Electric potential Questions in English

(B) The work done $W$ to bring a charge $Q$ from infinity to the origin is given by $W = V \cdot Q$,where $V$ is the electric potential at the origin due to the four charges.
The distances of the four charges from the origin $(0, 0)$ are:
$r_1 = \sqrt{0^2 + 2^2} = 2$
$r_2 = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$
$r_3 = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$
$r_4 = \sqrt{0^2 + (-2)^2} = 2$
The potential $V$ at the origin is:
$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q}{r_1} + \frac{Q}{r_2} + \frac{Q}{r_3} + \frac{Q}{r_4} \right)$
$V = \frac{Q}{4\pi \varepsilon_0} \left( \frac{1}{2} + \frac{1}{2\sqrt{5}} + \frac{1}{2\sqrt{5}} + \frac{1}{2} \right)$
$V = \frac{Q}{4\pi \varepsilon_0} \left( 1 + \frac{2}{2\sqrt{5}} \right) = \frac{Q}{4\pi \varepsilon_0} \left( 1 + \frac{1}{\sqrt{5}} \right)$
Thus,the work done is $W = V \cdot Q = \frac{Q^2}{4\pi \varepsilon_0} \left( 1 + \frac{1}{\sqrt{5}} \right)$.

(B) For a uniformly charged spherical shell of radius $r_0$:
$1$. The electric potential $V$ inside the shell $(r < r_0)$ is constant and equal to the potential at the surface,$V = \frac{kq}{r_0}$.
$2$. The electric potential $V$ outside the shell $(r > r_0)$ decreases as $V = \frac{kq}{r}$,which is a hyperbolic variation.
$3$. The graph shows a constant value for $r < r_0$ and a hyperbolic decrease for $r > r_0$,which matches the behavior of the potential of a uniformly charged spherical shell.

(D) Let the radius of the solid sphere be $r_1$ and the radius of the hollow shell be $r_2$.
In the first condition, the solid sphere has charge $Q$ and the shell is uncharged.
The potential of the solid sphere is $V_{in} = \frac{kQ}{r_1} + \frac{k(0)}{r_2} = \frac{kQ}{r_1}$.
The potential of the hollow shell is $V_{out} = \frac{kQ}{r_2} + \frac{k(0)}{r_2} = \frac{kQ}{r_2}$.
The potential difference is $V = V_{in} - V_{out} = kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.
In the second condition, the shell is given a charge of $-4Q$.
The potential of the solid sphere is $V'_{in} = \frac{kQ}{r_1} + \frac{k(-4Q)}{r_2} = \frac{kQ}{r_1} - \frac{4kQ}{r_2}$.
The potential of the hollow shell is $V'_{out} = \frac{kQ}{r_2} + \frac{k(-4Q)}{r_2} = -\frac{3kQ}{r_2}$.
The new potential difference is $V' = V'_{in} - V'_{out} = \left( \frac{kQ}{r_1} - \frac{4kQ}{r_2} \right) - \left( -\frac{3kQ}{r_2} \right)$.
$V' = \frac{kQ}{r_1} - \frac{4kQ}{r_2} + \frac{3kQ}{r_2} = \frac{kQ}{r_1} - \frac{kQ}{r_2} = kQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = V$.
Thus, the potential difference remains $V$.

(D) The potential at a distance $z$ on the axis of a ring of radius $R$ and charge $q$ is $V = \frac{1}{4\pi \epsilon_0} \frac{q}{\sqrt{R^2 + z^2}}$.
Here,$R = 3a$. At $z = 4a$,the potential energy is $U_i = qV = \frac{q^2}{4\pi \epsilon_0 \sqrt{(3a)^2 + (4a)^2}} = \frac{q^2}{4\pi \epsilon_0 (5a)}$.
At the origin $(z = 0)$,the potential energy is $U_f = qV = \frac{q^2}{4\pi \epsilon_0 (3a)}$.
By the law of conservation of energy,$K_i + U_i = K_f + U_f$. For the charge to just cross the origin,$K_f = 0$.
$\frac{1}{2}mv^2 + \frac{q^2}{4\pi \epsilon_0 (5a)} = 0 + \frac{q^2}{4\pi \epsilon_0 (3a)}$.
$\frac{1}{2}mv^2 = \frac{q^2}{4\pi \epsilon_0 a} (\frac{1}{3} - \frac{1}{5}) = \frac{q^2}{4\pi \epsilon_0 a} (\frac{2}{15})$.
$v^2 = \frac{2}{m} \cdot \frac{2}{15} \cdot \frac{q^2}{4\pi \epsilon_0 a}$.
$v = \sqrt{\frac{2}{m}} \left( \frac{2}{15} \frac{q^2}{4\pi \epsilon_0 a} \right)^{1/2}$.

