Electric potential Questions in English

(A) 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: $Q = 5\,\mu C = 5 \times 10^{-6}\,C$ and $W = 10\,mJ = 10 \times 10^{-3}\,J$.
Rearranging the formula to find the potential difference: $(V_B - V_A) = \frac{W}{Q}$.
Substituting the values: $(V_B - V_A) = \frac{10 \times 10^{-3}}{5 \times 10^{-6}} = 2 \times 10^3\,V = 2\,kV$.

(C) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by the formula:
$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r}$
From this expression,it is clear that the potential $V$ is inversely proportional to the distance $r$ from the charge $(V \propto \frac{1}{r})$.
Therefore,option $(c)$ is correct.

(B) The Earth is considered a good conductor. Due to its massive size,it can accept or provide an almost unlimited amount of charge without a significant change in its potential. By convention,the electric potential of the Earth is taken to be $0 \ V$.

(C) For a charged conducting sphere,the electric field inside the conductor is zero $(E = 0)$.
Since the electric field is the negative gradient of the electric potential $(E = -dV/dr)$,if $E = 0$,then $dV/dr = 0$,which implies that the potential $(V)$ is constant throughout the interior of the conductor.
Therefore,the electric potential at any point inside the conducting sphere is the same as the potential on its surface.

(C) For a conducting sphere of radius $R$ carrying a charge $Q$,the electric potential is constant throughout its interior and is equal to the potential at its surface.
Since the distance $R/2$ is less than the radius $R$,the point lies inside the conducting sphere.
Therefore,the potential at $R/2$ is the same as the potential at the surface,which is given by $V = \frac{Q}{4\pi \varepsilon_0 R}$.

(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$. The total potential is the algebraic sum of potentials due to individual charges.
Let the point where potential is zero be at position $x$. The potential is given by:
$V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1}{x} + \frac{q_2}{|x - 6|} \right] = 0$
Case $1$: Point between the charges $(0 < x < 6)$:
$\frac{2 \times 10^{-6}}{x} + \frac{-1 \times 10^{-6}}{6 - x} = 0$
$\frac{2}{x} = \frac{1}{6 - x} \implies 12 - 2x = x \implies 3x = 12 \implies x = 4$.
Case $2$: Point outside the charges $(x > 6)$:
$\frac{2 \times 10^{-6}}{x} + \frac{-1 \times 10^{-6}}{x - 6} = 0$
$\frac{2}{x} = \frac{1}{x - 6} \implies 2x - 12 = x \implies x = 12$.
Thus,the potential is zero at $x = 4$ and $x = 12$.

(B) Given: Mass $m = 2\,g = 2 \times 10^{-3}\,kg$, Charge $q = 2\,\mu C = 2 \times 10^{-6}\,C$, Final velocity $v = 10\,m/s$, Initial velocity $u = 0\,m/s$.
According to the work-energy theorem, the work done by the electric field is equal to the change in kinetic energy:
$W = \Delta K$
$qV = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$
Since $u = 0$, we have:
$qV = \frac{1}{2}mv^2$
$V = \frac{mv^2}{2q}$
Substituting the values:
$V = \frac{(2 \times 10^{-3}\,kg) \times (10\,m/s)^2}{2 \times (2 \times 10^{-6}\,C)}$
$V = \frac{2 \times 10^{-3} \times 100}{4 \times 10^{-6}}$
$V = \frac{2 \times 10^{-1}}{4 \times 10^{-6}} = 0.5 \times 10^5\,V = 50,000\,V = 50\,kV$.
Therefore, the required potential difference is $50\,kV$.

(C) positive charge naturally moves from a region of higher potential to a region of lower potential.
Since the potential inside the circle $S$ is positive and the potential outside is negative,the electric field lines will point from the inside to the outside.
Therefore,a free positive charge placed inside $S$ will experience an electric force directed towards the outside of $S$.
As a result,the charge will move from the region of positive potential to the region of negative potential,meaning it must cross the circle $S$ at some time.

