Electric potential Questions in English

(C) The kinetic energy $(KE)$ acquired by a charged particle of charge $q$ when accelerated through a potential difference $V$ is given by the formula:
$KE = q \times V$
Here,the charge of a proton is $q = +e$.
The accelerating potential difference is $V = 1 \ kV$.
Substituting these values into the formula:
$KE = 1e \times 1 \ kV$
$KE = 1 \ keV$
The mass of the particle does not affect the kinetic energy gained when accelerated through a potential difference,as the energy depends only on the charge and the potential difference.

(C) For a charged conducting sphere,the potential inside the sphere is constant and equal to the potential at its surface.
Given radius $R = 10 \, cm$.
The potential at any point inside the sphere (including the center) is equal to the potential at the surface: $V_{surface} = \frac{KQ}{R} = \frac{KQ}{10} = V$.
Thus,$KQ = 10V$.
For a point outside the sphere at a distance $r = 15 \, cm$,the potential is given by $V_{outside} = \frac{KQ}{r}$.
Substituting the values: $V_{outside} = \frac{10V}{15} = \frac{2}{3}V$.

(D) The linear charge density of the rod is $\lambda = \frac{Q}{L}$.
Consider a small element of length $dx$ at a distance $x$ from point $O$.
The charge of this element is $dq = \lambda dx = \frac{Q}{L} dx$.
The potential $dV$ at point $O$ due to this element is $dV = \frac{k dq}{x} = \frac{1}{4\pi \epsilon_0} \frac{Q}{L} \frac{dx}{x}$.
To find the total potential $V$,integrate from $x = L$ to $x = 2L$:
$V = \int_{L}^{2L} \frac{Q}{4\pi \epsilon_0 L} \frac{dx}{x} = \frac{Q}{4\pi \epsilon_0 L} [\ln x]_{L}^{2L}$.
$V = \frac{Q}{4\pi \epsilon_0 L} (\ln 2L - \ln L) = \frac{Q}{4\pi \epsilon_0 L} \ln\left(\frac{2L}{L}\right) = \frac{Q \ln 2}{4\pi \epsilon_0 L}$.

(B) The kinetic energy gained by a particle of charge $Q$ and mass $m$ when accelerated through a potential difference $V$ is given by $K = QV$.
Since the particles start from rest,$K = \frac{1}{2}mv^2$.
Equating the two,we get $\frac{1}{2}mv^2 = QV$,which implies $v = \sqrt{\frac{2QV}{m}}$.
Since $m$ and $V$ are the same for both particles,the speed $v$ is proportional to the square root of the charge: $v \propto \sqrt{Q}$.
Therefore,the ratio of their speeds is $\frac{v_A}{v_B} = \sqrt{\frac{Q_A}{Q_B}}$.
Substituting the given values $Q_A = q$ and $Q_B = 4q$,we get $\frac{v_A}{v_B} = \sqrt{\frac{q}{4q}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio $V_A / V_B$ is $1:2$.

(C) For a hollow conducting sphere,the electric field inside the sphere is zero.
This implies that the electric potential is 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 $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.
Since $r = R/3$ is inside the sphere,the potential at this point is also $\frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.

(B) The charge of $100$ electrons is $q = 100 \times (-1.6 \times 10^{-19} \ C) = -1.6 \times 10^{-17} \ C$.
The work done $W$ in moving a charge $q$ from point $P$ to point $Q$ is given by $W = q(V_Q - V_P)$.
Substituting the given values:
$W = (-1.6 \times 10^{-17} \ C) \times (-4 \ V - 10 \ V)$
$W = (-1.6 \times 10^{-17}) \times (-14)$
$W = 22.4 \times 10^{-17} \ J$
$W = 2.24 \times 10^{-16} \ J$.

(C) The total electric potential $V$ at point $P$ is the sum of the potential due to the charge $q$ on the shell and the potential due to the charge $Q$ at the center.
For a point inside a conducting shell,the potential due to the shell is constant and equal to the potential at its surface,which is $V_{shell} = \frac{1}{4\pi \epsilon_0} \frac{q}{R}$.
The potential due to the point charge $Q$ at a distance $r = R/2$ is $V_{point} = \frac{1}{4\pi \epsilon_0} \frac{Q}{R/2} = \frac{1}{4\pi \epsilon_0} \frac{2Q}{R}$.
Therefore,the total potential $V = V_{shell} + V_{point} = \frac{q}{4\pi \epsilon_0 R} + \frac{2Q}{4\pi \epsilon_0 R} = \frac{1}{4\pi \epsilon_0 R} (q + 2Q)$.

