Electric potential Questions in English

(D) The electron volt (symbol $eV$) is a unit of energy equal to approximately $1.602 \times 10^{-19} \text{ J}$.
By definition,it is the amount of energy gained by the charge of a single electron when it is moved across an electric potential difference of $1 \text{ V}$.
Since it represents the work done or energy transferred,it is a unit of energy.
Hence,the correct option is $D$.

(B) The electron volt $(eV)$ 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 $1\,volt$.
Since the charge of an electron is $q = 1.6 \times 10^{-19}\,C$ and the potential difference is $V = 1\,V$,the energy $E$ is given by:
$E = q \times V = (1.6 \times 10^{-19}\,C) \times (1\,V) = 1.6 \times 10^{-19}\,J$.
Therefore,the correct option is $B$.

(A) Electric potential $V$ is defined as the work done $W$ per unit charge $Q$,given by the formula $V = \frac{W}{Q}$.
The dimensional formula for work $W$ is $[M L^2 T^{-2}]$.
The dimensional formula for charge $Q$ is $[Q]$.
Substituting these into the formula:
$V = \frac{[M L^2 T^{-2}]}{[Q]} = [M L^2 T^{-2} Q^{-1}]$.
Therefore,the correct option is $A$.

(B) The kinetic energy $K.E.$ acquired by a charged particle accelerated through a potential difference $V$ is given by the formula $K.E. = qV$.
For an electron,the charge $q = e = 1.6 \times 10^{-19} \, C$.
Given the potential difference $V = 1 \, V$.
Therefore,the energy acquired is $K.E. = (1.6 \times 10^{-19} \, C) \times (1 \, V) = 1.6 \times 10^{-19} \, J$.
By definition,this amount of energy is known as $1 \, eV$ (electron volt).

(A) When a particle of charge $q$ is accelerated through a potential difference $V$,the work done on the particle by the electric field is given by $W = qV$.
According to the work-energy theorem,this work done is equal to the change in the kinetic energy of the particle.
Assuming the particle starts from rest,its final kinetic energy $K$ is equal to the work done.
Therefore,$K = qV$.

(A) For a hollow charged spherical conductor of radius $R$ and charge $Q$,the electric field $E$ inside the conductor is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -dV/dr)$,if $E = 0$,then $dV/dr = 0$.
This implies that the potential $V$ is constant throughout the interior of the conductor.
The value of this constant potential is equal to the potential at the surface,which is $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.

(B) Electric charge flows between two conductors only when there is a potential difference between them.
When two charged spheres are connected by a thin wire,charge will flow from the sphere at a higher potential to the sphere at a lower potential.
This flow of charge continues until both spheres reach the same electrical potential.
Once the potentials are equal,the potential difference becomes zero,and consequently,no further current (charge flow) will occur.
Therefore,no current will flow if they have the same potential.

(B) 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 $10\, V$,the potential at the centre is also $10\, V$.

(B) The potential $V$ at the center $O$ of the square due to four identical charges $q$ at the corners is given by the sum of potentials due to each charge.
$V = 4 \times \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r}$,where $r$ is the distance from each corner to the center.
For a square of side $a = 8 \, cm = 8 \times 10^{-2} \, m$,the distance $r$ from the corner to the center is half the diagonal length:
$r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} = \frac{8 \times 10^{-2}}{\sqrt{2}} \, m$.
Given $q = \frac{10}{3} \times 10^{-9} \, C$ and $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2$.
Substituting the values:
$V = 4 \times (9 \times 10^9) \times \frac{\frac{10}{3} \times 10^{-9}}{\frac{8 \times 10^{-2}}{\sqrt{2}}}$
$V = 4 \times 9 \times 10^9 \times \frac{10}{3} \times 10^{-9} \times \frac{\sqrt{2}}{8 \times 10^{-2}}$
$V = 120 \times \frac{\sqrt{2}}{8 \times 10^{-2}} = 15 \times 10^2 \times \sqrt{2} = 1500\sqrt{2} \, V$.

(D) The electric potential of a conductor depends on the reference point chosen for zero potential.
If a positively charged conductor is placed in the vicinity of other charges,its potential is determined by the superposition of the potential due to its own charge and the potential due to external charges.
By choosing an appropriate reference point or placing the conductor in an external electric field,the potential of a positively charged conductor can be made positive,zero,or even negative.
Therefore,the correct option is $(d)$.

