Electric potential Questions in English

(B) Let the charges on the rings be $Q_1 = 10 \text{ C}$ and $Q_2 = 5 \text{ C}$. The distance between the centres $O$ and $O'$ is $R$.
The potential at the centre $O$ of the first ring is due to its own charge $Q_1$ and the charge $Q_2$ on the second ring:
$V_O = \frac{k Q_1}{R} + \frac{k Q_2}{\sqrt{R^2 + R^2}} = \frac{k Q_1}{R} + \frac{k Q_2}{\sqrt{2}R} = \frac{k}{R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} \right)$
The potential at the centre $O'$ of the second ring is due to its own charge $Q_2$ and the charge $Q_1$ on the first ring:
$V_{O'} = \frac{k Q_2}{R} + \frac{k Q_1}{\sqrt{R^2 + R^2}} = \frac{k Q_2}{R} + \frac{k Q_1}{\sqrt{2}R} = \frac{k}{R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} \right)$
The potential difference is $\Delta V = V_O - V_{O'}$:
$\Delta V = \frac{k}{R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} - Q_2 - \frac{Q_1}{\sqrt{2}} \right) = \frac{k}{R} \left( Q_1(1 - \frac{1}{\sqrt{2}}) - Q_2(1 - \frac{1}{\sqrt{2}}) \right)$
$\Delta V = \frac{k}{R} (Q_1 - Q_2) \left( 1 - \frac{1}{\sqrt{2}} \right)$
Substituting $Q_1 = 10 \text{ C}$,$Q_2 = 5 \text{ C}$,and $k = \frac{1}{4 \pi \varepsilon_0}$:
$W = q \Delta V = \frac{q}{4 \pi \varepsilon_0 R} (10 - 5) \left( 1 - \frac{1}{\sqrt{2}} \right) = \frac{5 q}{4 \pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} \right) \text{ J}$.

(C) The line integral of the electric field $\vec{E}$ along a path is defined as the potential difference $V$ between two points.
Mathematically,$V = -\int \vec{E} \cdot d\vec{l}$.
The $SI$ unit of electric potential $V$ is the volt $(V)$,which is defined as the work done per unit charge.
Therefore,$1 \ V = 1 \ J C^{-1}$.
Thus,the unit of the physical quantity obtained by the line integral of the electric field is $J C^{-1}$ (or Volts).

(A) For a charged hollow metal sphere,the electric field inside the sphere is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -dV/dr)$,if $E = 0$,then the potential $V$ must be constant throughout the interior of the sphere.
Therefore,the potential at any point inside the sphere,including the centre,is equal to the potential on its surface.
Given that the potential on the surface is $80 \ V$,the potential at the centre of the sphere is $80 \ V$.

(D) For a conducting sphere,the electric potential $V$ at the surface is given by $V = \frac{kQ}{R}$,where $k$ is Coulomb's constant,$Q$ is the charge on the sphere,and $R$ is the radius of the sphere.
Since both spheres are made of copper (a conductor) and have the same size,they have the same radius $R$.
Given that both spheres are charged to the same potential $V$,we have $V_x = V_y = V$.
Using the formula $V = \frac{kQ}{R}$,we get $V = \frac{kQ_x}{R}$ and $V = \frac{kQ_y}{R}$.
Equating the two,we find $\frac{kQ_x}{R} = \frac{kQ_y}{R}$,which implies $Q_x = Q_y$.
Therefore,the charge on both spheres is equal,regardless of whether the sphere is hollow or solid,because the charge on a conductor resides entirely on its outer surface.

(C) The electric potential at the surface of shell $A$ is the sum of the potentials due to all three shells $A$,$B$,and $C$.
$V_A = V_a + V_b + V_c$
Since the point on the surface of shell $A$ is inside shells $B$ and $C$,the potential due to shells $B$ and $C$ at this point is equal to the potential at their respective surfaces.
$V_A = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q_a}{a} + \frac{q_b}{b} + \frac{q_c}{c} \right]$
Given surface charge densities: $\sigma_a = \sigma$,$\sigma_b = -\sigma$,$\sigma_c = \sigma$.
The charges are $q_a = 4 \pi a^2 \sigma$,$q_b = 4 \pi b^2(-\sigma)$,and $q_c = 4 \pi c^2 \sigma$.
Substituting these into the potential formula:
$V_A = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{4 \pi a^2 \sigma}{a} + \frac{4 \pi b^2(-\sigma)}{b} + \frac{4 \pi c^2 \sigma}{c} \right]$
$V_A = \frac{1}{4 \pi \varepsilon_0} [4 \pi a \sigma - 4 \pi b \sigma + 4 \pi c \sigma]$
$V_A = \frac{\sigma}{\varepsilon_0} (a - b + c)$

(B) The work done $W$ in moving a charge $q$ between two points with potential difference $\Delta V$ is given by the formula: $W = q \Delta V$.
Here,$W = 20 \ J$,$q = 4 \ C$,initial potential $V_i = 10 \ V$,and final potential $V_f = V$.
The potential difference is $\Delta V = V_f - V_i = V - 10$.
Substituting the values into the formula:
$20 = 4(V - 10)$
Divide both sides by $4$:
$5 = V - 10$
Adding $10$ to both sides:
$V = 15 \ V$.