(C) The potential energy $U$ of a charge $q$ in an electric field $V$ is given by $U = qV$.
In the $CGS$ system, the electrostatic constant $k = 1$.
The potential $V_A$ at the position of the $10 \, \text{esu}$ charge due to the other two charges is:
$V_A = \frac{q_B}{r_1} + \frac{q_C}{r_2}$
Given $q_B = 40 \, \text{esu}$, $r_1 = 2 \, \text{cm}$, $q_C = -20 \, \text{esu}$, and $r_2 = 4 \, \text{cm}$.
$V_A = \frac{40}{2} + \frac{-20}{4} = 20 - 5 = 15 \, \text{statvolts}$.
The potential energy $U_A$ of the $10 \, \text{esu}$ charge is:
$U_A = q_A \times V_A = 10 \times 15 = 150 \, \text{ergs}$.

(A) The potential energy of the charge $q$ at the center of the ring is $U_i = \frac{kQq}{R}$.
When the charge is at infinity,the potential energy is $U_f = 0$.
By the law of conservation of mechanical energy,the initial potential energy is converted into the final kinetic energy of the charge.
$U_i + K_i = U_f + K_f$
$\frac{kQq}{R} + 0 = 0 + \frac{1}{2}mv^2$
$\frac{1}{2}mv^2 = \frac{kQq}{R}$
$v^2 = \frac{2kQq}{mR}$
$v = \sqrt{\frac{2kQq}{mR}}$

(A) The potential difference $\Delta V$ between two points is defined as the work done $W$ per unit charge $q$ to move the charge between those points.
Formula: $\Delta V = \frac{W}{q}$
Given:
Work done $W = 2 \, J$
Charge $q = 20 \, C$
Calculation:
$\Delta V = \frac{2 \, J}{20 \, C} = 0.1 \, V$
Therefore,the potential difference between the two points is $0.1 \, V$.

(D) The length of the body diagonal of a cube with side $b$ is $\sqrt{3}b$.
Therefore,the distance from the center of the cube to each corner is $r = \frac{\sqrt{3}b}{2}$.
The electrostatic potential energy $U$ of a charge $q_2$ at a distance $r$ from a charge $q_1$ is given by $U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$.
Since there are $8$ such charges of $(-q)$ at the corners and a charge $(+q)$ at the center,the total potential energy is:
$U = 8 \times \left[ \frac{1}{4\pi\varepsilon_0} \cdot \frac{(-q)(q)}{\frac{\sqrt{3}b}{2}} \right]$
$U = 8 \times \left[ \frac{1}{4\pi\varepsilon_0} \cdot \frac{-2q^2}{\sqrt{3}b} \right]$
$U = \frac{-16q^2}{4\sqrt{3}\pi\varepsilon_0 b} = \frac{-4q^2}{\sqrt{3}\pi\varepsilon_0 b}$.

(B) The work done by the electric field on a charge $q$ accelerated through a potential difference $V$ is equal to its kinetic energy.
$W = qV = \frac{1}{2}mv^2$
Since the mass $m$ and potential difference $V$ are the same for both particles,we have $v^2 \propto q$,which implies $v \propto \sqrt{q}$.
Therefore,the ratio of their speeds is:
$\frac{v_A}{v_B} = \sqrt{\frac{q_A}{q_B}} = \sqrt{\frac{q}{9q}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Thus,the ratio of their speeds is $1 : 3$.

(B) The work done $W$ in moving a charge $q$ from point $P$ to point $Q$ is given by $W = q(V_Q - V_P)$.
Here,the charge $q$ is the total charge of $100$ electrons,so $q = -100e = -100 \times 1.6 \times 10^{-19} \ C = -1.6 \times 10^{-17} \ C$.
The potential difference is $V_Q - V_P = (-4 \ V) - (10 \ V) = -14 \ V$.
Substituting these values into the work formula:
$W = (-1.6 \times 10^{-17} \ C) \times (-14 \ V) = 22.4 \times 10^{-17} \ J = 2.24 \times 10^{-16} \ J$.

(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.
The electric field $E$ due to a point charge $q$ at a distance $r$ is given by $E = \frac{kq}{r^2}$.
Dividing the two equations: $\frac{V}{E} = \frac{kq/r}{kq/r^2} = r$.
Substituting the given values: $r = \frac{600}{200} = 3 \, m$.
Now,using $V = \frac{kq}{r}$,we solve for $q$: $q = \frac{V \cdot r}{k}$.
Given $k = 9 \times 10^9 \, N \cdot m^2/C^2$,$V = 600 \, V$,and $r = 3 \, m$:
$q = \frac{600 \times 3}{9 \times 10^9} = \frac{1800}{9 \times 10^9} = 200 \times 10^{-9} \, C = 0.2 \times 10^{-6} \, C = 0.2 \, \mu C$.