(C) The electrostatic potential energy of a system of point charges is given by the sum of the potential energies of all distinct pairs of charges: $U = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i q_j}{r_{ij}}$.
For an equilateral triangle with three identical charges $q = 10\,\mu C = 10^{-5}\,C$ and side length $r = 10\,cm = 0.1\,m$,there are three identical pairs.
$U_{system} = 3 \times \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r} \right)$.
Substituting the values: $U_{system} = 3 \times (9 \times 10^9) \times \frac{(10^{-5})^2}{0.1}$.
$U_{system} = 27 \times 10^9 \times \frac{10^{-10}}{0.1} = 27 \times 10^9 \times 10^{-9} = 27\,J$.

(C) The distance of each corner from the center $O$ of the square is $r = \frac{a}{\sqrt{2}}$.
The electric potential $V_O$ at the center $O$ due to the four charges $Q$ at the corners is:
$V_O = 4 \times \left( \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r} \right) = 4 \times \left( \frac{Q}{4\pi \varepsilon_0 (a/\sqrt{2})} \right) = \frac{4\sqrt{2} Q}{4\pi \varepsilon_0 a} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 a}$.
The work done $W$ in moving a charge $q = -Q$ from the center to infinity is given by $W = q(V_{\infty} - V_O)$.
Since the potential at infinity $V_{\infty} = 0$,we have:
$W = (-Q)(0 - V_O) = Q V_O$.
Substituting the value of $V_O$:
$W = Q \times \left( \frac{\sqrt{2} Q}{\pi \varepsilon_0 a} \right) = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 a}$.

(A) The side of the square is $a = 2\,m$. The distance from any corner to the centre of the square is $r = \frac{a}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\,m$.
Potential at the centre due to one charge $q = 50\,\mu C$ is $V_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} = 9 \times 10^9 \times \frac{50 \times 10^{-6}}{\sqrt{2}} = \frac{45 \times 10^4}{\sqrt{2}}\,V$.
Since there are four identical charges at the corners,the total potential at the centre is $V = 4 \times V_1 = 4 \times \frac{45 \times 10^4}{\sqrt{2}} = 180 \times 10^4 \times \frac{1}{\sqrt{2}} = 90\sqrt{2} \times 10^4\,V$.
The work done $W$ to bring a charge $q' = 50\,\mu C$ from infinity to the centre is $W = q'V$.
$W = (50 \times 10^{-6}) \times (90\sqrt{2} \times 10^4) = 4500\sqrt{2} \times 10^{-2} = 45\sqrt{2} \approx 45 \times 1.414 = 63.63\,J \approx 64\,J$.

(A) The electric potential $V$ of a charged conducting sphere of radius $R$ carrying a charge $q$ is given by the formula $V = \frac{kq}{R}$,where $k$ is Coulomb's constant.
Since the charges $q$ on both spheres are equal,the potential $V$ is inversely proportional to the radius $R$ $(V \propto \frac{1}{R})$.
Therefore,the sphere with the smaller radius will have a higher potential compared to the sphere with the larger radius.

(A) The work done $W$ in moving a charge $q$ between two points with a potential difference $V$ is given by the formula: $W = qV$.
Given:
Charge $q = 5\,C$
Work done $W = 10\,J$
We need to find the potential difference $V$.
Using the formula $V = W / q$,we get:
$V = 10\,J / 5\,C = 2\,V$.
Therefore,the potential difference between the two points is $2\,V$.

(D) The electric field $E$ at the surface of a sphere of radius $r$ with potential $V$ is given by $E = \frac{V}{r}$.
Given $V = 8000\,V$ and $r = 1\,cm = 0.01\,m$.
So,$E = \frac{8000}{0.01} = 8 \times 10^5\,V/m$.
The energy density $u_e$ is given by the formula $u_e = \frac{1}{2} \varepsilon_0 E^2$.
Substituting the values: $u_e = \frac{1}{2} \times 8.854 \times 10^{-12} \times (8 \times 10^5)^2$.
$u_e = 0.5 \times 8.854 \times 10^{-12} \times 64 \times 10^{10}$.
$u_e = 283.328 \times 10^{-2} = 2.833\,J/m^3$.
Thus,the correct option is $D$.