(D) The length of the body diagonal of a cube with side $a$ is $\sqrt{3} a$. The distance $r$ of each vertex from the centre $O$ of the cube is half of the body diagonal: $r = \frac{\sqrt{3} a}{2}$.
Since there are $8$ identical charges $q$ at the vertices,the total electric potential $V$ at the centre $O$ is the sum of the potentials due to each individual charge:
$V = 8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{q}{r}$
Substituting the value of $r$:
$V = 8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{q}{\frac{\sqrt{3} a}{2}}$
$V = 8 \times \frac{1}{4 \pi \varepsilon_0} \times \frac{2q}{\sqrt{3} a}$
$V = \frac{4 q}{\sqrt{3} \pi \varepsilon_0 a}$

(D) The electric potential $V$ at the surface of a nucleus is given by the formula: $V = \frac{k Q}{r}$.
Here,$k = 9 \times 10^9 \ N \ m^2/C^2$,$Q = Ze = 50 \times 1.6 \times 10^{-19} \ C$,and $r = 9 \times 10^{-15} \ m$.
Substituting these values into the formula:
$V = \frac{(9 \times 10^9) \times (50 \times 1.6 \times 10^{-19})}{9 \times 10^{-15}}$
$V = \frac{9 \times 10^9 \times 80 \times 10^{-19}}{9 \times 10^{-15}}$
$V = 80 \times 10^{-10} \times 10^{15} \ V$
$V = 80 \times 10^5 \ V = 8 \times 10^6 \ V$.
Therefore,the correct option is $D$.

(C) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side length $r = 9 \ cm = 0.09 \ m$.
Each vertex has a charge $q = 5 \mu C = 5 \times 10^{-6} \ C$.
The electric potential $V$ at the centre due to a single charge $q$ at distance $r$ is $V_i = \frac{kq}{r}$.
Since there are $6$ identical charges at the vertices,the total potential $V$ at the centre is the sum of potentials due to each charge:
$V = 6 \times \frac{kq}{r}$
Substituting the values:
$V = 6 \times \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.09}$
$V = 6 \times \frac{45 \times 10^3}{9 \times 10^{-2}}$
$V = 6 \times 5 \times 10^5$
$V = 30 \times 10^5 = 3 \times 10^6 \ V$.

(B) The electric potential $V$ at a point is defined as the electric potential energy $U$ per unit charge $q$.
Formula: $V = \frac{U}{q}$
Given:
Charge $q = 2 \mu C = 2 \times 10^{-6} \text{ C}$
Potential Energy $U = 3 \times 10^{-5} \text{ J}$
Calculation:
$V = \frac{3 \times 10^{-5}}{2 \times 10^{-6}}$
$V = 1.5 \times 10^1$
$V = 15 \text{ V}$
Therefore,the electric potential at the point is $15 \text{ V}$.

(C) For a positive charge $+q$,the electric potential at a distance $r$ is given by $V = \frac{kq}{r}$. Since $V \propto \frac{1}{r}$,the potential is higher closer to the charge.
For the positive charge,point $Q$ is closer to the charge than point $P$,so $r_Q < r_P$. This implies $V_Q > V_P$,therefore $V_Q - V_P > 0$ (positive).
For a negative charge $-q$,the electric potential is given by $V = \frac{k(-q)}{r} = -\frac{kq}{r}$. Here,the potential is more negative (lower) closer to the charge.
For the negative charge,point $A$ is closer to the charge than point $B$,so $r_A < r_B$. This implies $V_A < V_B$ (since $V_A$ is more negative than $V_B$). Therefore,$V_B - V_A > 0$ (positive).
Thus,both potential differences are positive.

(C) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by the formula:
$V = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}$
Given values:
Charge $q = 3 \text{ nC} = 3 \times 10^{-9} \text{ C}$
Distance $r = 9 \text{ cm} = 9 \times 10^{-2} \text{ m}$
Coulomb's constant $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \text{ N m}^2/\text{C}^2$
Substituting these values into the formula:
$V = (9 \times 10^{9}) \times \frac{3 \times 10^{-9}}{9 \times 10^{-2}}$
$V = \frac{9 \times 3 \times 10^{0}}{9 \times 10^{-2}}$
$V = 3 \times 10^{2} \text{ V} = 300 \text{ V}$
Therefore, the electric potential is $300 \text{ V}$.

(C) For a charged spherical shell of radius $R = 10 \,cm$ carrying charge $q$:
$1$. Inside the shell $(r < R)$, the electric potential is constant and equal to the potential at the surface: $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}$.
$2$. Outside the shell $(r > R)$, the potential varies inversely with distance: $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$.
Given distances are $r_{1} = 5 \,cm$, $r_{2} = 10 \,cm$, and $r_{3} = 15 \,cm$.
Since $r_{1} < R$ and $r_{2} = R$, we have $V_{1} = V_{2} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{10}$.
For $r_{3} = 15 \,cm$, $V_{3} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{15}$.
Comparing the values, since $15 > 10$, it follows that $\frac{q}{15} < \frac{q}{10}$, therefore $V_{3} < V_{1} = V_{2}$.
Thus, the correct relation is $V_{1} = V_{2} > V_{3}$.

(A) The work done $W$ in moving a charge $q$ from point $B$ to point $A$ is given by the formula: $W = q(V_A - V_B)$.
Given:
Charge $q = 5 \ mC = 5 \times 10^{-3} \ C$.
Potential at $A$,$V_A = -3 \ V$.
Potential at $B$,$V_B = 5 \ V$.
Substituting the values:
$W = 5 \times 10^{-3} \times (-3 - 5)$
$W = 5 \times 10^{-3} \times (-8)$
$W = -40 \times 10^{-3} \ J$
$W = -0.04 \ J$.