(C) The kinetic energy $K$ gained by a particle of charge $Q$ accelerated through a potential difference $V$ is given by $K = QV$.
Since the particle starts from rest,its final kinetic energy is $K = \frac{1}{2}mv^2$,where $v$ is the velocity.
The relation between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$.
Substituting $K = QV$ into the momentum equation,we get $p = \sqrt{2mQV}$.

(A) The work done $W$ in moving a charge $q$ between two points with potentials $V_1$ and $V_2$ is given by the formula $W = q(V_2 - V_1)$.
Here,the charge is an electron,so $q = -e$.
The initial potential $V_1 = -60\, V$ and the final potential $V_2 = -20\, V$.
Substituting these values into the formula:
$W = -e(-20\, V - (-60\, V))$
$W = -e(-20\, V + 60\, V)$
$W = -e(40\, V)$
$W = -40\, eV$.
Note: The question asks for the work done by an external agent to move the electron. Since the electron is moved from a lower potential to a higher potential,the external work is positive if we consider the change in potential energy. However,using the standard definition $W = q\Delta V$,we get $-40\, eV$. Given the options provided,the magnitude is $40\, eV$.

(D) The change in electric potential energy is given by the formula $\Delta U = q \Delta V$,where $q$ is the charge and $\Delta V$ is the change in potential.
When a charge is moved from a low potential region to a high potential region,the change in potential $\Delta V = V_{final} - V_{initial}$ is positive.
If the charge $q$ is positive,then $\Delta U = (+q)(+\Delta V) > 0$,meaning the potential energy increases.
If the charge $q$ is negative,then $\Delta U = (-q)(+\Delta V) < 0$,meaning the potential energy decreases.
Therefore,the change in electric potential energy depends on the sign of the charge $q$.

(B) The electrostatic force is a conservative force. For any closed path in an electrostatic field,the total work done is zero.
Given that the work done in moving a charge from $A$ to $B$ is $W_{AB} = 2 \, J$.
The work done in moving a charge from $B$ to $C$ is $W_{BC} = -3 \, J$.
According to the property of conservative forces,the sum of work done along a closed path $A \rightarrow B \rightarrow C \rightarrow A$ is zero:
$W_{AB} + W_{BC} + W_{CA} = 0$
Substituting the given values:
$2 + (-3) + W_{CA} = 0$
$-1 + W_{CA} = 0$
$W_{CA} = 1 \, J$
Therefore,the work done to bring the charge from $C$ to $A$ is $1 \, J$.

(D) 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 $q = 100 \times (-e) = 100 \times (-1.6 \times 10^{-19} \, C) = -1.6 \times 10^{-17} \, C$.
The potential difference is $V_Q - V_P = -4 \, V - 10 \, V = -14 \, V$.
Therefore,the work done is $W = (-1.6 \times 10^{-17} \, C) \times (-14 \, V) = 22.4 \times 10^{-17} \, J = 2.24 \times 10^{-16} \, J$.

(A) The surface charge density $\sigma$ is defined as the charge per unit area,$\sigma = \frac{q}{4 \pi R^2}$.
Since the surface charge density is equal for both spheres,we have $\sigma_1 = \sigma_2 = \sigma$.
Thus,the charge on the spheres is $q_1 = \sigma (4 \pi R_1^2)$ and $q_2 = \sigma (4 \pi R_2^2)$.
The electric potential $V$ at the surface of a charged sphere is given by $V = \frac{kq}{R}$.
Substituting the expression for $q$,we get $V = \frac{k (\sigma 4 \pi R^2)}{R} = k \sigma 4 \pi R$.
Since $k$,$\sigma$,and $4 \pi$ are constants,the potential is directly proportional to the radius: $V \propto R$.
Therefore,the ratio of their potentials is $\frac{V_1}{V_2} = \frac{R_1}{R_2}$.