(D) Let the centres of the two rings be $O_1$ and $O_2$. The potential at any point on the axis of a charged ring of radius $R$ and charge $Q$ at a distance $x$ from its centre is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{\sqrt{R^2 + x^2}}$.
For the first ring (charge $+q$) at centre $O_1$:
Potential due to itself is $V_{1, self} = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}$.
Potential due to the second ring (charge $-q$) at distance $d$ is $V_{1, other} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}}$.
So,$V_{O_1} = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.
For the second ring (charge $-q$) at centre $O_2$:
Potential due to itself is $V_{2, self} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{R}$.
Potential due to the first ring (charge $+q$) at distance $d$ is $V_{2, other} = \frac{1}{4\pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}}$.
So,$V_{O_2} = \frac{q}{4\pi \varepsilon_0} \left[ -\frac{1}{R} + \frac{1}{\sqrt{R^2 + d^2}} \right] = -V_{O_1}$.
The potential difference is $\Delta V = V_{O_1} - V_{O_2} = V_{O_1} - (-V_{O_1}) = 2V_{O_1}$.
$\Delta V = 2 \cdot \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right] = \frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.

(B) The work done on a charge $q$ moving between two points with a potential difference $\Delta V$ is given by $W = q \Delta V$.
Since the electron starts from rest,the work done by the electric field is equal to the change in its kinetic energy.
The potential difference is $\Delta V = V_{final} - V_{initial} = 70\, V - 50\, V = 20\, V$.
The charge of an electron is $q = 1.6 \times 10^{-19} \, C$.
Therefore,the kinetic energy $K.E. = q \times \Delta V = (1.6 \times 10^{-19} \, C) \times (20 \, V) = 3.2 \times 10^{-18} \, J$.

(B) According to the figure,there is no external electric field present.
Work done in moving a charge $q$ in an electric field is given by $W = q \Delta V$.
Since there is no external charge or field,the electric potential $V$ is zero everywhere in the region.
Therefore,the potential difference between any two points is zero.
Consequently,the work done in moving a point charge from point $P$ to any point $A$,$B$,or $C$ is zero.
Thus,$W_A = W_B = W_C = 0$.

(B) For a hollow metallic sphere of radius $R$ carrying a charge $Q$,the electric field inside the sphere is zero.
Since the electric field $E = -\frac{dV}{dr} = 0$,the potential $V$ must be constant throughout the interior of the sphere.
The potential at the surface of the sphere is given by $V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{R}$.
Since the potential is constant inside,the potential at the centre is the same as the potential at the surface,which is $V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{R}$.

(D) Current flows between two conductors only when there is a potential difference between them.
When two charged bodies are connected,charge flows from the body at a higher potential to the body at a lower potential.
If both bodies are at the same potential,there is no potential difference,and therefore,no current will flow between them.
Thus,the correct condition is that they must have the same potential.

(B) Let the radius of each small drop be $r$ and the charge be $q$. The potential of a small drop is $v = \frac{kq}{r}$.
When $n = 8$ drops combine to form a big drop of radius $R$,the volume remains constant: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$,which gives $R = n^{1/3} r = 8^{1/3} r = 2r$.
The total charge on the big drop is $Q = nq = 8q$.
The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left( \frac{kq}{r} \right) = 4v$.
Therefore,the ratio of the potential of the big drop to the small drop is $\frac{V}{v} = 4:1$.

(A) Let the radius of each small drop be $r$ and the charge on each be $q$.
The potential of a small drop is $v = \frac{kq}{r}$.
When $n = 27$ drops unite to form a big drop of radius $R$,the volume remains conserved:
$\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3 \implies R^3 = n r^3 \implies R = n^{1/3} r$.
For $n = 27$,$R = (27)^{1/3} r = 3r$.
The total charge on the big drop is $Q = nq = 27q$.
The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left( \frac{kq}{r} \right) = n^{2/3} v$.
Substituting $n = 27$,we get $V = (27)^{2/3} v = (3^3)^{2/3} v = 3^2 v = 9v$.
Thus,the potential increases by a factor of $9$.