(A) The electric potential at point $A$ is $V_A$.
Since the electric field $E$ is uniform and directed towards the right,the potential decreases in the direction of the electric field.
The horizontal distance between $A$ and $B$ is $x$. Therefore,the potential difference between $A$ and $B$ is $\Delta V = V_B - V_A = -E \cdot x$.
Thus,the potential at $B$ is $V_B = V_A - Ex$.
The potential energy $U$ of a charge $q$ at a point with potential $V$ is given by $U = qV$.
Therefore,the potential energy of the charge at point $B$ is $U_B = q V_B = q(V_A - Ex)$.

(A) Charge on the first sphere,$q_1 = 1.8 \mu C = 1.8 \times 10^{-6} \ C$.
Charge on the second sphere,$q_2 = 2.8 \mu C = 2.8 \times 10^{-6} \ C$.
Distance between the two spheres,$r = 40 \ cm = 0.4 \ m$.
Distance from the mid-point to each sphere,$r_1 = r_2 = 20 \ cm = 0.2 \ m$.
The total electric potential $V$ at the mid-point is the algebraic sum of the potentials due to individual charges:
$V = V_1 + V_2 = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right)$
Substituting the values,where $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$:
$V = 9 \times 10^9 \times \left( \frac{1.8 \times 10^{-6}}{0.2} + \frac{2.8 \times 10^{-6}}{0.2} \right)$
$V = 9 \times 10^9 \times \frac{10^{-6}}{0.2} \times (1.8 + 2.8)$
$V = 9 \times 10^3 \times \frac{4.6}{0.2}$
$V = 9 \times 10^3 \times 23 = 207 \times 10^3 \ V = 2.07 \times 10^5 \ V$.
Rounding to two significant figures,$V \approx 2.1 \times 10^5 \ V$.

(A) The electric potential at the centre of a charged spherical shell of radius $R$ with charge $q$ is $V = \frac{q}{4 \pi \varepsilon_{0} R}$.
For the two concentric spheres,the total potential $V$ at the common centre is the sum of potentials due to each sphere:
$V = V_{1} + V_{2} = \frac{q_{1}}{4 \pi \varepsilon_{0} R} + \frac{q_{2}}{4 \pi \varepsilon_{0} r}$.
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 $q_{1} = 4 \pi R^{2} \sigma$ and $q_{2} = 4 \pi r^{2} \sigma$.
Substituting these values into the potential equation:
$V = \frac{4 \pi R^{2} \sigma}{4 \pi \varepsilon_{0} R} + \frac{4 \pi r^{2} \sigma}{4 \pi \varepsilon_{0} r}$.
$V = \frac{R \sigma}{\varepsilon_{0}} + \frac{r \sigma}{\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}(R + r)$.

(A) The potential $V_{1}$ of the smaller sphere (radius $r$,charge $q$) is due to its own charge and the potential due to the outer sphere (radius $R$,charge $Q$).
$V_{1} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}$
The potential $V_{2}$ of the bigger sphere (radius $R$,charge $Q$) is due to its own charge and the potential due to the inner sphere (which acts as a point charge at the center for points outside it).
$V_{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$
The potential difference between the spheres is $V = V_{1} - V_{2}$.
$V = \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} \right) - \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R} \right)$
$V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$
$V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{r} - \frac{q}{R} \right)$

(C) Let the radius of each small drop be $r$ and its charge be $q$. The potential of a small drop is $V' = \frac{kq}{r}$.
When $8$ drops combine to form a large drop of radius $R$ and charge $Q$, the volume remains constant: $\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3$, which gives $R = 2r$.
The total charge of the large drop is $Q = 8q$.
The potential of the large drop is $V = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left( \frac{kq}{r} \right) = 4V'$.
Given $V = 20 \,V$, we have $20 = 4V'$.
Therefore, $V' = \frac{20}{4} = 5 \,V$.

(D) Let $n = 27$ be the number of small droplets.
Let $r = 10^{-3} \,m$ be the radius of each small droplet.
Let $q = 10^{-12} \,C$ be the charge on each small droplet.
The volume of the big drop is equal to the sum of the volumes of the $27$ small droplets: $\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi r^3$.
Thus, $R^3 = 27 r^3$, which implies $R = 3r = 3 \times 10^{-3} \,m$.
The total charge on the big drop is $Q = n \times q = 27 \times 10^{-12} \,C$.
The potential of the big drop is given by $V = \frac{kQ}{R}$, where $k = 9 \times 10^9 \,N \cdot m^2/C^2$.
Substituting the values: $V = \frac{9 \times 10^9 \times 27 \times 10^{-12}}{3 \times 10^{-3}}$.
$V = \frac{9 \times 27 \times 10^{-3}}{3 \times 10^{-3}} = 3 \times 27 = 81 \,V$.

(C) The work done in moving a charge $Q$ from point $C$ to point $D$ in an electrostatic field is given by $W = Q(V_D - V_C)$,where $V_C$ and $V_D$ are the electric potentials at points $C$ and $D$ respectively.
The potential at any point is due to the charges at $A$ $(+q)$ and $B$ $(-q)$.
Distance $AB = 2d$,$BC = d$,$BD = d$. Thus,$AC = AB - BC = 2d - d = d$.
Potential at $C$ $(V_C)$:
$V_C = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AC} + \frac{-q}{BC} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{d} - \frac{q}{d} \right) = 0$.
Potential at $D$ $(V_D)$:
$V_D = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AD} + \frac{-q}{BD} \right)$.
Since $AD = AB + BD = 2d + d = 3d$ and $BD = d$,
$V_D = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{3d} - \frac{q}{d} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q - 3q}{3d} \right) = \frac{-2q}{12 \pi \varepsilon_0 d} = \frac{-q}{6 \pi \varepsilon_0 d}$.
Work done $W = Q(V_D - V_C) = Q \left( \frac{-q}{6 \pi \varepsilon_0 d} - 0 \right) = \frac{-qQ}{6 \pi \varepsilon_0 d}$.

