials due to each individual charge.
Since the distance from the centre to each vertex of a regular hexagon is equal to the side length $r = 10 \text{ cm} = 0.1 \text{ m}$,the potential $V$ at the centre is given by:
$V = 6 \times \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \right)$
Given $q = 1 \mu\text{C} = 1 \times 10^{-6} \text{ C}$,$r = 0.1 \text{ m}$,and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$:
$V = 6 \times \left( 9 \times 10^9 \times \frac{1 \times 10^{-6}}{0.1} \right)$
$V = 6 \times 9 \times 10^9 \times 10^{-5}$
$V = 54 \times 10^4 = 5.4 \times 10^5 \text{ volt}$.

(B) The electric potential $V$ due to a charged arc at its centre is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$,where $Q$ is the total charge on the arc and $R$ is its radius.
For a half ring of radius $R$,the total charge $Q = \lambda \times (\pi R)$.
Thus,the potential at the centre due to the first half ring is $V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda (\pi R_1)}{R_1} = \frac{\lambda}{4 \varepsilon_0}$.
Similarly,the potential at the centre due to the second half ring is $V_2 = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda (\pi R_2)}{R_2} = \frac{\lambda}{4 \varepsilon_0}$.
The net potential at the centre is $V_{net} = V_1 + V_2 = \frac{\lambda}{4 \varepsilon_0} + \frac{\lambda}{4 \varepsilon_0} = \frac{\lambda}{2 \varepsilon_0}$.

(A) The electric potential $V$ at any point at a distance $r$ from a point charge $Q$ is given by $V = \frac{kQ}{r}$.
Since the charge $10 \mu C$ is at the centre of the square,all four corners $(A, B, C, D)$ are at the same distance $r$ from the centre.
Therefore,the electric potential at corner $A$ $(V_A)$ and corner $B$ $(V_B)$ are equal,i.e.,$V_A = V_B$.
The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.
Since $V_A = V_B$,the potential difference $(V_B - V_A) = 0$.
Thus,the work done $W = 2 \mu C \times 0 = 0$.

(B) Let the square be $ABCD$ with side length $2L$. The charges are placed as follows: $A(+q), B(+q), C(-q), D(-q)$. Point $P$ is the midpoint of side $AB$.
The distances from point $P$ to the charges are:
$AP = L$
$BP = L$
$DP = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$
$CP = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$
The total electric potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = V_A + V_B + V_C + V_D$
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{CP} + \frac{-q}{DP} \right)$
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{L} + \frac{q}{L} - \frac{q}{L\sqrt{5}} - \frac{q}{L\sqrt{5}} \right)$
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{2q}{L} - \frac{2q}{L\sqrt{5}} \right)$
$V = \frac{1}{4 \pi \varepsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$

(A) The electric potential energy $U$ of a charge $q$ in an electric potential $V$ is given by $U = qV$.
When a unit positive charge $(q = 1)$ is moved from a region of low potential $(V_L)$ to a region of high potential $(V_H)$,the change in potential energy is $\Delta U = q(V_H - V_L)$.
Since $V_H > V_L$,the term $(V_H - V_L)$ is positive.
Therefore,$\Delta U > 0$,which means the potential energy increases.
Alternatively,work must be done by an external agent against the electric field to move a positive charge from low potential to high potential,and this work is stored as potential energy in the system.

(C) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{K q_1 q_2}{r}$.
Initial potential energy $U_i$ at distance $d$ is $U_i = \frac{K q_1 q_2}{d}$.
When charge $q_2$ is moved towards $q_1$ by a distance $x$,the new separation becomes $r' = d - x$.
The final potential energy $U_f$ is $U_f = \frac{K q_1 q_2}{d - x}$.
The increase in potential energy $\Delta U$ is given by $\Delta U = U_f - U_i$.
Substituting the values:
$\Delta U = \frac{K q_1 q_2}{d - x} - \frac{K q_1 q_2}{d}$
$\Delta U = K q_1 q_2 \left( \frac{1}{d - x} - \frac{1}{d} \right)$
$\Delta U = K q_1 q_2 \left( \frac{d - (d - x)}{d(d - x)} \right)$
$\Delta U = \frac{K q_1 q_2 x}{d(d - x)}$.