(A) Let the radius of each small drop be $r = 2 \, cm = 0.02 \, m$ and the charge on each be $q = 10^{-9} \, C$.
The potential of a small drop is $v = \frac{kq}{r} = \frac{9 \times 10^9 \times 10^{-9}}{0.02} = \frac{9}{0.02} = 450 \, V$.
When $n = 64$ drops combine to form a bigger drop of radius $R$,the volume remains constant: $\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3$,so $R = n^{1/3}r$.
The total charge on the big drop is $Q = nq$.
The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \times \frac{kq}{r} = n^{2/3}v$.
Substituting the values: $V = (64)^{2/3} \times 450 = (4^3)^{2/3} \times 450 = 4^2 \times 450 = 16 \times 450 = 7200 \, V$.
Thus,$V = 7.2 \times 10^3 \, V$.

(A) Let the radius of each small drop be $r$ and its charge be $q$. The potential of a small drop is $v = \frac{kq}{r} = 10\,V$.
Let the radius of the big drop be $R$ and its charge be $Q$. Since the volume is conserved,the volume of $64$ small drops equals the volume of the big drop: $\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$,which gives $R^3 = 64r^3$,so $R = 4r$.
The total charge of the big drop is $Q = 64q$.
The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \times \frac{kq}{r}$.
Substituting $v = \frac{kq}{r} = 10\,V$,we get $V = 16 \times 10 = 160\,V$.

(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is $V = \frac{kq}{r}$.
When $n = 8$ drops are joined to form a bigger drop of radius $R$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which gives $R = n^{1/3}r = 8^{1/3}r = 2r$.
The total charge on the bigger drop is $Q = nq = 8q$.
The potential of the bigger drop is $V' = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left( \frac{kq}{r} \right) = n^{2/3}V$.
Substituting $n = 8$,we get $V' = 8^{2/3}V = (2^3)^{2/3}V = 2^2 V = 4V$.
Therefore,the potential on the bigger drop is $4$ times that of the smaller drop.

(C) The electric potential $V$ of a conductor is given by the relation $V = Q/C$.
Here,$Q$ represents the amount of charge given to the conductor.
$C$ represents the capacitance of the conductor,which is determined by its geometry,size,and the surrounding medium.
Therefore,the potential $V$ depends on both the amount of charge $Q$ and the capacitance $C$ (which depends on geometry and size).
Thus,the correct option is $(c)$.

(D) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively. Given that $q_1 + q_2 = Q$.
Since surface charge densities are equal,$\sigma = \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}$.
This implies $\frac{q_1}{r^2} = \frac{q_2}{R^2}$,so $q_1 = q_2 \frac{r^2}{R^2}$.
Substituting this into the total charge equation: $q_2 \frac{r^2}{R^2} + q_2 = Q \implies q_2 \left( \frac{r^2 + R^2}{R^2} \right) = Q \implies q_2 = \frac{Q R^2}{R^2 + r^2}$.
Similarly,$q_1 = \frac{Q r^2}{R^2 + r^2}$.
The potential at the common centre is the sum of potentials due to both spheres: $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.
Substituting the values of $q_1$ and $q_2$: $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right)$.
$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r}{R^2 + r^2} + \frac{Q R}{R^2 + r^2} \right) = \frac{Q(R + r)}{4\pi \varepsilon_0(R^2 + r^2)}$.

(B) The work done $W$ in moving a charge $q$ from the centre of the first ring $(O_1)$ to the centre of the second ring $(O_2)$ is given by $W = q(V_{O_2} - V_{O_1})$.
The potential at the centre of the first ring $(O_1)$ due to both rings is:
$V_{O_1} = \frac{Q_1}{4\pi \varepsilon_0 R} + \frac{Q_2}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} \right)$.
The potential at the centre of the second ring $(O_2)$ due to both rings is:
$V_{O_2} = \frac{Q_2}{4\pi \varepsilon_0 R} + \frac{Q_1}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} \right)$.
The potential difference is:
$V_{O_2} - V_{O_1} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} - Q_1 - \frac{Q_2}{\sqrt{2}} \right) = \frac{1}{4\pi \varepsilon_0 R} \left( (Q_2 - Q_1) - \frac{1}{\sqrt{2}}(Q_2 - Q_1) \right) = \frac{Q_2 - Q_1}{4\pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} \right) = \frac{(Q_2 - Q_1)(\sqrt{2} - 1)}{\sqrt{2} \cdot 4\pi \varepsilon_0 R}$.
Therefore,the work done is $W = \frac{q(Q_2 - Q_1)(\sqrt{2} - 1)}{\sqrt{2} \cdot 4\pi \varepsilon_0 R}$.