(C) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \ N \ m^2/C^2$.
Given charge $q = 10^{-3} \mu C = 10^{-9} \ C$.
For point $A$ at $(\sqrt{2} m, \sqrt{2} m)$,the distance $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = 2 \ m$.
The potential at $A$ is $V_A = \frac{kq}{r_A} = \frac{9 \times 10^9 \times 10^{-9}}{2} = 4.5 \ V$.
For point $B$ at $(2 m, 0 m)$,the distance $r_B = \sqrt{2^2 + 0^2} = 2 \ m$.
The potential at $B$ is $V_B = \frac{kq}{r_B} = \frac{9 \times 10^9 \times 10^{-9}}{2} = 4.5 \ V$.
The potential difference between $A$ and $B$ is $V_A - V_B = 4.5 \ V - 4.5 \ V = 0 \ V$.

(B) The side of the square is $a = \sqrt{2} \text{ m}$. The diagonal of the square is $d = a\sqrt{2} = \sqrt{2} \times \sqrt{2} = 2 \text{ m}$.
The distance from the centre of the square to each vertex is $r = d/2 = 2/2 = 1 \text{ m}$.
The electric potential $V$ at the centre due to a point charge $q$ is given by $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.
The net potential at the centre is the algebraic sum of the potentials due to individual charges: $V_{net} = \frac{k}{r} (q_1 + q_2 + q_3 + q_4)$.
Substituting the given values: $V_{net} = \frac{9 \times 10^9}{1} \times (12 - 20 + 32 - 15) \times 10^{-9} \text{ V}$.
$V_{net} = 9 \times (9) \text{ V} = 81 \text{ V}$.

(A) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all unique pairs of charges.
For three charges $q_1, q_2, q_3$ at distances $r_{12}, r_{23}, r_{31}$,the potential energy is $U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$.
In this case,$q_1 = q_2 = q_3 = q$ and $r_{12} = r_{23} = r_{31} = L$.
Substituting these values,we get:
$U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q^2}{L} + \frac{q^2}{L} + \frac{q^2}{L} \right)$
$U = \frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{L}$

(C) Given charges at the corners of a square of side $a = 1 \text{ m}$ are:
$q_1 = +1 \times 10^{-8} \text{ C}$
$q_2 = -2 \times 10^{-8} \text{ C}$
$q_3 = +3 \times 10^{-8} \text{ C}$
$q_4 = +2 \times 10^{-8} \text{ C}$
The distance $r$ from each corner to the centre of the square is half the diagonal length:
$r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}} = \frac{1}{\sqrt{2}} \text{ m}$
The electric potential $V$ at the centre is the algebraic sum of the potentials due to each charge:
$V = \frac{k q_1}{r} + \frac{k q_2}{r} + \frac{k q_3}{r} + \frac{k q_4}{r} = \frac{k}{r} (q_1 + q_2 + q_3 + q_4)$
Substituting the values:
$V = \frac{9 \times 10^9}{1/\sqrt{2}} (1 - 2 + 3 + 2) \times 10^{-8}$
$V = 9 \sqrt{2} \times 10^1 \times (4)$
$V = 36 \sqrt{2} \times 10 \approx 36 \times 1.414 \times 10 = 509.04 \text{ V}$
Rounding to the nearest given option, $V \approx 510 \text{ V}$.

(C) Let the radius of each small sphere be $r$ and the charge on each be $q$. The potential on each small sphere is given by $V_s = \frac{kq}{r} = 60 \text{ mV} = 0.06 \text{ V}$.
When $125$ small spheres coalesce to form a big sphere of radius $R$, the volume remains conserved:
$125 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3 \Rightarrow R^3 = 125 r^3 \Rightarrow R = 5r$.
The total charge on the big sphere is $Q = 125q$.
The potential on the big sphere is $V_B = \frac{kQ}{R} = \frac{k(125q)}{5r} = 25 \times (\frac{kq}{r})$.
Substituting the value of $V_s$: $V_B = 25 \times 0.06 \text{ V} = 1.5 \text{ V}$.

(B) The potential at point $A$ due to $+q$ is $V_{A(+q)} = \frac{1}{4\pi\epsilon_0} \frac{q}{x}$ and due to $-q$ is $V_{A(-q)} = \frac{1}{4\pi\epsilon_0} \frac{-q}{x+y}$.
Thus,$V_A = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x} - \frac{1}{x+y} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{y}{x(x+y)} \right)$.
Similarly,the potential at point $B$ is $V_B = \frac{1}{4\pi\epsilon_0} \frac{q}{x+y} + \frac{1}{4\pi\epsilon_0} \frac{-q}{x} = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x+y} - \frac{1}{x} \right) = -V_A$.
Therefore,$V_A - V_B = V_A - (-V_A) = 2V_A = 2 \times \frac{q}{4\pi\epsilon_0} \left( \frac{y}{x(x+y)} \right)$.
Given $q = 10^{-6} \text{ C}$,$x = 0.02 \text{ m}$,$y = 0.03 \text{ m}$,and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$.
$V_A - V_B = 2 \times 9 \times 10^9 \times 10^{-6} \times \frac{0.03}{0.02(0.02+0.03)} = 18 \times 10^3 \times \frac{0.03}{0.02 \times 0.05} = 18 \times 10^3 \times \frac{0.03}{0.001} = 18 \times 10^3 \times 30 = 5.4 \times 10^5 \text{ V}$.