(A) The kinetic energy gained by a charged particle moving through a potential difference is given by the formula: $\Delta KE = q \times \Delta V$.
An $\alpha$-particle has a charge of $q = +2e$,where $e$ is the elementary charge.
The potential difference between the two points is $\Delta V = V_1 - V_2 = 70\,V - 50\,V = 20\,V$.
Substituting these values into the formula:
$\Delta KE = 2e \times 20\,V = 40\,eV$.
Therefore,the kinetic energy gained by the $\alpha$-particle is $40\,eV$.

(B) For a charged spherical conductor of radius $R = 10\,cm$,the potential inside the sphere (at $r < R$) is constant and equal to the potential at the surface.
Given,potential at $r_1 = 5\,cm$ is $V$. Since $5\,cm < 10\,cm$,this is the potential at the surface: $V = \frac{kQ}{R} = \frac{kQ}{10}$.
For a point outside the sphere (at $r_2 = 15\,cm$),the potential is given by $V_{out} = \frac{kQ}{r_2} = \frac{kQ}{15}$.
Taking the ratio: $\frac{V_{out}}{V} = \frac{kQ/15}{kQ/10} = \frac{10}{15} = \frac{2}{3}$.
Therefore,$V_{out} = \frac{2}{3}\,V$.

(A) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r}$.
Let the two charges be $+q$ and $-q$,separated by a distance $2d$. The midpoint $O$ is at a distance $d$ from both charges.
The potential at the midpoint $O$ is the algebraic sum of the potentials due to both charges:
$V_O = V_{+q} + V_{-q}$
$V_O = \frac{1}{4\pi \varepsilon_0} \frac{q}{d} + \frac{1}{4\pi \varepsilon_0} \frac{(-q)}{d}$
$V_O = \frac{1}{4\pi \varepsilon_0} \left( \frac{q}{d} - \frac{q}{d} \right) = 0$
Therefore,the potential at the midpoint is zero.

(B) The kinetic energy $K$ gained by a charged particle accelerated through a potential difference $V$ is given by the formula $K = Q \cdot V$.
For a proton,the charge $Q = +e = 1.6 \times 10^{-19} \, C$.
The potential difference $V = 50,000 \, V$.
Substituting these values into the formula:
$K = (1.6 \times 10^{-19} \, C) \times (50,000 \, V)$
$K = 1.6 \times 10^{-19} \times 5 \times 10^4 \, J$
$K = 8 \times 10^{-15} \, J$.
Therefore,the energy increases by $8 \times 10^{-15} \, J$.

(B) The electric potential $V$ of a charged conducting sphere of radius $R$ carrying charge $Q$ is given by $V = k \frac{Q}{R}$,where $k = \frac{1}{4 \pi \epsilon_0}$.
Since both spheres are charged to the same potential $V$,we have:
$V_1 = V_2$
$k \frac{Q_1}{R_1} = k \frac{Q_2}{R_2}$
Dividing both sides by $k$,we get:
$\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$
Rearranging the terms to find the ratio of charges $\frac{Q_1}{Q_2}$:
$\frac{Q_1}{Q_2} = \frac{R_1}{R_2}$
Thus,the ratio of charges on the spheres is $R_1 : R_2$.

(C) The length of each side of the square is $a = \sqrt{2}\,m$. The diagonal of the square is $d = a\sqrt{2} = \sqrt{2} \times \sqrt{2} = 2\,m$. The distance of the centre from each corner is $r = d/2 = 2/2 = 1\,m$.
The electric potential $V$ at the centre due to multiple charges is given by the sum of potentials due to individual charges: $V = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i}{r_i}$.
Here,$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$ and $r = 1\,m$ for all charges.
$V = (9 \times 10^9) \times \left[ \frac{10 \times 10^{-6}}{1} + \frac{5 \times 10^{-6}}{1} + \frac{-3 \times 10^{-6}}{1} + \frac{8 \times 10^{-6}}{1} \right]$
$V = 9 \times 10^9 \times (10 + 5 - 3 + 8) \times 10^{-6}$
$V = 9 \times 10^9 \times 20 \times 10^{-6}$
$V = 180 \times 10^3 = 1.8 \times 10^5\,V$.