(C) Let the corners of the square be $A, B, C, D$ in order,with charges $+q$ at $A$ and $B$,and $-q$ at $C$ and $D$. The side length is $2r$. Point $P$ is the midpoint of side $CD$.
Thus,the distances are: $DP = PC = r$.
The distances from $A$ and $B$ to $P$ are: $AP = BP = \sqrt{AD^2 + DP^2} = \sqrt{(2r)^2 + r^2} = \sqrt{5r^2} = r\sqrt{5}$.
The electric potential $V_P$ at point $P$ is the sum of potentials due to all four charges:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{CP} + \frac{-q}{DP} \right]$
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{r\sqrt{5}} + \frac{q}{r\sqrt{5}} - \frac{q}{r} - \frac{q}{r} \right]$
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{2q}{r\sqrt{5}} - \frac{2q}{r} \right]$
$V_P = \frac{1}{4 \pi \varepsilon_0} \frac{2q}{r} \left[ \frac{1}{\sqrt{5}} - 1 \right]$

(B) The potential at the surface of the sphere is $V_s = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
The potential at a distance $5R$ from the centre is $V_p = \frac{1}{4\pi\epsilon_0} \frac{q}{5R}$.
The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\epsilon_0} (\frac{1}{R} - \frac{1}{5R}) = \frac{q}{4\pi\epsilon_0} (\frac{4}{5R}) = \frac{q}{5\pi\epsilon_0 R}$.
From this,we can express the charge $q$ as $q = \frac{5\pi\epsilon_0 RV}{1}$.
The electric field intensity $E$ at a distance $r = 5R$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{(5R)^2} = \frac{q}{100\pi\epsilon_0 R^2}$.
Substituting the value of $q$ in terms of $V$: $E = \frac{1}{100\pi\epsilon_0 R^2} \cdot (5\pi\epsilon_0 RV) = \frac{5}{100} \frac{V}{R} = \frac{V}{20R}$.

(D) The electric potential $dV$ due to a small charge element $dq$ at the centre of the ring is given by $dV = \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{R}$.
Since the ring is a half-ring,the total charge $q$ is the product of the linear charge density $\sigma$ and the length of the half-ring $(L = \pi R)$.
Thus,$q = \sigma \cdot \pi R$.
The total potential $V$ at the centre is the integral of $dV$ over the entire half-ring:
$V = \int dV = \int \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{R} = \frac{1}{4 \pi \epsilon_0 R} \int dq$.
Substituting the total charge $q = \sigma \pi R$:
$V = \frac{1}{4 \pi \epsilon_0 R} \cdot (\sigma \pi R) = \frac{\sigma}{4 \epsilon_0}$.

(A) Given that the two spherical conductors are at the same potential $V$.
For a spherical conductor of radius $r$ and charge $q$,the potential is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}$.
Since $V_1 = V_2$,we have $\frac{q_1}{r_1} = \frac{q_2}{r_2}$,which implies $\frac{q_1}{q_2} = \frac{r_1}{r_2}$.
The surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi r^2}$.
Therefore,the ratio of surface charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4 \pi r_1^2)}{q_2 / (4 \pi r_2^2)} = \frac{q_1}{q_2} \cdot \frac{r_2^2}{r_1^2}$.
Substituting $\frac{q_1}{q_2} = \frac{r_1}{r_2}$,we get $\frac{\sigma_1}{\sigma_2} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$.
Given $r_1 = 4 \ cm$ and $r_2 = 5 \ cm$,the ratio is $\frac{\sigma_1}{\sigma_2} = \frac{5}{4}$.

(A) The kinetic energy gained by a particle of charge $q$ and mass $m$ falling through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^2$.
For particle $A$ with charge $q$:
$\frac{1}{2}mv_A^2 = qV$ (Equation $1$)
For particle $B$ with charge $4q$:
$\frac{1}{2}mv_B^2 = 4qV$ (Equation $2$)
Dividing Equation $1$ by Equation $2$:
$\frac{v_A^2}{v_B^2} = \frac{qV}{4qV} = \frac{1}{4}$
Taking the square root on both sides:
$\frac{v_A}{v_B} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Therefore,the ratio of their speeds $v_A : v_B$ is $1:2$.

(C) The electric potential $V$ in an electric field $\overrightarrow{E}$ is given by the relation $dV = -\overrightarrow{E} \cdot d\overrightarrow{r}$.
This implies that the electric potential decreases as we move in the direction of the electric field.
In the given figure,point $B$ is at the leftmost position,meaning it is at the highest potential among the three points.
Points $A$ and $C$ are further along the direction of the electric field compared to point $B$,so their potentials will be lower than that of point $B$.
Therefore,the electric potential is maximum at point $B$.