(D) The electric potential $V$ at the origin due to a system of point charges is given by the sum $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.
Substituting the given positions and charges:
$V = \frac{1}{4\pi\varepsilon_0} \left[ \left( \frac{q}{x_0} + \frac{q}{3x_0} + \frac{q}{5x_0} + \dots \right) - \left( \frac{q}{2x_0} + \frac{q}{4x_0} + \frac{q}{6x_0} + \dots \right) \right]$
$V = \frac{q}{4\pi\varepsilon_0 x_0} \left[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \right]$
Using the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$, for $x=1$, we get $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
Therefore, $V = \frac{q \ln 2}{4\pi\varepsilon_0 x_0}$.

(A) The line integral of the electric field from infinity to a point $P$ is defined as the electric potential $V$ at that point.
$\int_{\infty}^{P} -\vec{E} \cdot d\vec{l} = V_P - V_{\infty}$.
Since the potential at infinity $V_{\infty} = 0$,the integral represents the potential at the centre of the ring.
For a ring of radius $R$ with total charge $q$,the potential at the centre is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
Given $q = 1.11 \times 10^{-10}\,C$,$R = 0.5\,m$,and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$.
$V = (9 \times 10^9) \times \frac{1.11 \times 10^{-10}}{0.5} = \frac{9.99 \times 10^{-1}}{0.5} = \frac{0.999}{0.5} \approx 2\,V$.

(B) Initially,according to figure $(i)$,the potential energy of $Q$ is $U_i = \frac{2kqQ}{a}$ ... $(i)$
According to figure $(ii)$,when charge $Q$ is displaced by a small distance $x$,its potential energy becomes:
$U_f = kqQ \left[ \frac{1}{a+x} + \frac{1}{a-x} \right] = kqQ \left[ \frac{(a-x) + (a+x)}{a^2 - x^2} \right] = \frac{2kqQa}{a^2 - x^2}$ ... $(ii)$
Hence,the change in potential energy is:
$\Delta U = U_f - U_i = 2kqQ \left[ \frac{a}{a^2 - x^2} - \frac{1}{a} \right] = 2kqQ \left[ \frac{a^2 - (a^2 - x^2)}{a(a^2 - x^2)} \right] = \frac{2kqQx^2}{a(a^2 - x^2)}$
Since $x << a$,we can approximate $a^2 - x^2 \approx a^2$. Therefore:
$\Delta U \approx \frac{2kqQx^2}{a(a^2)} = \frac{2kqQ}{a^3} x^2$
Thus,$\Delta U \propto x^2$.

(A) Let the radius of the solid sphere be $a$ and the radius of the hollow shell be $b$.
Initially,the solid sphere has charge $Q$ and the shell has charge $0$.
The potential at the surface of the solid sphere is $V_{\text{sphere}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a} + \frac{1}{4\pi\varepsilon_0} \frac{0}{b} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a}$.
The potential at the surface of the hollow shell is $V_{\text{shell}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi\varepsilon_0} \frac{0}{b} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b}$.
The initial potential difference is $V = V_{\text{sphere}} - V_{\text{shell}} = \frac{Q}{4\pi\varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right)$.
Now,the shell is given a charge of $-3Q$.
The new potential at the surface of the solid sphere is $V'_{\text{sphere}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a} + \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{b}$.
The new potential at the surface of the hollow shell is $V'_{\text{shell}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{b} = \frac{1}{4\pi\varepsilon_0} \left( -\frac{2Q}{b} \right)$.
The new potential difference is $V' = V'_{\text{sphere}} - V'_{\text{shell}} = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} - (-\frac{2Q}{b}) \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{a} - \frac{Q}{b} \right) = V$.