(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.
Since the triangle is equilateral with side length $0.2\, m$,the distance from both charges at $A$ and $B$ to point $C$ is $r = 0.2\, m$.
The total potential at $C$ is the sum of potentials due to both charges:
$V_C = V_A + V_B = \frac{1}{4\pi \varepsilon_0} \frac{q_A}{r} + \frac{1}{4\pi \varepsilon_0} \frac{q_B}{r}$
Given $q_A = q_B = 4\,\mu C = 4 \times 10^{-6}\, C$ and $r = 0.2\, m$:
$V_C = 2 \times \left( 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.2} \right)$
$V_C = 2 \times (9 \times 10^9 \times 20 \times 10^{-6})$
$V_C = 2 \times 18 \times 10^4 = 36 \times 10^4\, V$.

(B) The electric potential $V$ at the centre of a circle due to a charge $q$ at distance $R$ is given by $V = \frac{kq}{R}$. Since there are $10$ electrons,each with charge $-e$,the total potential at the centre is $V = 10 \times \frac{k(-e)}{R} = -\frac{10ke}{R}$. Since $V \ne 0$,the potential is non-zero.
For the electric field,because the electrons are equally spaced around the circle,the electric field vector produced by each electron is cancelled by the electric field vector produced by the electron diametrically opposite to it. Therefore,the net electric field $\vec{E}$ at the centre is $0$.

(B) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by the formula:
$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r}$
Given:
$Q = 100\,\mu C = 100 \times 10^{-6}\,C = 10^{-4}\,C$
$r = 9\,m$
$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$
Substituting the values:
$V = (9 \times 10^9) \times \frac{10^{-4}}{9}$
$V = 10^9 \times 10^{-4} = 10^5\,V$
Therefore,the correct option is $B$.

(B) The electric potential $V$ at a point due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r}$.
In the given isosceles triangle,the distance from vertex $A$ to both vertices $B$ and $C$ is $r = \sqrt{a^2 + b^2}$ by the Pythagorean theorem.
The potential at vertex $A$ is the algebraic sum of the potentials due to charges at $B$ and $C$:
$V_A = V_B + V_C$
$V_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{\sqrt{a^2 + b^2}} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{-q}{\sqrt{a^2 + b^2}}$
$V_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q - q}{\sqrt{a^2 + b^2}} = 0$.

(C) Let the distance of point $P$ from the $-6\,\mu C$ charge be $x$ meters. The distance of point $P$ from the $12\,\mu C$ charge is $(0.2 + x)$ meters.
The electric potential $V$ at point $P$ due to both charges is given by:
$V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{-6 \times 10^{-6}}{x} + \frac{12 \times 10^{-6}}{0.2 + x} \right] = 0$
Since $V = 0$,we have:
$\frac{6 \times 10^{-6}}{x} = \frac{12 \times 10^{-6}}{0.2 + x}$
Dividing both sides by $6 \times 10^{-6}$:
$\frac{1}{x} = \frac{2}{0.2 + x}$
Cross-multiplying gives:
$0.2 + x = 2x$
$x = 0.2\, m$
Thus,the distance of point $P$ from the $-6\,\mu C$ charge is $0.20\, m$.

(C) The electric potential $V$ of a charged spherical conductor (like a soap bubble) is given by the formula $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r}$,where $q$ is the charge and $r$ is the radius.
Since the charge $q$ on the soap bubble remains constant when its radius is changed,we have $V \propto \frac{1}{r}$.
Given the initial potential $V_1 = 16\,V$ and initial radius $r_1 = r$.
The new radius is $r_2 = 2r$.
Using the proportionality $V_1 r_1 = V_2 r_2$,we get:
$16 \times r = V_2 \times (2r)$
$V_2 = \frac{16}{2} = 8\,V$.
Therefore,the new potential of the bubble will be $8\,V$.