(D) The potential at the surface of the conductor is $V_s = \frac{q}{4\pi\varepsilon_0 r}$.
The potential at a distance $3r$ from the center is $V_p = \frac{q}{4\pi\varepsilon_0 (3r)}$.
The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\varepsilon_0 r} - \frac{q}{12\pi\varepsilon_0 r} = \frac{q}{4\pi\varepsilon_0 r} \left(1 - \frac{1}{3}\right) = \frac{2q}{12\pi\varepsilon_0 r} = \frac{q}{6\pi\varepsilon_0 r}$.
Thus,$\frac{q}{4\pi\varepsilon_0} = \frac{3}{2} Vr$.
The electric intensity $E$ at a distance $3r$ is $E = \frac{q}{4\pi\varepsilon_0 (3r)^2} = \frac{q}{4\pi\varepsilon_0 (9r^2)}$.
Substituting the value of $\frac{q}{4\pi\varepsilon_0}$,we get $E = \frac{3Vr}{2(9r^2)} = \frac{V}{6r}$.

(D) Let the square be $ABCD$ with side length $2L$. Charges $+q$ are at $A$ and $B$,and charges $-q$ are at $D$ and $C$. Point $P$ is the midpoint of $AB$.
Distance $AP = PB = L$.
The distance from $P$ to $D$ and $C$ can be found using the Pythagorean theorem in $\triangle ADP$ and $\triangle BCP$:
$PD = PC = \sqrt{(2L)^2 + L^2} = \sqrt{4L^2 + L^2} = \sqrt{5}L$.
The electric potential $V$ at point $P$ is the sum of potentials due to all four charges:
$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{PD} + \frac{-q}{PC} \right)$
$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{L} + \frac{q}{L} - \frac{q}{\sqrt{5}L} - \frac{q}{\sqrt{5}L} \right)$
$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{2q}{L} - \frac{2q}{\sqrt{5}L} \right)$
$V = \frac{1}{4 \pi \epsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$

(A) For a spherical charged shell,the electric potential at any point inside the shell is constant and equal to the potential at its surface.
Given that the radius of the shell is $R = 10 \ cm$ and the potential on the surface is $V_{surface} = 100 \ V$.
Since $2 \ cm < 10 \ cm$,the point is inside the shell.
Therefore,the potential at $2 \ cm$ from the centre is the same as the potential on the surface,which is $100 \ V$.

(C) The energy gained by a charged particle accelerated through a potential difference $V$ is given by the formula: $E = qV$.
Here,the charge of an electron is $q = e = 1.6 \times 10^{-19} \ C$ and the potential difference is $V = 2.5 \ V$.
Therefore,the energy gained is $E = e \times 2.5 \ V = 2.5 \ eV$.
Since $1 \ eV$ is the energy gained by an electron when accelerated through a potential difference of $1 \ V$,the energy gained is $2.5 \ eV$.

(A) The length of the diagonal of a square with side $a$ is $a \sqrt{2}$.
Given side $a = 2 \sqrt{2} \text{ m}$,the diagonal length is $d = (2 \sqrt{2}) \times \sqrt{2} = 4 \text{ m}$.
The distance $r$ of each vertex from the center (intersection of diagonals) is half the diagonal length:
$r = \frac{d}{2} = \frac{4}{2} = 2 \text{ m}$.
The electric potential $V$ at the center due to four identical charges $q = 1 \mu C = 10^{-6} \text{ C}$ at distance $r$ is given by:
$V = 4 \times \frac{k q}{r}$,where $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.
Substituting the values:
$V = 4 \times \frac{9 \times 10^9 \times 10^{-6}}{2}$
$V = 2 \times 9 \times 10^3 = 18 \times 10^3 \text{ V}$.

(C) The electrical potential energy $U$ of a charged sphere can be calculated by considering the work done in bringing small charge elements $dq$ from infinity to the surface of the sphere.
Alternatively,we can use the average potential method.
Initially,when there is no charge on the sphere,the potential $V_1 = 0$.
When the total charge $Q$ is on the sphere,the potential $V_2 = \frac{k Q}{R}$.
The average potential $V$ during the charging process is $V = \frac{V_1 + V_2}{2} = \frac{0 + \frac{k Q}{R}}{2} = \frac{k Q}{2 R}$.
The total potential energy $U$ is the work done to charge the sphere,which is given by $U = \int_0^Q V(q) dq = \int_0^Q \frac{k q}{R} dq = \frac{k}{R} \left[ \frac{q^2}{2} \right]_0^Q = \frac{1}{2} \frac{k Q^2}{R}$.