(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: $Q = 0.01 \ C$ and $W = 15 \ J$.
Substituting the values into the formula: $15 = 0.01 \times (V_B - V_A)$.
Solving for the potential difference: $(V_B - V_A) = \frac{15}{0.01} = 1500 \ V$.

(A) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula: $U = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r}$.
Given:
$q_1 = 7\ \mu C = 7 \times 10^{-6}\ C$
$q_2 = -2\ \mu C = -2 \times 10^{-6}\ C$
The distance $r$ between the points $(-9\ cm, 0, 0)$ and $(9\ cm, 0, 0)$ is $r = 9\ cm - (-9\ cm) = 18\ cm = 0.18\ m$.
Substituting the values:
$U = (9 \times 10^9) \times \frac{(7 \times 10^{-6}) \times (-2 \times 10^{-6})}{0.18}$
$U = (9 \times 10^9) \times \frac{-14 \times 10^{-12}}{0.18}$
$U = \frac{-126 \times 10^{-3}}{0.18} = -0.7\ J$.

(C) The electrostatic potential energy $U$ of a system of three point charges is given by the formula:
$U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1q_2}{r_{12}} + \frac{q_2q_3}{r_{23}} + \frac{q_3q_1}{r_{31}} \right)$
Given: $q_1 = q_2 = q_3 = 10 \ \mu C = 10 \times 10^{-6} \ C = 10^{-5} \ C$,and side $a = 10 \ cm = 0.1 \ m = 10^{-1} \ m$.
Since the triangle is equilateral,$r_{12} = r_{23} = r_{31} = a = 10^{-1} \ m$.
Substituting the values:
$U = 3 \times \left( \frac{k \cdot q^2}{a} \right)$
$U = 3 \times \left( \frac{9 \times 10^9 \times (10^{-5})^2}{10^{-1}} \right)$
$U = 3 \times \left( \frac{9 \times 10^9 \times 10^{-10}}{10^{-1}} \right)$
$U = 3 \times 9 \times 10^9 \times 10^{-10} \times 10^1$
$U = 27 \times 10^0 = 27 \ J$.

(B) The work done $W$ in bringing a charge $q_3$ from infinity to a point is equal to the change in potential energy of the system,which is $W = q_3 \times V$,where $V$ is the electric potential at the point $x = 2 \ m$ due to charges $q_1$ and $q_2$.
The potential $V$ at $x = 2 \ m$ is given by:
$V = \frac{1}{4\pi \epsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right)$
Here,$r_1$ is the distance from $q_1$ to $x = 2 \ m$,so $r_1 = 2 \ m$. $r_2$ is the distance from $q_2$ to $x = 2 \ m$,so $r_2 = 1 \ m$.
Substituting the values:
$W = q_3 \times \frac{1}{4\pi \epsilon_0} \left( \frac{-1 \times 10^{-6}}{2} + \frac{1 \times 10^{-6}}{1} \right)$
$W = (9 \times 10^9) \times (1 \times 10^{-6}) \times \left( -0.5 \times 10^{-6} + 1 \times 10^{-6} \right)$
$W = 9 \times 10^3 \times (0.5 \times 10^{-6})$
$W = 4.5 \times 10^{-3} \ J$

(D) The potential at point $A$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:
$V_A = \frac{k(1 \times 10^{-6})}{2} + \frac{k(-1 \times 10^{-6})}{3} = k \times 10^{-6} \left( \frac{1}{2} - \frac{1}{3} \right) = k \times 10^{-6} \left( \frac{1}{6} \right)$
$V_A = \frac{9 \times 10^9 \times 10^{-6}}{6} = 1.5 \times 10^3 \ V$
The potential at point $B$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:
$V_B = \frac{k(1 \times 10^{-6})}{3} + \frac{k(-1 \times 10^{-6})}{2} = k \times 10^{-6} \left( \frac{1}{3} - \frac{1}{2} \right) = k \times 10^{-6} \left( -\frac{1}{6} \right)$
$V_B = -\frac{9 \times 10^9 \times 10^{-6}}{6} = -1.5 \times 10^3 \ V$
The potential difference between $A$ and $B$ is:
$V_A - V_B = 1.5 \times 10^3 - (-1.5 \times 10^3) = 3.0 \times 10^3 \ V = 3000 \ V$

(B) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.
For point $A$ at $(\sqrt{2}, \sqrt{2})$,the distance from the origin $(0, 0)$ is $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.
For point $B$ at $(2, 0)$,the distance from the origin $(0, 0)$ is $r_B = \sqrt{2^2 + 0^2} = 2$.
Since the distances $r_A$ and $r_B$ are equal,the electric potentials at points $A$ and $B$ are equal:
$V_A = \frac{kq}{r_A} = \frac{kq}{2}$
$V_B = \frac{kq}{r_B} = \frac{kq}{2}$
Therefore,the potential difference between points $A$ and $B$ is:
$V_A - V_B = \frac{kq}{2} - \frac{kq}{2} = 0 \ V$.