(A) The work done $W$ in moving a charge $q_0$ from point $B$ to $A$ is given by $W = q_0(V_A - V_B)$.
Given: $q_1 = -5 \times 10^{-6}\,C$,$q_2 = +2.0 \times 10^{-6}\,C$,$q_0 = +3.0 \times 10^{-6}\,C$,$k = 10^{10}\,N\cdot m^2/C^2$.
From the geometry of the rectangle:
At point $A$: distance from $q_1$ is $15\,cm = 0.15\,m$,distance from $q_2$ is $5\,cm = 0.05\,m$.
$V_A = k \left( \frac{q_1}{0.15} + \frac{q_2}{0.05} \right) = 10^{10} \left( \frac{-5 \times 10^{-6}}{0.15} + \frac{2 \times 10^{-6}}{0.05} \right) = 10^4 \left( -\frac{5}{0.15} + \frac{2}{0.05} \right) = 10^4 \left( -33.33 + 40 \right) = 6.67 \times 10^4\,V$.
At point $B$: distance from $q_1$ is $5\,cm = 0.05\,m$,distance from $q_2$ is $15\,cm = 0.15\,m$.
$V_B = k \left( \frac{q_1}{0.05} + \frac{q_2}{0.15} \right) = 10^{10} \left( \frac{-5 \times 10^{-6}}{0.05} + \frac{2 \times 10^{-6}}{0.15} \right) = 10^4 \left( -100 + 13.33 \right) = -86.67 \times 10^4\,V$.
$W = 3.0 \times 10^{-6} \times (6.67 \times 10^4 - (-86.67 \times 10^4)) = 3.0 \times 10^{-6} \times (93.34 \times 10^4) = 2.8\,J$.

(C) For a hollow conducting sphere,the electric field inside the sphere is zero because the charge resides only on the outer surface.
Since the electric field $E = -\frac{dV}{dr} = 0$,the potential $V$ must be constant throughout the interior of the sphere.
The potential at any point inside the sphere is equal to the potential at its surface.
The potential at the surface of a sphere of radius $R$ with charge $Q$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.
Therefore,at a distance $r = \frac{R}{3}$ from the centre,the potential is $\frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.

(D) The work done $W$ in moving a charge $q_0$ from point $A$ to point $C$ in the presence of a source charge $Q$ at $B$ is given by $W = q_0(V_C - V_A)$.
Here,the source charge is $Q$ at $B$. The distance $BA = l$ and $BC = l$.
The electric potential at point $A$ due to charge $Q$ is $V_A = \frac{1}{4\pi \varepsilon_0} \frac{Q}{l}$.
The electric potential at point $C$ due to charge $Q$ is $V_C = \frac{1}{4\pi \varepsilon_0} \frac{Q}{l}$.
Since $V_A = V_C$,the potential difference $\Delta V = V_C - V_A = 0$.
Therefore,the work done $W = (-q) \times (0) = 0$.

(A) When a particle of charge $q$ is accelerated through a potential difference $V$,the work done on the particle by the electric field is given by $W = qV$.
According to the work-energy theorem,this work done is equal to the change in the kinetic energy of the particle.
Therefore,the kinetic energy gained by the particle is $KE = qV$.

(B) Given that the electric potential of both spheres is the same,i.e.,$V_A = V_B$.
Using the formula for the potential of a charged sphere $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r}$,we have:
$\frac{1}{4\pi\epsilon_0} \cdot \frac{Q_A}{a} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_B}{b} \implies \frac{Q_A}{Q_B} = \frac{a}{b}$......$(i)$
Surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{4\pi r^2}$.
Therefore,the ratio of surface charge densities is:
$\frac{\sigma_A}{\sigma_B} = \frac{Q_A}{4\pi a^2} \cdot \frac{4\pi b^2}{Q_B} = \frac{Q_A}{Q_B} \cdot \frac{b^2}{a^2}$
Substituting the ratio from equation $(i)$:
$\frac{\sigma_A}{\sigma_B} = \frac{a}{b} \cdot \frac{b^2}{a^2} = \frac{b}{a}$.

(A) For a charged conducting sphere,the charge $Q$ resides entirely on its outer surface.
Inside the conductor,the electric field $E = 0$.
Since the electric field is zero inside,the potential $V$ remains constant throughout the interior and is equal to the potential at the surface.
The potential at the surface of a sphere of radius $R$ is given by $V = \frac{Q}{4\pi\epsilon_0 R}$.
Therefore,at any point $x$ inside the sphere (where $x < R$),the potential is $V = \frac{Q}{4\pi\epsilon_0 R}$.

(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the electric field.
$W = q(V_A - V_B) = \Delta K$
$q(V_A - V_B) = \frac{1}{2}m(v_B^2 - v_A^2)$
Given: $m = 1\, g = 10^{-3}\, kg$,$q = 10^{-8}\, C$,$V_A = 600\, V$,$V_B = 0\, V$,$v_B = 20\, cm/s = 0.2\, m/s$.
Substituting the values:
$10^{-8}(600 - 0) = \frac{1}{2} \times 10^{-3} \times ((0.2)^2 - v_A^2)$
$6 \times 10^{-6} = 0.5 \times 10^{-3} \times (0.04 - v_A^2)$
$12 \times 10^{-3} = 0.04 - v_A^2$
$v_A^2 = 0.04 - 0.012 = 0.028$
$v_A = \sqrt{0.028} \approx 0.1673\, m/s \approx 16.73\, cm/s \approx 16.8\, cm/s$.