(B) For a hollow spherical shell of radius $R$ carrying charge $Q$:
$1$. Inside the shell $(r \le R)$,the electric field is zero,which implies that the potential is constant and equal to the potential at the surface: $V_{inside} = \frac{Q}{4\pi \varepsilon_0 R}$.
$2$. Outside the shell $(r \ge R)$,the shell behaves like a point charge placed at the centre,so the potential varies as $V_{out} = \frac{Q}{4\pi \varepsilon_0 r}$,which means $V \propto \frac{1}{r}$.
$3$. Comparing this with the given options,Graph $B$ correctly shows a constant potential for $r \le R$ and a hyperbolic decrease for $r > R$.

(B) The kinetic energy $K$ gained by a charged particle accelerated through a potential difference $V$ is given by the formula $K = qV$.
Here,the charge of the particle is $q = 2e$,where $e$ is the elementary charge of an electron.
The potential difference is $V = 5 \; V$.
Substituting these values into the formula:
$K = (2e) \times (5 \; V)$
$K = 10 \; eV$.
Therefore,the final kinetic energy of the particle is $10 \; eV$.

(D) The energy $E$ acquired by a charged particle accelerated through a potential difference $V$ is given by the formula $E = qV$.
Here,the charge of the electron is $q = 1.6 \times 10^{-19} \text{ C}$ and the potential difference is $V = 100,000 \text{ V} = 10^5 \text{ V}$.
Substituting these values into the formula:
$E = (1.6 \times 10^{-19} \text{ C}) \times (10^5 \text{ V})$
$E = 1.6 \times 10^{-14} \text{ J}$.
Therefore,the correct option is $D$.

(A) When an electron of charge $e$ is accelerated through a potential difference $V$,the work done on the electron by the electric field is $W = eV$.
This work done is converted into the kinetic energy $(K)$ of the electron.
Therefore,$K = eV$.
As the potential difference $V$ increases,the kinetic energy of the electron increases.
Thus,the energy of electrons can be increased by allowing them to fall through an electric potential.

(B) The work done on a charge $q$ moving through a potential difference $V$ is given by the formula $W = qV$.
Here,the charge of the electron is $q = e$ and the potential difference is $V$.
Therefore,the energy acquired by the electron is $K = e \times V = eV$ joules.

(D) The correct option is $D$.
The electron volt (symbol $eV$) is a unit of energy.
By definition,$1 \ eV$ is the amount of kinetic energy gained by a single electron accelerating from rest through an electric potential difference of $1 \ V$.
Mathematically,$1 \ eV = 1.602 \times 10^{-19} \ J$,where $J$ is the $SI$ unit of energy,the Joule.

(C) For a charged conducting sphere of radius $R$,the charge $Q$ resides entirely on its surface.
Inside the sphere (for distance $r < R$),the electric field is zero.
Since the electric field is zero inside the conductor,the potential remains constant and is equal to the potential at the surface.
The potential at the surface of the sphere is given by $V = \frac{Q}{4\pi \epsilon_0 R}$.
Therefore,at any point inside the sphere,including the distance $R/2$ from the center,the potential is $V = \frac{Q}{4\pi \epsilon_0 R}$.