(B) The electric potential $V$ at a point due to a system of point charges is the algebraic sum of the potentials due to each charge individually: $V = \sum \frac{kq_i}{r_i}$.
In square $ABCD$ of side $a$:
$1$. The distance from $A$ to $D$ is $r_A = a$.
$2$. The distance from $B$ to $D$ is the diagonal of the square,$r_B = a\sqrt{2}$.
$3$. The distance from $C$ to $D$ is $r_C = a$.
The potential at $D$ is $V_D = \frac{kq_A}{r_A} + \frac{kq_B}{r_B} + \frac{kq_C}{r_C}$.
Substituting the values: $V_D = \frac{kq}{a} + \frac{k(\sqrt{2}q)}{a\sqrt{2}} + \frac{k(2q)}{a}$.
$V_D = \frac{kq}{a} + \frac{kq}{a} + \frac{2kq}{a} = \frac{kq}{a}(1 + 1 + 2) = \frac{4kq}{a}$.

(C) The electric field $E$ on the surface of a conducting sphere of radius $R$ and charge $q$ is given by $E = \frac{kq}{R^2}$.
Given that the electric fields near the surfaces are equal:
$\frac{kq_1}{r_1^2} = \frac{kq_2}{r_2^2}$
$\frac{q_1}{q_2} = \frac{r_1^2}{r_2^2} \quad -(1)$
The electric potential $V$ on the surface of a conducting sphere is given by $V = \frac{kq}{R}$.
The ratio of their potentials is:
$\frac{V_1}{V_2} = \frac{kq_1 / r_1}{kq_2 / r_2} = \frac{q_1}{q_2} \cdot \frac{r_2}{r_1}$
Substituting the value of $\frac{q_1}{q_2}$ from equation $(1)$:
$\frac{V_1}{V_2} = \left( \frac{r_1^2}{r_2^2} \right) \cdot \left( \frac{r_2}{r_1} \right) = \frac{r_1}{r_2}$
Thus,the ratio of their electric potentials is $(r_1 / r_2)$.

(B) The potential of the first ball is $V_1 = K \left( \frac{q_1}{r} + \frac{q_2}{a} \right)$ and the potential of the second ball is $V_2 = K \left( \frac{q_2}{r} + \frac{q_1}{a} \right)$,where $K$ is the electrostatic constant.
We have the system of equations:
$V_1 = K \left( \frac{q_1}{r} + \frac{q_2}{a} \right) \implies \frac{V_1}{K} = \frac{q_1}{r} + \frac{q_2}{a} \quad (1)$
$V_2 = K \left( \frac{q_2}{r} + \frac{q_1}{a} \right) \implies \frac{V_2}{K} = \frac{q_2}{r} + \frac{q_1}{a} \quad (2)$
Multiply equation $(1)$ by $a$ and equation $(2)$ by $r$:
$a \frac{V_1}{K} = \frac{aq_1}{r} + q_2$
$r \frac{V_2}{K} = \frac{rq_2}{r} + \frac{rq_1}{a} = q_2 + \frac{rq_1}{a}$
Subtracting the two equations:
$\frac{rV_2 - aV_1}{K} = q_1 \left( \frac{r}{a} - \frac{a}{r} \right) = q_1 \left( \frac{r^2 - a^2}{ar} \right)$
Solving for $q_1$:
$q_1 = \frac{1}{K} \left( \frac{rV_2 - aV_1}{r^2 - a^2} \right) ra$
Similarly,solving for $q_2$:
$q_2 = \frac{1}{K} \left( \frac{rV_1 - aV_2}{r^2 - a^2} \right) ra$

(D) The electric potential $V$ at a point due to a system of point charges is given by $V = \sum \frac{kq_i}{r_i}$.
For point $E$,which is the midpoint of side $BC$,the distances are $AE = \sqrt{(2a)^2 + a^2} = a\sqrt{5}$,$BE = a$,$CE = a$,and $DE = \sqrt{(2a)^2 + a^2} = a\sqrt{5}$.
The potential at $E$ is $V_E = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{a\sqrt{5}} + \frac{q}{a} - \frac{q}{a} - \frac{q}{a\sqrt{5}} \right) = 0$.
Since $V_E = 0$,the ratio of the potential at $E$ to that at $F$ is $\frac{V_E}{V_F} = \frac{0}{V_F} = 0$ (assuming $V_F \neq 0$).