(A) The electric potential $V$ at the center of a square of side $a$ with charges $q$ at each of the $4$ corners is given by $V = 4 \times \frac{Kq}{r}$,where $r$ is the distance from the center to any corner.
For a square of side $a$,the diagonal is $a\sqrt{2}$,so the distance $r$ from the center to a corner is $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
For the first square,$a_1 = 1 \ m$,so $r_1 = \frac{1}{\sqrt{2}} \ m$. The potential is $V_1 = 4 \times \frac{K \times 2}{1/\sqrt{2}} = 8\sqrt{2}K$.
For the second square,$a_2 = 2 \ m$,so $r_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \ m$. The potential is $V_2 = 4 \times \frac{K \times 2}{\sqrt{2}} = \frac{8K}{\sqrt{2}} = 4\sqrt{2}K$.
Therefore,the ratio is $\frac{V_2}{V_1} = \frac{4\sqrt{2}K}{8\sqrt{2}K} = \frac{1}{2}$.

(A) The electric potential $V$ at a point due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \ N \ m^2 \ C^{-2}$.
Since the triangle is equilateral with side $a = 0.2 \ m$,the distance from both corners $A$ and $B$ to corner $C$ is $r = 0.2 \ m$.
The total potential at corner $C$ is the algebraic sum of the potentials due to charges at $A$ and $B$: $V_C = V_A + V_B$.
$V_C = \frac{k q_A}{r} + \frac{k q_B}{r} = \frac{k}{r} (q_A + q_B)$.
Substituting the values: $V_C = \frac{9 \times 10^9}{0.2} (8 \times 10^{-6} + 8 \times 10^{-6}) \ V$.
$V_C = \frac{9 \times 10^9}{0.2} (16 \times 10^{-6}) \ V$.
$V_C = 9 \times 10^9 \times 80 \times 10^{-6} \ V$.
$V_C = 720 \times 10^3 \ V = 7.2 \times 10^5 \ V$.

(C) The work done $W$ to bring a charge $q$ from infinity to a distance $r$ from a charge $Q$ is given by the change in potential energy: $W = U_f - U_i$.
Since the initial distance $r_i = \infty$,the initial potential energy $U_i = 0$.
The final potential energy is $U_f = \frac{kQq}{r_f}$.
Given:
$q = 6 \times 10^{-6} \ C$
$Q = 30 \times 10^{-6} \ C$
$r_f = 0.75 \ m$
$k = 9 \times 10^9 \ N \ m^2 \ C^{-2}$
Substituting the values:
$W = \frac{(9 \times 10^9) \times (30 \times 10^{-6}) \times (6 \times 10^{-6})}{0.75}$
$W = \frac{9 \times 30 \times 6 \times 10^{-3}}{0.75}$
$W = \frac{1620 \times 10^{-3}}{0.75} = \frac{1.62}{0.75} = 2.16 \ J$.

(A) Given that,the point charges are placed on the vertices of a square of side $a = 2.8 \,m$. Let $r$ be the distance of the centre $O$ from each corner.
The diagonal of the square is $d = \sqrt{a^2 + a^2} = a\sqrt{2}$.
The distance $r$ from the centre to each corner is half of the diagonal:
$r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} = \frac{2.8}{\sqrt{2}} \,m$.
The electric potential $V$ at a point due to a point charge $q$ is given by $V = \frac{Kq}{r}$,where $K = 9 \times 10^9 \,N \cdot m^2/C^2$.
The total electric potential $V_0$ at the centre $O$ due to all four charges $q_1, q_2, q_3,$ and $q_4$ is:
$V_0 = V_1 + V_2 + V_3 + V_4 = \frac{K}{r}(q_1 + q_2 + q_3 + q_4)$.
Given charges: $q_1 = +20 \,nC$,$q_2 = +40 \,nC$,$q_3 = -34 \,nC$,$q_4 = +16 \,nC$.
Sum of charges: $\sum q = (20 + 40 - 34 + 16) \,nC = 42 \,nC = 42 \times 10^{-9} \,C$.
Substituting the values:
$V_0 = \frac{9 \times 10^9 \times 42 \times 10^{-9}}{2.8 / \sqrt{2}} = \frac{9 \times 42 \times \sqrt{2}}{2.8} = \frac{378 \times 1.414}{2.8} = 135 \times 1.414 = 190.89 \,V$.

(B) Given that,two charges $q_1 = 20 \mu C$ and $q_2 = 10 \mu C$ are separated by a distance $d = 50 \ m$.
Work done $W$ in bringing a unit positive charge $q = 1 \ C$ from infinity to the midpoint $M$ is given by $W = qV$,where $V$ is the electric potential at point $M$.
The potential at $M$ due to charges $q_1$ and $q_2$ is $V = \frac{K q_1}{(d/2)} + \frac{K q_2}{(d/2)} = \frac{2K}{d} (q_1 + q_2)$.
Substituting the values:
$W = 1 \times \frac{2 \times (9 \times 10^9)}{50} \times (20 + 10) \times 10^{-6} \ J$
$W = \frac{18 \times 10^9}{50} \times 30 \times 10^{-6} \ J$
$W = \frac{540 \times 10^3}{50} \ J = 10.8 \times 10^3 \ J$.