(A) The electric potential $V$ at the surface of the outer shell (radius $R_2 = 30 \, cm = 0.3 \, m$) is the sum of the potentials due to both the inner shell (charge $q_1 = 0.5 \, \mu C$) and the outer shell (charge $q_2 = 3 \, \mu C$).
The potential due to the inner shell at its surface and outside is $V_1 = \frac{k q_1}{R_2}$.
The potential due to the outer shell at its surface is $V_2 = \frac{k q_2}{R_2}$.
Total potential $V = V_1 + V_2 = \frac{k}{R_2} (q_1 + q_2)$.
Substituting the values: $k = 9 \times 10^9 \, N \cdot m^2/C^2$,$q_1 = 0.5 \times 10^{-6} \, C$,$q_2 = 3 \times 10^{-6} \, C$,and $R_2 = 0.3 \, m$.
$V = \frac{9 \times 10^9}{0.3} (0.5 \times 10^{-6} + 3 \times 10^{-6})$
$V = (30 \times 10^9) \times (3.5 \times 10^{-6})$
$V = 105 \times 10^3 \, V = 1.05 \times 10^5 \, V = 10.5 \times 10^4 \, V$.

(A) Let the radius of each small drop be $r$ and the charge be $q$. The potential of a small drop is $V_{tiny} = \frac{kq}{r} = 10 \ V$.
When $8$ small drops combine to form a large drop of radius $R$,the volume remains conserved:
$8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$8r^3 = R^3 \Rightarrow R = 2r$.
The total charge on the large drop is $Q = 8q$.
The potential of the large drop is $V_{big} = \frac{kQ}{R} = \frac{k(8q)}{2r}$.
$V_{big} = 4 \times (\frac{kq}{r}) = 4 \times 10 \ V = 40 \ V$.

(B) For a hollow metal sphere,the electric charge resides entirely on its outer surface.
Inside the hollow sphere,the electric field is zero everywhere.
Since the electric field is zero inside the sphere,the electric potential is constant throughout the interior and is equal to the potential at the surface.
The potential at the surface (and thus at any point inside) is given by $V = \frac{kQ}{R}$,where $k = 9 \times 10^9 \ N \cdot m^2/C^2$,$Q = 3.2 \times 10^{-19} \ C$,and $R = 10 \ cm = 0.1 \ m$.
Substituting the values: $V = \frac{9 \times 10^9 \times 3.2 \times 10^{-19}}{0.1}$.
$V = \frac{28.8 \times 10^{-10}}{0.1} = 28.8 \times 10^{-9} \ V$.

(C) The kinetic energy $(KE)$ gained by a charged particle when accelerated through a potential difference $(V)$ is given by the formula:
$KE = q \times V$
Where:
$q$ is the charge of the particle.
$V$ is the potential difference.
For a proton,the charge $q = e$ (elementary charge).
Given,$V = 1 \,V$.
Substituting these values:
$KE = e \times 1 \,V = 1 \,eV$.
Therefore,the kinetic energy of the proton is $1 \,eV$.

(B) Let the point where the electric potential is zero be at a distance $x$ from point $A$ along the line $AB$.
The potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.
The total potential at the point is the algebraic sum of the potentials due to $q_1$ and $q_2$.
$V_{total} = \frac{k q_1}{x} + \frac{k q_2}{50 - x} = 0$.
Substituting the values: $\frac{4 \times 10^{-8}}{x} + \frac{-6 \times 10^{-8}}{50 - x} = 0$.
$\frac{4}{x} = \frac{6}{50 - x}$.
$4(50 - x) = 6x$.
$200 - 4x = 6x$.
$10x = 200$.
$x = 20 \ cm$.
Thus,the potential is zero at a distance of $20 \ cm$ from point $A$.

(B) For a charged spherical shell,the electric field inside the shell 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 shell.
Therefore,the potential at any point inside the shell is equal to the potential at its surface.
Given that the potential at the surface is $10 \ V$,the potential at the center of the shell is also $10 \ V$.

(B) The resting membrane potential of a nerve fiber is the electrical potential difference across the cell membrane when the neuron is at rest.
This potential is maintained by the unequal distribution of ions (primarily $Na^+$,$K^+$,and $Cl^-$) across the membrane.
The typical value for the resting membrane potential in a nerve cell is approximately $-70 \ mV$.
Converting this to volts,we get $-70 \times 10^{-3} \ V = -0.07 \ V$.
Comparing this value with the given options,$0.1 \ V$ is the closest order of magnitude.