(A) Let $Q_1$ be the charge on the inner shell of radius $r$ and $Q_2$ be the charge on the outer shell of radius $2r$.
The potential at the surface of the inner shell is given by:
$V_{in} = \frac{kQ_1}{r} + \frac{kQ_2}{2r} = 10\,V$ --- $(1)$
The potential at the surface of the outer shell is given by:
$V_{out} = \frac{kQ_1}{2r} + \frac{kQ_2}{2r} = 5\,V$ --- $(2)$
The potential at the centre of the shells is the sum of potentials due to both shells:
$V_{centre} = \frac{kQ_1}{r} + \frac{kQ_2}{2r}$
Comparing this expression with equation $(1)$, we see that the potential at the centre is exactly equal to the potential at the surface of the inner shell.
Therefore, $V_{centre} = 10\,V$.

(B) The electric field on the surface of a charged spherical conductor is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}}$.
Since the charge $q$ is constant,$E \propto \frac{1}{R^{2}}$.
When the radius is halved $(R' = R/2)$,the new electric field $E'$ becomes $E' = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(R/2)^{2}} = 4 \times \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}} \right) = 4E$.
The electric potential of the conductor is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}$.
Since the charge $q$ is constant,$V \propto \frac{1}{R}$.
When the radius is halved $(R' = R/2)$,the new potential $V'$ becomes $V' = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(R/2)} = 2 \times \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R} \right) = 2V$.
Thus,the new values are $4E$ and $2V$.

(B) The electric potential $V$ at a point is defined as the work done by an external agent in moving a unit positive charge from infinity to that point without any change in kinetic energy.
Formula: $V = \frac{W_{ext}}{q}$
Given:
Work done $W_{ext} = 20\,\mu J = 20 \times 10^{-6}\,J$
Charge $q = 2\,\mu C = 2 \times 10^{-6}\,C$
Calculation:
$V = \frac{20 \times 10^{-6}}{2 \times 10^{-6}} = 10\,V$
Therefore,the potential at that point is $10\,V$.

(B) For a conducting sphere of radius $R$ carrying a charge $Q$,the charge resides entirely on its outer surface.
Inside the conductor,the electric field $E$ is zero everywhere because the charges are in electrostatic equilibrium.
However,the electric potential $V$ inside the conductor is constant and equal to the potential at its surface.
The potential at the surface is given by $V = \frac{Q}{4\pi \epsilon_0 R}$.
Therefore,at the centre of the sphere,the electric potential is $\frac{Q}{4\pi \epsilon_0 R}$ and the electric field is $0$.

(C) For a hollow metallic sphere,the electric field inside 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.
Therefore,the potential at any point inside the sphere,including the centre,is equal to the potential at its surface.
Given that the potential at the surface is $80 \; V$,the potential at the centre is also $80 \; V$.

(B) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let charges $Q$ be placed at each vertex. Let $P$ be the midpoint of side $BC$.
The distance from $P$ to the vertices $B$ and $C$ is $r_1 = r_2 = d/2$.
The distance from $P$ to vertex $A$ is the altitude of the triangle,$h = \frac{\sqrt{3}}{2}d$.
The total potential at $P$ is the sum of potentials due to the three charges: $V_P = V_A + V_B + V_C$.
$V_P = \frac{1}{4\pi \epsilon_0} \left( \frac{Q}{h} + \frac{Q}{d/2} + \frac{Q}{d/2} \right) = \frac{Q}{4\pi \epsilon_0} \left( \frac{1}{\frac{\sqrt{3}}{2}d} + \frac{2}{d} + \frac{2}{d} \right)$.
$V_P = \frac{Q}{4\pi \epsilon_0 d} \left( \frac{2}{\sqrt{3}} + 4 \right) = \frac{Q}{2\pi \epsilon_0 d} \left( \frac{1}{\sqrt{3}} + 2 \right)$.

(D) The electric field $E$ is directed from a region of higher potential to a region of lower potential.
An electron carries a negative charge $q = -e$.
The force on a charged particle in an electric field is given by $F = qE$.
Since the charge is negative,the force on the electron is in the direction opposite to the electric field.
Therefore,an electron moves from a region of lower potential to a region of higher potential.
Thus,the Assertion is incorrect,and the Reason is correct.