(D) Electric potential on the surface of the sphere is given by $V = k \frac{Q}{R}$,where $k = 9 \times 10^9 \,N \cdot m^2/C^2$.
Given $V = 200 \,V$ and $R = 5 \,cm = 0.05 \,m$.
$200 = 9 \times 10^9 \times \frac{Q}{0.05} \implies Q = \frac{200 \times 0.05}{9 \times 10^9} = \frac{10}{9} \times 10^{-9} \,C$.
The potential at a distance $r$ from the centre of the sphere is $V(r) = k \frac{Q}{r}$.
Potential at point $A$ $(r_A = 15 \,cm = 0.15 \,m)$: $V_A = 9 \times 10^9 \times \frac{Q}{0.15} = \frac{9 \times 10^9}{0.15} \times \frac{10}{9} \times 10^{-9} = \frac{10}{0.15} = 66.67 \,V$.
Potential at point $B$ $(r_B = 10 \,cm = 0.10 \,m)$: $V_B = 9 \times 10^9 \times \frac{Q}{0.10} = \frac{9 \times 10^9}{0.10} \times \frac{10}{9} \times 10^{-9} = \frac{10}{0.10} = 100 \,V$.
Work done in moving a charge $q = 5 \,C$ from $A$ to $B$ is $W = q(V_B - V_A)$.
$W = 5 \times (100 - 66.67) = 5 \times 33.33 = 166.65 \,J \approx 166.7 \,J$.

(C) The charge on the spherical conductor is $q = +1 \,nC = 10^{-9} \,C$.
The distance of the point from the center of the spherical conductor is $r = 0.5 \,m$.
The formula for the electric potential $V$ at a distance $r$ from a point charge (or outside a spherical conductor) is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}$.
Substituting the given values:
$V = (9 \times 10^9 \,N \cdot m^2/C^2) \times \frac{10^{-9} \,C}{0.5 \,m}$.
$V = 9 \times \frac{1}{0.5} \,V$.
$V = 9 \times 2 \,V = 18 \,V$.
Therefore,the electric potential is $+18 \,V$.

(D) The potential difference $V$ between two points is defined as the work done $W$ per unit charge $q$ in moving the charge from one point to the other.
Formula: $V = \frac{W}{q}$
Given values:
Charge,$q = 20 C$
Work done,$W = 2 J$
Substituting the values into the formula:
$V = \frac{2 J}{20 C} = 0.1 V$
Therefore,the potential difference between the two points is $0.1 V$.

(C) The potential difference between two points in a uniform electric field is given by the formula $\Delta V = -\vec{E} \cdot \vec{d}$,where $\vec{d}$ is the displacement vector from the initial point to the final point.
For the potential difference $V_A - V_B$,the displacement vector is $\vec{BA}$.
The magnitude of the displacement is $d = |\vec{BA}| = 2 \ m$.
The angle $\theta$ between the electric field $\vec{E}$ and the displacement vector $\vec{BA}$ is $60^\circ$ (as it is an equilateral triangle).
Therefore,$V_A - V_B = -E d \cos(\theta)$.
Substituting the given values: $V_A - V_B = -(10 \ N \ C^{-1}) \times (2 \ m) \times \cos(60^\circ)$.
$V_A - V_B = -20 \times 0.5 = -10 \ V$.

(B) For a charged spherical conductor of radius $R = 10 \,cm$, the potential inside the conductor (at any distance $r < R$) is constant and equal to the potential at the surface.
Given that the potential at $r = 5 \,cm$ is $V$, the potential at the surface $(r = 10 \,cm)$ is also $V$.
The potential at a distance $r$ from the centre for $r > R$ is given by $V(r) = \frac{kQ}{r}$, where $kQ$ is the potential at the surface times the radius $(V \times R)$.
Thus, $V(r) = \frac{V \times R}{r}$.
Substituting the values $R = 10 \,cm$ and $r = 15 \,cm$:
$V(15) = \frac{V \times 10}{15} = \frac{2}{3} V$.

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$, where $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.
For point $A$, the distance from $+q$ is $x = 2 \text{ cm} = 0.02 \text{ m}$ and the distance from $-q$ is $(x+y) = 2+3 = 5 \text{ cm} = 0.05 \text{ m}$.
$V_A = \frac{kq}{x} + \frac{k(-q)}{x+y} = kq \left( \frac{1}{0.02} - \frac{1}{0.05} \right) = (9 \times 10^9)(10^{-6}) (50 - 20) = 9 \times 10^3 \times 30 = 2.7 \times 10^5 \text{ V}$.
For point $B$, the distance from $+q$ is $(x+y) = 5 \text{ cm} = 0.05 \text{ m}$ and the distance from $-q$ is $x = 2 \text{ cm} = 0.02 \text{ m}$.
$V_B = \frac{kq}{x+y} + \frac{k(-q)}{x} = kq \left( \frac{1}{0.05} - \frac{1}{0.02} \right) = (9 \times 10^9)(10^{-6}) (20 - 50) = 9 \times 10^3 \times (-30) = -2.7 \times 10^5 \text{ V}$.
The potential difference is $V_A - V_B = 2.7 \times 10^5 - (-2.7 \times 10^5) = 5.4 \times 10^5 \text{ V}$.