(B) The electric potential at a distance $x$ from the center of a charged ring of radius $R$ along its axis is $V = \frac{q}{4\pi \epsilon_0 \sqrt{x^2 + R^2}}$.
Let the centers of the rings be $C_1$ and $C_2$,separated by distance $d$. The potential at $C_1$ is the sum of the potential due to the ring at $C_1$ (charge $+q$) and the potential due to the ring at $C_2$ (charge $-q$):
$V_1 = \frac{q}{4\pi \epsilon_0 R} + \frac{-q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}}$.
The potential at $C_2$ is the sum of the potential due to the ring at $C_2$ (charge $-q$) and the potential due to the ring at $C_1$ (charge $+q$):
$V_2 = \frac{-q}{4\pi \epsilon_0 R} + \frac{q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}}$.
The potential difference between the centers is $\Delta V = V_1 - V_2$:
$\Delta V = \left( \frac{q}{4\pi \epsilon_0 R} - \frac{q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}} \right) - \left( -\frac{q}{4\pi \epsilon_0 R} + \frac{q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}} \right)$
$\Delta V = \frac{q}{4\pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} + \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} \right)$
$\Delta V = \frac{q}{4\pi \epsilon_0} \cdot 2 \left( \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} \right) = \frac{q}{2\pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} \right)$.

(C) For a uniformly charged sphere of radius $a$ and total charge $Q$,the electric potential $V$ at a distance $r$ from the center,where $r > a$,is given by the formula:
$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r}$
Here,$\frac{1}{4\pi\epsilon_0}$ and $Q$ are constants.
Therefore,the electric potential $V$ is inversely proportional to the distance $r$ from the center of the sphere $(V \propto \frac{1}{r})$.
Thus,the correct option is $C$.

(D) $1$. Option $A$ is incorrect because Gauss's law is a direct consequence of the inverse square law $(1/r^2)$. If the field varied as $1/r^n$ where $n \neq 2$,the flux would depend on the shape of the surface,not just the enclosed charge.
$2$. Option $B$ is incorrect because while Gauss's law is always true,it is only useful for calculating field distributions when there is high symmetry (like spherical,cylindrical,or planar symmetry). $A$ dipole lacks such symmetry.
$3$. Option $C$ is incorrect because for two point charges,the electric field is zero at a point between them only if the charges have the same sign (repulsive force). If they had opposite signs,the field would be zero outside the region between them.
$4$. Option $D$ is correct. By definition,the potential difference $V_B - V_A$ is the work done by an external agent in moving a unit positive charge from $A$ to $B$ without acceleration.

(A) For a conducting sphere,the charge $Q$ resides entirely on its outer surface.
Since the electric field $E$ inside a conducting sphere is zero,the potential $V$ remains constant throughout the interior.
The potential inside the sphere is equal to the potential at its surface.
Therefore,at any distance $x < R$ from the center,the potential is $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.

(A) For a hollow metallic 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 the center $(V_C)$ is equal to the potential at the surface $(V_S)$.
Given $V_S = 10\, V$,the potential at the center is $V_C = 10\, V$.

(D) The electric potential $V$ due to a point charge $q$ at a distance $x$ is given by $V = \frac{kq}{x}$.
For two charges $q_1 = -20 \,\mu C$ and $q_2 = +40 \,\mu C$ separated by distance $r$,the total potential $V = V_1 + V_2 = 0$.
This implies $\frac{kq_1}{x_1} + \frac{kq_2}{x_2} = 0$,or $\frac{|q_1|}{x_1} = \frac{|q_2|}{x_2}$.
Since $|q_1| < |q_2|$,the point where potential is zero must be closer to the smaller charge $(-20 \,\mu C)$.
$1$. In Region $A$ (to the left of $-20 \,\mu C$): Let the distance from $-20 \,\mu C$ be $x$. Then $\frac{20}{x} = \frac{40}{r+x} \implies 20r + 20x = 40x \implies 20x = 20r \implies x = r$. Thus,potential is zero in Region $A$.
$2$. In Region $B$ (between the charges): Let the distance from $-20 \,\mu C$ be $x$. Then $\frac{20}{x} = \frac{40}{r-x} \implies 20r - 20x = 40x \implies 60x = 20r \implies x = r/3$. Thus,potential is zero in Region $B$.
$3$. In Region $C$ (to the right of $+40 \,\mu C$): Since the magnitude of the positive charge is larger,the potential will never be zero here.
Therefore,the potential is zero in both Region $A$ and Region $B$.