(D) The electric field $E$ at the surface of a conducting sphere of radius $r$ is given by $E = \frac{V}{r}$,where $V$ is the potential of the sphere.
Given,the maximum electric field $E_{max} = 10^7\,V/m$ and the radius $r = 0.10\,m$.
To find the maximum potential $V_{max}$,we use the relation $V_{max} = E_{max} \times r$.
Substituting the values: $V_{max} = 10^7\,V/m \times 0.10\,m$.
$V_{max} = 10^7 \times 10^{-1} = 10^6\,V$.
Thus,the maximum potential to which the sphere can be charged is $10^6\,V$.

(A) Let the distance from the charge $+q$ where the potential is zero be $x$ (in meters).
The total potential $V$ at this point is the sum of the potentials due to both charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{x} + \frac{-3q}{1-x} \right) = 0$
Since $\frac{1}{4 \pi \varepsilon_0} \neq 0$,we have:
$\frac{q}{x} - \frac{3q}{1-x} = 0$
$\frac{1}{x} = \frac{3}{1-x}$
$1 - x = 3x$
$1 = 4x$
$x = \frac{1}{4} \, m = 0.25 \, m = 25 \, cm$.
Thus,the distance is $25 \, cm$.

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

(A) Let the inner shell $B$ have radius $R_1$ and charge $Q_1$,and the outer shell $A$ have radius $R_2$ and charge $Q_2$.
The potential at the surface of the outer shell $A$ is $V_A = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1 + Q_2}{R_2}$.
The potential at the surface of the inner shell $B$ is $V_B = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{R_1} + \frac{Q_2}{R_2} \right)$.
The potential difference between the shells is $V_B - V_A = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{R_1} + \frac{Q_2}{R_2} - \frac{Q_1 + Q_2}{R_2} \right) = \frac{1}{4 \pi \varepsilon_0} Q_1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
This shows that the potential difference depends only on the charge of the inner shell $Q_1$. Thus,the Assertion is correct.
The potential due to the charge of the outer shell $Q_2$ at any point inside it (including the surface of the inner shell) is constant and equal to $\frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R_2}$. Thus,the Reason is also correct and explains why the $Q_2$ term cancels out in the potential difference calculation.

(C) Electric potential $(V)$ at a point is defined as the work done per unit positive charge in bringing it from infinity to that point,given by $V = W/q$.
Electric potential energy $(U)$ is the total work done in assembling a system of charges,given by $U = qV$.
Since $U$ and $V$ have different physical dimensions ($[ML^2T^{-3}A^{-1}]$ for $V$ and $[ML^2T^{-2}A^{-1}]$ for $U$),they are distinct physical quantities.
The Assertion is correct.
The Reason states that $U = V$,which is dimensionally inconsistent and physically incorrect.
Therefore,the Assertion is correct but the Reason is incorrect.

(N/A) The electric potential $V$ at a distance $r$ due to a charge $Q$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$.
Given $Q = 4 \times 10^{-7} \; C$ and $r = 9 \; cm = 0.09 \; m$.
$V = (9 \times 10^{9} \; N \cdot m^{2} \cdot C^{-2}) \times \frac{4 \times 10^{-7} \; C}{0.09 \; m} = 4 \times 10^{4} \; V$.
$(b)$ The work done $W$ in bringing a charge $q$ from infinity to a point $P$ is $W = qV$.
Given $q = 2 \times 10^{-9} \; C$.
$W = (2 \times 10^{-9} \; C) \times (4 \times 10^{4} \; V) = 8 \times 10^{-5} \; J$.
No,the work done is path-independent because the electrostatic force is a conservative force. The work done depends only on the initial and final positions.

(N/A) Let us take the origin $O$ at the location of the positive charge. The line joining the two charges is taken to be the $x$-axis; the negative charge is taken to be on the right side of the origin.
Let $P$ be the required point on the $x$-axis where the potential is zero.
If $x$ is the $x$-coordinate of $P$,obviously $x$ must be positive. If $x$ lies between $O$ and $A$,we have:
$\frac{1}{4 \pi \varepsilon_{0}} \left[ \frac{3 \times 10^{-8}}{x \times 10^{-2}} - \frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}} \right] = 0$
where $x$ is in $cm$. That is,
$\frac{3}{x} - \frac{2}{15-x} = 0$
which gives $x = 9 \; cm$.
If $x$ lies on the extended line $OA$,the required condition is:
$\frac{3}{x} - \frac{2}{x-15} = 0$
which gives $x = 45 \; cm$.
Thus,the electric potential is zero at $9 \; cm$ and $45 \; cm$ away from the positive charge on the side of the negative charge.