(B) The work done by the electric field on a charge $q$ accelerated through a potential difference $V$ is given by $W = qV$.
According to the work-energy theorem,this work is equal to the change in kinetic energy: $W = \frac{1}{2}mv^2$.
Equating the two,we get $qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.
Since both particles have the same mass $m$ and are accelerated through the same potential difference $V$,the velocity $v$ is directly proportional to the square root of the charge: $v \propto \sqrt{q}$.
Therefore,the ratio of their velocities is $\frac{v_A}{v_B} = \sqrt{\frac{q_A}{q_B}}$.
Substituting the given values $q_A = q$ and $q_B = 4q$,we get $\frac{v_A}{v_B} = \sqrt{\frac{q}{4q}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio $v_A/v_B = 1:2$.

(A) The electric potential $V$ at the center of a circle due to multiple point charges is the algebraic sum of the potentials due to each individual charge.
$V = \sum \frac{k q_i}{r}$,where $k = 9 \times 10^9 \ N \cdot m^2/C^2$ and $r = 0.4 \ m$.
Sum of charges $Q_{total} = (+5 - 5 + 7 - 5 + 11 + 7 + 15 - 7) \ \mu C = 28 \ \mu C = 28 \times 10^{-6} \ C$.
$V = \frac{9 \times 10^9 \times 28 \times 10^{-6}}{0.4} = \frac{9 \times 28 \times 10^3}{0.4} = \frac{252 \times 10^3}{0.4} = 630 \times 10^3 \ V = 63 \times 10^4 \ V$.

(D) The electric potential $V$ at a point due to a charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$.
From the figure,the distance from both charges $+q$ (at $B$) and $-q$ (at $C$) to point $A$ is $r = \sqrt{a^2 + b^2}$.
The total potential at $A$ is the algebraic sum of the potentials due to both charges:
$V_A = V_B + V_C$
$V_A = \frac{1}{4\pi\varepsilon_0} \frac{q}{\sqrt{a^2 + b^2}} + \frac{1}{4\pi\varepsilon_0} \frac{(-q)}{\sqrt{a^2 + b^2}}$
$V_A = \frac{1}{4\pi\varepsilon_0} \left( \frac{q - q}{\sqrt{a^2 + b^2}} \right) = 0$.

(A) For a charged spherical shell of radius $R$ and charge $Q$:
$1$. The potential at any point on the surface (point $B$) is $V_B = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
$2$. The potential at any point inside the shell (including the center $A$ and point $C$ at $R/2$) is constant and equal to the potential at the surface.
$3$. Therefore,$V_A = V_B = V_C = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
Thus,the correct option is $A$.

(D) The electric potential $V$ at the center of a square due to four identical charges $Q$ at its corners is given by the sum of potentials due to each charge.
Each charge is at a distance $r$ from the center,where $r$ is half the diagonal of the square.
Diagonal of the square $= a\sqrt{2}$.
So,$r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
The potential at the center is $V = 4 \times \frac{kQ}{r} = 4 \times \frac{kQ}{a/\sqrt{2}} = \frac{4\sqrt{2}kQ}{a}$.
Given $Q = \frac{10}{3} \times 10^{-9} \ C$,$a = 8 \times 10^{-2} \ m$,and $k = 9 \times 10^9 \ N \cdot m^2/C^2$.
$V = \frac{4 \times \sqrt{2} \times 9 \times 10^9 \times (\frac{10}{3} \times 10^{-9})}{8 \times 10^{-2}}$.
$V = \frac{4 \times \sqrt{2} \times 3 \times 10}{8 \times 10^{-2}} = \frac{120\sqrt{2}}{8 \times 10^{-2}} = 15 \times 10^2 \sqrt{2} = 1500\sqrt{2} \ V$.

(B) For a hollow conducting sphere (or a charged spherical shell),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 center,is equal to the potential at its surface.
Given that the potential at the surface is $10 \ V$,the potential at the center is also $10 \ V$.

(D) Let $q_1$ and $q_2$ be the charges on the inner and outer shells respectively. Given $Q = q_1 + q_2$ ... $(i)$
Since surface charge densities are equal,$\sigma_1 = \sigma_2$,so $\frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2} \implies \frac{q_1}{q_2} = \frac{r^2}{R^2}$ ... $(ii)$
From $(i)$ and $(ii)$,we get $q_1 = \frac{Q r^2}{R^2 + r^2}$ and $q_2 = \frac{Q R^2}{R^2 + r^2}$.
The potential at the center is the sum of potentials due to both shells: $V = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q_1}{r} + \frac{q_2}{R} \right]$.
Substituting the values of $q_1$ and $q_2$: $V = \frac{1}{4\pi \varepsilon_0} \left[ \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right] = \frac{1}{4\pi \varepsilon_0} \left[ \frac{Qr}{R^2 + r^2} + \frac{QR}{R^2 + r^2} \right]$.
Thus,$V = \frac{Q(R + r)}{4\pi \varepsilon_0(R^2 + r^2)}$.