(C) Given, the side length of the regular hexagon is $r = 5 \, cm = 5 \times 10^{-2} \, m$.
In a regular hexagon, the distance from each vertex to the centre is equal to the side length of the hexagon. Thus, $r = 5 \times 10^{-2} \, m$.
The charge at each vertex is $q = 10 \, \mu C = 10 \times 10^{-6} \, C = 10^{-5} \, C$.
The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$, where $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
The potential due to one charge at the centre is:
$V_1 = \frac{9 \times 10^9 \times 10^{-5}}{5 \times 10^{-2}} = \frac{9}{5} \times 10^6 = 1.8 \times 10^6 \, V$.
Since there are $6$ identical charges at equal distances from the centre, the total potential $V_{\text{total}}$ is:
$V_{\text{total}} = 6 \times V_1 = 6 \times 1.8 \times 10^6 \, V = 10.8 \times 10^6 \, V = 1.08 \times 10^7 \, V$.

(C) The potential of the spherical shell is given as $V = 15 \times 10^6 \,V$.
The dielectric strength of the surrounding medium, which represents the maximum electric field intensity $E$ the medium can withstand before breakdown, is $E = 5 \times 10^7 \,V/m$.
For a spherical shell of radius $r$, the potential is related to the electric field at its surface by $V = E \times r$.
Therefore, the minimum radius $r$ required is $r = \frac{V}{E}$.
Substituting the given values: $r = \frac{15 \times 10^6 \,V}{5 \times 10^7 \,V/m} = 0.3 \,m$.
The diameter $d$ of the shell is $d = 2r = 2 \times 0.3 \,m = 0.6 \,m$.
Converting to centimeters, $d = 0.6 \times 100 \,cm = 60 \,cm$.

(D) The charges on the shells are $q_A = \sigma(4\pi a^2)$,$q_B = -\sigma(4\pi b^2)$,and $q_C = \sigma(4\pi c^2)$.
The potential at the surface of shell $A$ is $V_A = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{\sigma}{\epsilon_0} [a - b + c]$.
The potential at the surface of shell $C$ is $V_C = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{c} + \frac{q_B}{c} + \frac{q_C}{c}] = \frac{\sigma}{\epsilon_0 c} [a^2 - b^2 + c^2]$.
Given $V_A = V_C$,we equate the expressions: $a - b + c = \frac{a^2 - b^2 + c^2}{c}$.
$c(a - b + c) = a^2 - b^2 + c^2 \implies ac - bc + c^2 = a^2 - b^2 + c^2$.
$ac - bc = a^2 - b^2 \implies c(a - b) = (a - b)(a + b)$.
Since $a \neq b$,we divide by $(a - b)$ to get $c = a + b$.
Given $a = 7 \ cm$ and $b = 17 \ cm$,$c = 7 + 17 = 24 \ cm$.

(B) The potential $V$ of a conducting sphere of radius $R$ with charge $q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
Given $V = 10 \,V$, $R_1 = 0.09 \,m$, and $R_2 = 0.01 \,m$.
The charges on the spheres are $q_1 = 4\pi\epsilon_0 R_1 V$ and $q_2 = 4\pi\epsilon_0 R_2 V$.
Substituting the values, $q_1 = \frac{0.09 \times 10}{9 \times 10^9} = 10^{-10} \,C$ and $q_2 = \frac{0.01 \times 10}{9 \times 10^9} = \frac{1}{9} \times 10^{-10} \,C$.
The distance between the centers is $d = 0.2 \,m$.
The force of repulsion is $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2} = (9 \times 10^9) \times \frac{10^{-10} \times (1/9) \times 10^{-10}}{(0.2)^2}$.
$F = \frac{10^{-11}}{0.04} = \frac{10^{-11}}{4 \times 10^{-2}} = 0.25 \times 10^{-9} \,N = \frac{10^{-9}}{4} \,N$.

(A) The potential $V$ of a charged metal sphere is given by the formula $V = \frac{k q}{r}$, where $k = 9 \times 10^9 \, N \cdot m^2/C^2$ is Coulomb's constant, $q$ is the charge, and $r$ is the radius of the sphere.
Given: Diameter $d = 18 \, cm$, so radius $r = 9 \, cm = 9 \times 10^{-2} \, m$.
Potential $V = 25 \, kV = 25 \times 10^3 \, V$.
Rearranging the formula for charge $q$: $q = \frac{V \cdot r}{k}$.
Substituting the values: $q = \frac{25 \times 10^3 \times 9 \times 10^{-2}}{9 \times 10^9}$.
$q = \frac{25 \times 10^1}{10^9} = 25 \times 10^{-8} \, C$.
Converting to microcoulombs: $q = 0.25 \times 10^{-6} \, C = 0.25 \, \mu C$.

(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 points $A, B$ and $C$ lie on a semicircle with the charge $q$ at its centre,all these points are at the same distance $r$ (the radius of the semicircle) from the charge $q$.
Therefore,the electric potential at points $A, B$ and $C$ is the same,i.e.,$V_A = V_B = V_C = V$.
The work done in moving a charge $q_0$ from point $P$ to a point $X$ is given by $W = q_0(V_X - V_P)$.
Thus,the work done to move the charge to points $A, B$ and $C$ are:
$W_A = q_0(V_A - V_P) = q_0(V - V_P)$
$W_B = q_0(V_B - V_P) = q_0(V - V_P)$
$W_C = q_0(V_C - V_P) = q_0(V - V_P)$
Since $V_A = V_B = V_C$,it follows that $W_A = W_B = W_C$.
Since point $P$ is at a different distance from the charge $q$ compared to the points on the semicircle,$V_P \neq V$,therefore $W_A = W_B = W_C \neq 0$.