(A-D) As $V \propto \frac{1}{r}$,the potential is higher closer to the positive charge. Thus,$V_{P} > V_{Q}$,so $(V_{P}-V_{Q})$ is positive. For the negative charge,the potential is more negative closer to the charge. Thus,$V_{B} > V_{A}$ (since $V_{A}$ is more negative),so $(V_{B}-V_{A})$ is positive.
$(b)$ Potential energy $U = qV$. For a negative charge $q < 0$,if $V_{P} > V_{Q}$,then $U_{P} < U_{Q}$. The difference $(U_{P}-U_{Q})$ is negative. Similarly,for the negative charge,since $V_{B} > V_{A}$,$U_{B} > U_{A}$,so $(U_{A}-U_{B})$ is negative.
$(c)$ $A$ positive charge experiences a repulsive force from the positive source charge. Moving it from $Q$ to $P$ (closer) requires work against the field. Thus,the work done by the field is negative.
$(d)$ $A$ negative charge is attracted to the positive source charge but repelled by the negative source charge. Moving a negative charge from $B$ to $A$ (closer to the negative source) requires work against the repulsive force. Thus,the work done by the external agency is positive.
$(e)$ As the negative charge moves from $B$ to $A$,it moves closer to the negative source charge,experiencing a stronger repulsive force. This force does negative work,causing the kinetic energy to decrease.

(N/A) Since the work done depends on the final arrangement of the charges,and not on how they are put together,we calculate the work needed for one way of placing the charges at $A, B, C$ and $D$. Suppose,first the charge $+q$ is brought to $A$,and then the charges $-q, +q,$ and $-q$ are brought to $B, C$ and $D$,respectively. The total work needed can be calculated in steps:
$(i)$ Work needed to bring charge $+q$ to $A$ when no charge is present elsewhere: This is $0$.
$(ii)$ Work needed to bring $-q$ to $B$ when $+q$ is at $A$. This is given by (charge at $B$) $\times$ (electrostatic potential at $B$ due to charge $+q$ at $A$):
$W_2 = -q \times \left(\frac{q}{4 \pi \varepsilon_{0} d}\right) = -\frac{q^{2}}{4 \pi \varepsilon_{0} d}$
$(iii)$ Work needed to bring charge $+q$ to $C$ when $+q$ is at $A$ and $-q$ is at $B$. This is given by (charge at $C$) $\times$ (potential at $C$ due to charges at $A$ and $B$):
$W_3 = +q \left(\frac{+q}{4 \pi \varepsilon_{0} d \sqrt{2}} + \frac{-q}{4 \pi \varepsilon_{0} d}\right) = \frac{q^2}{4 \pi \varepsilon_{0} d} \left(\frac{1}{\sqrt{2}} - 1\right)$
$(iv)$ Work needed to bring $-q$ to $D$ when $+q$ at $A, -q$ at $B,$ and $+q$ at $C$. This is given by (charge at $D$) $\times$ (potential at $D$ due to charges at $A, B$ and $C$):
$W_4 = -q \left(\frac{+q}{4 \pi \varepsilon_{0} d} + \frac{-q}{4 \pi \varepsilon_{0} d \sqrt{2}} + \frac{+q}{4 \pi \varepsilon_{0} d}\right) = -\frac{q^2}{4 \pi \varepsilon_{0} d} \left(2 - \frac{1}{\sqrt{2}}\right)$
Adding the work done in steps $(i), (ii), (iii)$ and $(iv)$,the total work required is:
$W = W_1 + W_2 + W_3 + W_4 = \frac{-q^{2}}{4 \pi \varepsilon_{0} d} \left(0 + 1 + (1 - \frac{1}{\sqrt{2}}) + (2 - \frac{1}{\sqrt{2}})\right) = \frac{-q^{2}}{4 \pi \varepsilon_{0} d} (4 - \sqrt{2})$
$(b)$ The extra work necessary to bring a charge $q_{0}$ to the point $E$ when the four charges are at $A, B, C$ and $D$ is $q_{0} \times$ (electrostatic potential at $E$ due to the charges at $A, B, C$ and $D$). The electrostatic potential at $E$ is zero since the potential due to $A$ and $C$ is cancelled by that due to $B$ and $D$. Hence,no work is required to bring any charge to point $E$.