(D) Let the radius of the inner sphere be $a$ and the radius of the outer shell be $b$.
Initially,the potential of the sphere is $V_{\text{sphere}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{a}$ and the potential of the shell is $V_{\text{shell}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{b}$.
The potential difference is $V = V_{\text{sphere}} - V_{\text{shell}} = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right)$.
Now,a charge of $-3Q$ is given to the shell.
The new potential of the sphere is $V'_{\text{sphere}} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} \right)$.
The new potential of the shell is $V'_{\text{shell}} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{b} - \frac{3Q}{b} \right) = \frac{1}{4\pi\epsilon_0} \left( -\frac{2Q}{b} \right)$.
The new potential difference is $V' = V'_{\text{sphere}} - V'_{\text{shell}} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} - (-\frac{2Q}{b}) \right) = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} + \frac{2Q}{b} \right) = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right) = V$.
Thus,the potential difference remains $V$.

(A) The charges on the shells are given by $q = \sigma \times 4\pi r^2$.
Thus,$q_a = \sigma(4\pi a^2)$,$q_b = -\sigma(4\pi b^2)$,and $q_c = \sigma(4\pi c^2)$.
The potential at any point is the sum of potentials due to all shells.
For shell $A$ (at radius $a$):
$V_A = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_a}{a} + \frac{q_b}{b} + \frac{q_c}{c} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{\sigma(4\pi a^2)}{a} + \frac{-\sigma(4\pi b^2)}{b} + \frac{\sigma(4\pi c^2)}{c} \right] = \frac{\sigma}{\varepsilon_0} (a - b + c)$.
For shell $B$ (at radius $b$):
$V_B = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_a}{b} + \frac{q_b}{b} + \frac{q_c}{c} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{\sigma(4\pi a^2)}{b} + \frac{-\sigma(4\pi b^2)}{b} + \frac{\sigma(4\pi c^2)}{c} \right] = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{b} - b + c \right)$.

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

(C) The work required to move a charge $q$ from a point with potential $V_0$ to infinity is given by $W = q(V_{\infty} - V_0)$. Since $V_{\infty} = 0$,the work required is $W = -qV_0$. Here,$q = -Q$,so $W = -(-Q)V_0 = QV_0$.
The potential $V_0$ at the center of the square due to the four charges $Q$ at the corners is the sum of the potentials due to each charge.
The distance from each corner to the center is $r = \frac{a}{\sqrt{2}}$.
$V_0 = 4 \times \left( \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{a/\sqrt{2}} \right) = \frac{4\sqrt{2} Q}{4\pi \varepsilon_0 a} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 a}$.
Therefore,the work required is $W = (-Q) \times (-V_0) = Q \times \left( \frac{\sqrt{2} Q}{\pi \varepsilon_0 a} \right) = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 a}$.

(B) The electric potential at the center $A$ of the first ring is the sum of the potential due to ring $1$ and ring $2$:
$V_A = \frac{Q_1}{4\pi \varepsilon_0 R} + \frac{Q_2}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} \right)$
The electric potential at the center $B$ of the second ring is:
$V_B = \frac{Q_2}{4\pi \varepsilon_0 R} + \frac{Q_1}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} \right)$
The work done $W$ to move a charge $q$ from $A$ to $B$ is given by $W = q(V_B - V_A)$:
$V_B - V_A = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} - Q_1 - \frac{Q_2}{\sqrt{2}} \right)$
$V_B - V_A = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2(1 - \frac{1}{\sqrt{2}}) - Q_1(1 - \frac{1}{\sqrt{2}}) \right)$
$V_B - V_A = \frac{1}{4\pi \varepsilon_0 R} (Q_2 - Q_1) \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)$
Since $W = q(V_B - V_A)$,we have:
$W = \frac{q(Q_2 - Q_1)(\sqrt{2} - 1)}{4\pi \varepsilon_0 R\sqrt{2}}$
Note: The magnitude of work is $\frac{q(Q_1 - Q_2)(\sqrt{2} - 1)}{4\pi \varepsilon_0 R\sqrt{2}}$.

(A) The change in kinetic energy $\Delta K$ of a charged particle moving in an electric field is given by $\Delta K = -\Delta U = -q\Delta V = q(V_i - V_f)$.
For an $\alpha$-particle,the charge $q = +2e$.
The potential difference is $\Delta V = V_f - V_i = 50\ V - 70\ V = -20\ V$.
Therefore,the change in kinetic energy is $\Delta K = -q\Delta V = -(2e) \times (-20\ V) = 40\ eV$.
Since the particle moves from a higher potential to a lower potential,its kinetic energy increases by $40\ eV$.