(D) The given graph shows a rectangular hyperbola,which represents an inverse relationship between the two variables,i.e.,$Y \propto 1/X$.
From the formula for electric potential,$V = Q/C$.
For a constant charge $Q$,the potential $V$ is inversely proportional to the capacitance $C$,i.e.,$V \propto 1/C$.
This relationship is represented by a rectangular hyperbola where $V$ is on the $Y$-axis and $C$ is on the $X$-axis.
Therefore,the graph represents electric potential $(V)$ versus capacitance $(C)$ for a constant charge $(Q)$.

(C) For a charged spherical shell of radius $R$ and charge $Q$:
$1$. Inside the shell $(r < R)$,the electric field is zero,which means the electric potential is constant and equal to the potential at the surface,i.e.,$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
$2$. Outside the shell $(r \geq R)$,the shell behaves like a point charge at its center,so the potential is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$,which means $V \propto \frac{1}{r}$.
Thus,the graph of $V$ versus $r$ should be a horizontal line for $r < R$ and a curve following $V \propto \frac{1}{r}$ for $r \geq R$. This corresponds to graph $C$.

(A) Volt is the electrical unit of voltage or potential difference $(V)$.
- $V$ is related to the change in potential energy $\Delta U$ as:
$V = \frac{\Delta U}{q}$
From the above formula,we get $1$ Volt of potential difference when a change of $1$ Joule of potential energy occurs in moving a charge of $1$ Coulomb.
Therefore,$1$ Volt $= 1$ Joule/Coulomb $(J/C)$.

(B) $1$. Electric potential is defined as the work done per unit positive charge to bring it from infinity to a point,which is equivalent to potential energy per unit charge. Thus,statement $A$ is correct.
$2$. Electrostatic force is a conservative force. The work done by a conservative force is independent of the path taken and depends only on the initial and final positions. Therefore,statement $B$ is incorrect because it claims the work depends on the path.
$3$. When a positive charge moves against the repulsive Coulomb force,external work is done on the system,which increases the potential energy of the system. Thus,statement $C$ is correct.
$4$. The elementary charge $(e \approx 1.6 \times 10^{-19} \ C)$ is a fundamental constant,while 'charge' is a physical property that can be any integer multiple of $e$. Thus,statement $D$ is correct.

(A) Let the radius of the inner solid sphere be $r$ and the radius of the outer shell be $R$.
According to Gauss's law,the electric field $E$ in the region between the sphere and the shell $(r < x < R)$ depends only on the charge enclosed by the Gaussian surface of radius $x$.
The charge enclosed is the charge on the inner sphere,which is $Q$.
Thus,$E = \frac{Q}{4 \pi \epsilon_0 x^2}$.
The potential difference $V$ between the surface of the inner sphere and the surface of the outer shell is given by $V = \int_{r}^{R} E \, dx = \int_{r}^{R} \frac{Q}{4 \pi \epsilon_0 x^2} \, dx$.
Evaluating the integral,$V = \frac{Q}{4 \pi \epsilon_0} \left[ -\frac{1}{x} \right]_{r}^{R} = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R} \right)$.
Since the potential difference $V$ depends only on the charge $Q$ of the inner sphere and the radii $r$ and $R$,changing the charge on the outer shell does not affect the potential difference between the two surfaces.
Therefore,the new potential difference remains $V$.

(D) The electrostatic force is a conservative force.
By definition,a force is conservative if the work done by it on a particle moving between two points is independent of the path taken.
Therefore,Statement-$1$ is true.
Furthermore,a fundamental property of conservative forces is that the net work done by them over any closed path is zero.
Since the electrostatic force is conservative,it follows this property.
Thus,Statement-$2$ is true and it provides the correct explanation for why the work done is path-independent (Statement-$1$).
Hence,the correct option is $D$.

(A) For a charged hollow conducting sphere of radius $R$ and charge $Q$,the electric field $E$ inside the sphere $(r < R)$ is zero.
Since the electric field $E = -\frac{dV}{dr}$,if $E = 0$,then the potential $V$ must be constant throughout the interior of the sphere.
The value of the potential inside the sphere is equal to the potential at the surface,which is $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
Therefore,the potential inside a charged hollow conducting sphere is constant.