(A) The work done $W$ by a force $\vec{F}$ over a displacement $d\vec{r}$ is given by the dot product: $W = \int \vec{F} \cdot d\vec{r}$.
When a point charge moves in a circular path around another charge,the electrostatic force is always directed radially (towards or away from the center),while the displacement vector is always tangential to the circular path.
Since the angle between the radial force and the tangential displacement is $90^{\circ}$,the dot product $\vec{F} \cdot d\vec{r} = F dr \cos 90^{\circ} = 0$.
Thus,the work done is zero. Both the assertion and the reason are true,and the reason correctly explains why the work done is zero.

(C) The work done to assemble a system of point charges is equal to the electrostatic potential energy of the system.
For a system of three charges $q_1, q_2,$ and $q_3$ placed at distances $r_{12}, r_{23},$ and $r_{13}$ from each other,the potential energy $U$ is given by:
$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right)$
Given the charges $q_1 = -q$,$q_2 = +q$,and $q_3 = -2q$,and the distance between each pair is $a$:
$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{(-q)(q)}{a} + \frac{(q)(-2q)}{a} + \frac{(-q)(-2q)}{a} \right)$
$U = \frac{1}{4 \pi \varepsilon_0 a} (-q^2 - 2q^2 + 2q^2)$
$U = \frac{-q^2}{4 \pi \varepsilon_0 a}$

(C) The work done $W$ in moving a charge $q$ from point $B$ to point $C$ is given by $W = q(V_C - V_B)$,where $V_C$ and $V_B$ are the electric potentials at points $C$ and $B$ respectively.
The potential at any point is the sum of potentials due to the three charges: one at $O$ (charge $q$),one at $A$ (charge $q$),and one at $B$ (charge $q$).
Potential at $B$ $(V_B)$: The distance from $O$ to $B$ is $R$,from $A$ to $B$ is $2R$,and the charge at $B$ is the one being moved,so we consider the potential due to the other two charges: $V_B = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{R} + \frac{q}{2R} \right) = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{2R}$.
Potential at $C$ $(V_C)$: The distance from $O$ to $C$ is $R$,from $A$ to $C$ is $\sqrt{R^2 + R^2} = R\sqrt{2}$,and from $B$ to $C$ is $R\sqrt{2}$. The charge at $C$ is the one being moved,so we consider the potential due to the other two charges: $V_C = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{R} + \frac{q}{R\sqrt{2}} \right) = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R} \left( 1 + \frac{1}{\sqrt{2}} \right)$.
Work done $W = q(V_C - V_B) = q \cdot \frac{1}{4 \pi \varepsilon_{0}} \left[ \frac{q}{R} (1 + \frac{1}{\sqrt{2}}) - \frac{3q}{2R} \right] = \frac{q^2}{4 \pi \varepsilon_{0} R} \left[ 1 + \frac{1}{\sqrt{2}} - 1.5 \right] = \frac{q^2}{4 \pi \varepsilon_{0} R} \left[ \frac{\sqrt{2}}{2} - 0.5 \right] = \frac{q^2}{4 \pi \varepsilon_{0} R} \left( \frac{\sqrt{2}-1}{2} \right)$.

(D) Let $n = 729$ be the number of small spheres,each of radius $r$ and potential $V_s = 3 \ V$.
The potential of a small sphere is given by $V_s = \frac{k q}{r} = 3 \ V$.
When $n$ small spheres combine to form a bigger sphere of radius $R$,the total charge $Q = nq$ and the volume remains conserved.
Volume of big sphere = $n \times$ Volume of small sphere $\implies \frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$.
Thus,$R^3 = n r^3$,which means $R = n^{1/3} r$.
For $n = 729$,$R = (729)^{1/3} r = 9r$.
The potential of the bigger sphere is $V_B = \frac{k Q}{R} = \frac{k (nq)}{n^{1/3} r} = n^{2/3} \times \frac{k q}{r} = n^{2/3} V_s$.
Substituting the values: $V_B = (729)^{2/3} \times 3 \ V = (9^3)^{2/3} \times 3 \ V = 9^2 \times 3 \ V = 81 \times 3 \ V = 243 \ V$.

(A) For a hollow conducting sphere,the electric charge resides entirely on its outer surface.
Inside the sphere,the electric field is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -\frac{dV}{dr})$,if $E = 0$,then the potential $V$ must be constant throughout the interior of the sphere.
Therefore,the potential at any point inside the sphere is equal to the potential at its surface.
Given that the potential at the surface is $V$,the potential at a distance $\frac{R}{3}$ from the centre (which is inside the sphere) is also $V$.

(B) Let the centres of the rings be $A$ and $B$. The distance between them is $d = \sqrt{3} R$.
Potential at centre $A$ due to ring $1$ (charge $q$) is $V_{A1} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.
Potential at centre $A$ due to ring $2$ (charge $-q$) is $V_{A2} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + 3R^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{2R}$.
Total potential at $A$ is $V_A = V_{A1} + V_{A2} = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{R} - \frac{q}{2R}) = \frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.
Similarly,potential at centre $B$ due to ring $2$ (charge $-q$) is $V_{B2} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{R}$.
Potential at centre $B$ due to ring $1$ (charge $q$) is $V_{B1} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.
Total potential at $B$ is $V_B = V_{B1} + V_{B2} = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{2R} - \frac{q}{R}) = -\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.
The magnitude of potential difference is $|V_A - V_B| = |\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R} - (-\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R})| = |\frac{1}{4 \pi \varepsilon_0} \frac{2q}{2R}| = \frac{q}{4 \pi \varepsilon_0 R}$.