(C) The work done $W$ in moving a charge $Q$ between two points with a potential difference $\Delta V$ is given by the formula:
$W = Q \cdot \Delta V$
Given:
Charge $Q = 20 \; C$
Work done $W = 2 \; J$
We need to find the potential difference $\Delta V$.
Rearranging the formula for $\Delta V$:
$\Delta V = \frac{W}{Q}$
Substituting the given values:
$\Delta V = \frac{2 \; J}{20 \; C} = 0.1 \; V$
Therefore,the potential difference between the points is $0.1 \; V$.

(B) The energy required to move a charge $q$ through a potential difference $V$ is given by the formula $W = qV$.
Given energy $W = 4 \times 10^{20} \text{ eV}$.
Since $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$,the energy in Joules is $W = (4 \times 10^{20}) \times (1.6 \times 10^{-19}) \text{ J} = 6.4 \times 10 = 64 \text{ J}$.
The charge $q = 0.25 \text{ C}$.
Using the formula $V = W / q$,we get $V = 64 / 0.25$.
$V = 64 \times 4 = 256 \text{ V}$.
Thus,the potential difference is $256 \text{ V}$.

(C) The potential energy $U$ of a system of point charges is given by the sum of potential energies of all unique pairs of charges: $U = \sum \frac{1}{4\pi \varepsilon_0} \frac{q_i q_j}{r_{ij}}$.
For an equilateral triangle with $3$ charges $q$ at each vertex and side length $l$,there are $3$ pairs of charges,each separated by distance $l$.
The potential energy for each pair is $U_{pair} = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{l}$.
Since there are $3$ such pairs,the total net potential energy is $U_{net} = 3 \times \frac{1}{4\pi \varepsilon_0} \frac{q^2}{l} = \frac{1}{4\pi \varepsilon_0} \frac{3q^2}{l}$.

(D) The length of the body diagonal of a cube with side $b$ is $\sqrt{3}b$.
The distance of the center of the cube from each vertex is $r = \frac{\sqrt{3}b}{2}$.
The electric potential energy $U$ of a charge $q$ at the center due to $8$ charges of $-q$ at the corners is given by the sum of individual potential energies:
$U = \sum_{i=1}^{8} \frac{1}{4\pi\varepsilon_0} \frac{(-q)(q)}{r}$
$U = 8 \times \left( \frac{1}{4\pi\varepsilon_0} \cdot \frac{-q^2}{\sqrt{3}b/2} \right)$
$U = 8 \times \left( \frac{-2q^2}{4\pi\varepsilon_0\sqrt{3}b} \right)$
$U = \frac{-16q^2}{4\pi\varepsilon_0\sqrt{3}b} = \frac{-4q^2}{\sqrt{3}\pi\varepsilon_0b}$

(C) The kinetic energy $(KE)$ gained by a charged particle when accelerated through a potential difference $(V)$ is given by the formula: $KE = qV$.
Here,the charge of a proton $(q)$ is equal to the elementary charge $(e)$.
The potential difference $(V)$ is given as $1\, kV$.
Substituting these values into the formula:
$KE = e \times 1\, kV = 1\, keV$.
Therefore,the kinetic energy of the proton is $1\, keV$.

(D) The total electrostatic potential at point $P$ is the sum of the potential due to the point charge $Q$ at the centre and the potential due to the spherical shell of charge $q$.
$1$. The potential due to the point charge $Q$ at a distance $r = \frac{R}{2}$ is given by $V_Q = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{R/2} = \frac{2Q}{4\pi \varepsilon_0 R}$.
$2$. For a spherical conducting shell of radius $R$ with charge $q$,the potential at any point inside the shell (including the centre) is constant and equal to the potential at its surface. Thus,the potential at distance $r = \frac{R}{2}$ due to the shell is $V_q = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{R}$.
$3$. The total potential at point $P$ is $V = V_Q + V_q = \frac{2Q}{4\pi \varepsilon_0 R} + \frac{q}{4\pi \varepsilon_0 R}$.
Therefore,the correct option is $(d)$.