(B) Let $\sigma$ be the surface charge density. The charges on the spheres are $q_1 = 4\pi r^2 \sigma$ and $q_2 = 4\pi R^2 \sigma$.
The total charge $Q = q_1 + q_2 = 4\pi \sigma (r^2 + R^2)$,so $\sigma = \frac{Q}{4\pi (r^2 + R^2)}$.
The potential at the center due to the inner sphere is $V_1 = \frac{1}{4\pi \varepsilon_0} \frac{q_1}{r} = \frac{1}{4\pi \varepsilon_0} \frac{4\pi r^2 \sigma}{r} = \frac{r \sigma}{\varepsilon_0}$.
The potential at the center due to the outer sphere is $V_2 = \frac{1}{4\pi \varepsilon_0} \frac{q_2}{R} = \frac{1}{4\pi \varepsilon_0} \frac{4\pi R^2 \sigma}{R} = \frac{R \sigma}{\varepsilon_0}$.
The total potential at the center is $V = V_1 + V_2 = \frac{\sigma}{\varepsilon_0} (r + R)$.
Substituting $\sigma = \frac{Q}{4\pi (r^2 + R^2)}$,we get $V = \frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{(R^2 + r^2)}$.

(A) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$,where $k = \frac{1}{4\pi\varepsilon_0}$.
At the midpoint $O$,the distance from both charges $+q$ and $-q$ is $d$.
The total electric potential $V_{total}$ at the midpoint is the algebraic sum of the potentials due to both charges:
$V_{total} = V_{+q} + V_{-q}$
$V_{total} = \frac{k(+q)}{d} + \frac{k(-q)}{d}$
$V_{total} = \frac{kq}{d} - \frac{kq}{d} = 0$.

(A) The electric field is given as $\vec{E} = E_0 \hat{i}$,where $E_0 > 0$.
$1$. For points $A(0,0)$ and $B(1,0)$: The potential difference is $V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} = -\int_0^1 E_0 dx = -E_0$.
Since $E_0 > 0$,$V_B - V_A < 0$,which implies $V_A > V_B$.
$2$. For points $A(0,0)$ and $C(0,1)$: The potential difference is $V_C - V_A = -\int_A^C \vec{E} \cdot d\vec{r}$. Since the electric field is in the $x$-direction and the displacement $d\vec{r}$ is in the $y$-direction,$\vec{E} \cdot d\vec{r} = 0$.
Therefore,$V_C - V_A = 0$,which implies $V_A = V_C$.
Combining these,we get $V_A > V_B$ and $V_A = V_C$.

(C) The distance of each corner of the cube from its center is $r = \frac{\sqrt{3}b}{2}$.
There are $8$ charges of magnitude $(-q)$ at the corners.
The potential energy $U$ of a charge $q_0$ at the center is given by $U = \sum \frac{1}{4\pi \epsilon_0} \frac{q_i q_0}{r}$.
Here,$q_i = -q$,$q_0 = q$,and $r = \frac{\sqrt{3}b}{2}$.
$U = 8 \times \left[ \frac{1}{4\pi \epsilon_0} \frac{(-q)(q)}{\frac{\sqrt{3}b}{2}} \right]$.
$U = 8 \times \left[ \frac{-q^2}{4\pi \epsilon_0} \times \frac{2}{\sqrt{3}b} \right]$.
$U = \frac{-16q^2}{4\sqrt{3}\pi \epsilon_0 b} = \frac{-4q^2}{\sqrt{3}\pi \epsilon_0 b}$.

(A) The electric potential $V$ at the origin due to a point charge $q$ at distance $x$ is given by $V = \frac{kq}{x}$.
For the given distribution,the charges are $q$ at $x=1$,$-q$ at $x=2$,$q$ at $x=3$,$-q$ at $x=4$,and so on.
The total potential at the origin is:
$V = kq \left( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right)$
Using the Taylor series expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$,for $x=1$,we get $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
Therefore,$V = kq \ln(2)$.