(C) From the figure,$AC = L$,$BC = L$. Since $C$ is the midpoint of $AB$ and $CD$ is a semicircle with diameter $CD$,the distance $BD = L$ (as $C, B, D$ are collinear and $CD$ is the diameter of the semicircle,$CB = BD = L$).
Potential at $C$ is given by:
$V_C = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AC} + \frac{-q}{BC} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{L} - \frac{q}{L} \right] = 0$
Potential at $D$ is given by:
$AD = AB + BD = 2L + L = 3L$
$V_D = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AD} + \frac{-q}{BD} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{3L} - \frac{q}{L} \right]$
$V_D = \frac{q}{4\pi\varepsilon_0 L} \left[ \frac{1}{3} - 1 \right] = \frac{q}{4\pi\varepsilon_0 L} \left( -\frac{2}{3} \right) = -\frac{q}{6\pi\varepsilon_0 L}$
Work done in moving charge $+Q$ from $C$ to $D$ is:
$W = Q(V_D - V_C) = Q \left( -\frac{q}{6\pi\varepsilon_0 L} - 0 \right) = -\frac{qQ}{6\pi\varepsilon_0 L}$

(D) The potential at any point on a shell is the sum of potentials due to all shells. The charge on a shell of radius $r$ and surface charge density $\sigma$ is $q = 4\pi r^2 \sigma$.
Thus,$q_A = 4\pi a^2 \sigma$,$q_B = -4\pi b^2 \sigma$,and $q_C = 4\pi c^2 \sigma$.
The potentials are:
$V_A = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [a\sigma - b\sigma + \frac{c^2\sigma}{c}] = \frac{\sigma}{\epsilon_0} [a - b + c]$
$V_B = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{b} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [\frac{a^2\sigma}{b} - b\sigma + \frac{c^2\sigma}{c}] = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} - b + c]$
$V_C = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{c} + \frac{q_B}{c} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [\frac{a^2\sigma}{c} - \frac{b^2\sigma}{c} + c\sigma] = \frac{\sigma}{\epsilon_0} [\frac{a^2 - b^2}{c} + c]$
Given $c = a + b$,then $c - b = a$ and $c - a = b$. Also $a^2 - b^2 = (a - b)(a + b) = (a - b)c$.
Substituting $c = a + b$ into $V_A$: $V_A = \frac{\sigma}{\epsilon_0} [a - b + (a + b)] = \frac{2a\sigma}{\epsilon_0}$.
Substituting $c = a + b$ into $V_C$: $V_C = \frac{\sigma}{\epsilon_0} [\frac{(a - b)(a + b)}{c} + c] = \frac{\sigma}{\epsilon_0} [\frac{(a - b)c}{c} + c] = \frac{\sigma}{\epsilon_0} [a - b + a + b] = \frac{2a\sigma}{\epsilon_0}$.
Since $V_A = V_C = \frac{2a\sigma}{\epsilon_0}$ and $V_B = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} - b + a + b] = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} + a]$,clearly $V_A = V_C \neq V_B$.

(D) Given that $AC = BC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively.
Therefore,$AE = EC = a$ and $BD = DC = a$.
In $\Delta ADC$,by the Pythagorean theorem,$(AD)^2 = (AC)^2 - (DC)^2 = (2a)^2 - (a)^2 = 4a^2 - a^2 = 3a^2$,so $AD = a\sqrt{3}$.
Similarly,in $\Delta BEC$,$(BE)^2 = (BC)^2 - (EC)^2 = (2a)^2 - (a)^2 = 3a^2$,so $BE = a\sqrt{3}$.
The electric potential at point $D$ due to the charges at $A, B,$ and $C$ is:
$V_D = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{BD} + \frac{q}{DC} + \frac{q}{AD} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{a} + \frac{1}{a} + \frac{1}{a\sqrt{3}} \right] = \frac{q}{4\pi\varepsilon_0 a} \left[ 2 + \frac{1}{\sqrt{3}} \right]$.
The electric potential at point $E$ due to the charges at $A, B,$ and $C$ is:
$V_E = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AE} + \frac{q}{EC} + \frac{q}{BE} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{a} + \frac{1}{a} + \frac{1}{a\sqrt{3}} \right] = \frac{q}{4\pi\varepsilon_0 a} \left[ 2 + \frac{1}{\sqrt{3}} \right]$.
Since $V_D = V_E$,the work done $W = Q(V_E - V_D) = 0$.