(A) The potential energy of a charge $q$ is given by $U = qV$.
Since the electron has a negative charge $(q = -e)$,its potential energy $U = -eV$ decreases as it moves from a lower potential region $V_1$ to a higher potential region $V_2$ (where $V_2 > V_1$).
According to the law of conservation of energy,the decrease in potential energy results in an increase in kinetic energy $(K = \frac{1}{2}mv^2)$.
Therefore,the velocity of the electron will increase.

(B) The kinetic energy $(K)$ gained by a charge $(q)$ accelerated through a potential difference $(V)$ is given by the formula:
$K = q \times V$
Given:
Charge $(q)$ = $0.5\, C$
Potential difference $(V)$ = $2,000\, V$
Substituting the values:
$K = 0.5\, C \times 2,000\, V$
$K = 1,000\, J$
Therefore,the kinetic energy attained by the pellet is $1,000\, J$.

(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:
$K = (2e) \times (5 \ V) = 10 \ eV$.
Since the particle was initially at rest,its final kinetic energy is equal to the work done on it by the electric field,which 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 $q = 1.6 \times 10^{-19} \text{ C}$.
The potential difference is $V = 1,00,000 \text{ V} = 10^5 \text{ V}$.
Substituting these values into the formula:
$E = (1.6 \times 10^{-19} \text{ C}) \times (10^5 \text{ V})$
$E = 1.6 \times 10^{-19+5} \text{ J}$
$E = 1.6 \times 10^{-14} \text{ J}$.
Therefore,the correct option is $D$.

(A) For a hollow conducting sphere of radius $R$,the electric potential at any point inside the sphere (where distance $r < R$) is constant and equal to the potential at the surface.
Given: Radius $R = 10\, cm = 0.1\, m$,Charge $q = 3.2 \times 10^{-19}\, C$.
The potential $V$ inside the sphere is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
Substituting the values: $V = (9 \times 10^9) \times \frac{3.2 \times 10^{-19}}{0.1}$.
$V = 9 \times 3.2 \times 10^{-9} \times 10 = 28.8 \times 10^{-9}\, V$ or $2.88 \times 10^{-8}\, V$.

(D) The potential energy $U$ of a system of point charges in the $CGS$ system (where $k = 1$) is given by $U = \sum \frac{q_i q_j}{r_{ij}}$.
Here,the charge $q_1 = 10\, e.s.u.$ is interacting with $q_2 = 40\, e.s.u.$ at a distance $r_1 = 2\, cm$ and $q_3 = 20\, e.s.u.$ at a distance $r_2 = 4\, cm$.
The total potential energy of the $10\, e.s.u.$ charge is the sum of the interaction energies with the other two charges:
$U = \frac{q_1 q_2}{r_1} + \frac{q_1 q_3}{r_2}$
$U = \frac{10 \times 40}{2} + \frac{10 \times 20}{4}$
$U = \frac{400}{2} + \frac{200}{4}$
$U = 200 + 50 = 250\, ergs$.

(A) The electric potential $V$ is defined as the work done $W$ per unit charge $q$ in moving a charge from infinity to a point.
Mathematically,$V = W / q$.
Rearranging this formula,we get $W = V \times q$.
Here,$W$ is measured in Joules $(J)$,$V$ is measured in Volts $(V)$,and $q$ is measured in Coulombs $(C)$.
Therefore,$1 \ Joule = 1 \ Volt \times 1 \ Coulomb$.
Thus,the correct relation is $Joule = coulomb \times volt$.

(B) The change in potential energy $(\Delta U)$ of a charge $q$ moved between two points is given by $\Delta U = q(V_f - V_i)$,where $V_f$ is the final potential and $V_i$ is the initial potential.
Since the charge is moved from a lower potential to a higher potential,$V_f > V_i$,which implies $(V_f - V_i) > 0$.
Given that $q$ is a positive charge,the product $q(V_f - V_i)$ is positive.
Therefore,the potential energy of the charge increases.

(B) For a charged hollow metal sphere,the electric potential inside the sphere is constant and equal to the potential at its surface.
Given,the radius of the sphere $R = 5\,cm$.
The potential on the surface $V_{surface} = 10\,V$.
Since $2\,cm < 5\,cm$,the point at $2\,cm$ from the centre lies inside the sphere.
Therefore,the potential at $2\,cm$ is the same as the potential on the surface,which is $10\,V$.