(C) Let the corners of the square be $P, Q, R, S$ in order. Let $P$ and $S$ have charges $+q$,and $Q$ and $R$ have charges $-q$. Point $A$ is the midpoint of side $PS$.
Since the side length is $2L$,the distance from $A$ to $P$ and $A$ to $S$ is $L$.
The distance from $A$ to $Q$ and $A$ to $R$ can be calculated using the Pythagorean theorem: $AQ = AR = \sqrt{(2L)^2 + L^2} = \sqrt{4L^2 + L^2} = L\sqrt{5}$.
The electric potential $V_A$ at point $A$ is the algebraic sum of potentials due to all four charges:
$V_A = \frac{1}{4\pi \varepsilon_0} [\frac{q}{AP} + \frac{q}{AS} + \frac{-q}{AQ} + \frac{-q}{AR}]$
$V_A = \frac{1}{4\pi \varepsilon_0} [\frac{q}{L} + \frac{q}{L} - \frac{q}{L\sqrt{5}} - \frac{q}{L\sqrt{5}}]$
$V_A = \frac{1}{4\pi \varepsilon_0} [\frac{2q}{L} - \frac{2q}{L\sqrt{5}}]$
$V_A = \frac{1}{4\pi \varepsilon_0} \frac{2q}{L} (1 - \frac{1}{\sqrt{5}})$

(A) Let the distance from the center of the square to each corner be $r$. The electric potential $V$ at the center due to a point charge $q_i$ at distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r}$.
Since the distance $r$ is the same for all four corners,the total potential at the center is the algebraic sum of the potentials due to each charge:
$V_{total} = \frac{1}{4\pi\epsilon_0} \left( \frac{-Q}{r} + \frac{-q}{r} + \frac{2q}{r} + \frac{2Q}{r} \right)$
$V_{total} = \frac{1}{4\pi\epsilon_0 r} (-Q - q + 2q + 2Q)$
$V_{total} = \frac{1}{4\pi\epsilon_0 r} (Q + q)$
For the potential at the center to be zero,we must have $V_{total} = 0$,which implies:
$Q + q = 0$
$Q = -q$

(B) The electric potential decreases in the direction of the electric field.
From the given figure,point $B$ is at the leftmost position,followed by point $C$,and then point $A$ in the direction of the electric field $\vec{E}$.
Therefore,the potential at these points follows the relation $V_B > V_C > V_A$.
Thus,the electric potential is maximum at point $B$.

(B) For a conducting sphere,the charge $Q$ resides entirely on its outer surface.
The electric field $E$ inside a conducting sphere is zero because the charges are distributed on the surface such that they cancel each other's field at every point inside.
The electric potential $V$ inside a conducting sphere is constant and equal to the potential at its surface.
Therefore,the potential at the centre is $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.
The electric field at the centre is $E = 0$.

(C) The length of the arc is given by $L = r\theta = r \times \frac{\pi}{3} = \frac{r\pi}{3}$.
Since the linear charge density is $\lambda$,the total charge $q$ on the arc is $q = \lambda \times L = \lambda \times \frac{r\pi}{3}$.
All points on the arc are at an equal distance $r$ from the centre.
The electric potential $V$ at the centre due to a charge $q$ at distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$.
Substituting the value of $q$,we get $V = \frac{1}{4\pi\varepsilon_0} \times \frac{(\lambda r\pi / 3)}{r}$.
Simplifying this,$V = \frac{\lambda r\pi}{12\pi\varepsilon_0 r} = \frac{\lambda}{12\varepsilon_0}$.

(B) The electric potential $V$ is defined as the work done per unit charge to move a charge from infinity to a point in an electric field.
Mathematically,$V = W / Q$.
The $SI$ unit of work $W$ is $Joule$ $(J)$ and the $SI$ unit of charge $Q$ is $coulomb$ $(C)$.
Therefore,the unit of potential $V$ is $Joule / coulomb$ $(J/C)$,which is equivalent to $1 \text{ volt}$.
Thus,the correct option is $B$.

(A) The electric potential at the surface of a uniformly charged sphere is given by $V_S = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R}$.
Substituting $q = \rho \cdot \frac{4}{3} \pi R^3$,we get $V_S = \frac{\rho R^2}{3 \epsilon_0}$.
The electric potential at the center of the sphere is given by $V_C = \frac{3}{2} \cdot \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R}$.
Substituting $q = \rho \cdot \frac{4}{3} \pi R^3$,we get $V_C = \frac{\rho R^2}{2 \epsilon_0}$.
The potential difference between the center and the surface is $\Delta V = V_C - V_S$.
$\Delta V = \frac{\rho R^2}{2 \epsilon_0} - \frac{\rho R^2}{3 \epsilon_0} = \frac{3 \rho R^2 - 2 \rho R^2}{6 \epsilon_0} = \frac{\rho R^2}{6 \epsilon_0}$.