(N/A) Given charges are $q_{1} = 5 \times 10^{-8} \; C$ and $q_{2} = -3 \times 10^{-8} \; C$. The distance between them is $d = 16 \; cm = 0.16 \; m$.
Case $1$: Point $P$ is between the charges at a distance $r$ from $q_{1}$.
The potential $V$ at $P$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_{1}}{r} + \frac{q_{2}}{d-r} \right)$.
Setting $V = 0$,we get $\frac{q_{1}}{r} = -\frac{q_{2}}{d-r}$.
Substituting values: $\frac{5 \times 10^{-8}}{r} = -\frac{-3 \times 10^{-8}}{0.16-r}$.
$5(0.16 - r) = 3r \Rightarrow 0.8 - 5r = 3r \Rightarrow 8r = 0.8 \Rightarrow r = 0.1 \; m = 10 \; cm$.
Case $2$: Point $P$ is outside the charges at a distance $s$ from $q_{1}$.
The potential $V$ at $P$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_{1}}{s} + \frac{q_{2}}{s-d} \right)$.
Setting $V = 0$,we get $\frac{q_{1}}{s} = -\frac{q_{2}}{s-d}$.
Substituting values: $\frac{5 \times 10^{-8}}{s} = -\frac{-3 \times 10^{-8}}{s-0.16}$.
$5(s - 0.16) = 3s \Rightarrow 5s - 0.8 = 3s \Rightarrow 2s = 0.8 \Rightarrow s = 0.4 \; m = 40 \; cm$.
Thus,the potential is zero at $10 \; cm$ from the positive charge between the charges and at $40 \; cm$ from the positive charge outside the system.

(D) The given figure shows six equal charges,$q$,at the vertices of a regular hexagon.
Charge,$q = 5 \; \mu C = 5 \times 10^{-6} \; C$
Side of the hexagon,$l = 10 \; cm = 0.1 \; m$
In a regular hexagon,the distance of each vertex from the centre $O$ is equal to the side length of the hexagon. Therefore,the distance $d = 10 \; cm = 0.1 \; m$.
The electric potential $V$ at the centre $O$ due to a single charge $q$ at distance $d$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d}$.
Since there are six identical charges at the vertices,the total potential at the centre is the sum of the potentials due to each charge:
$V_{total} = 6 \times \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d} \right)$
Using $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; N m^{2} C^{-2}$:
$V_{total} = 6 \times \frac{9 \times 10^{9} \times 5 \times 10^{-6}}{0.1}$
$V_{total} = \frac{270 \times 10^{3}}{0.1} = 2700 \times 10^{3} = 2.7 \times 10^{6} \; V$
Therefore,the potential at the centre of the hexagon is $2.7 \times 10^{6} \; V$.

(B) The charge located at the origin is $q = 8 \; mC = 8 \times 10^{-3} \; C$.
The small charge being moved is $q_1 = -2 \times 10^{-9} \; C$.
Since the electrostatic force is conservative,the work done is independent of the path taken and depends only on the initial and final positions.
Potential at point $P$ (at distance $d_1 = 3 \; cm = 0.03 \; m$): $V_P = \frac{1}{4 \pi \epsilon_0} \frac{q}{d_1} = 9 \times 10^9 \times \frac{8 \times 10^{-3}}{0.03} = 24 \times 10^8 \; V$.
Potential at point $Q$ (at distance $d_2 = 4 \; cm = 0.04 \; m$): $V_Q = \frac{1}{4 \pi \epsilon_0} \frac{q}{d_2} = 9 \times 10^9 \times \frac{8 \times 10^{-3}}{0.04} = 18 \times 10^8 \; V$.
Work done $W = q_1 (V_Q - V_P) = (-2 \times 10^{-9}) \times (18 \times 10^8 - 24 \times 10^8)$.
$W = (-2 \times 10^{-9}) \times (-6 \times 10^8) = 12 \times 10^{-1} = 1.2 \; J$.
Rounding to the nearest provided option,the work done is $1.27 \; J$.

(C) The length of the side of the cube is $b$.
There is a charge $q$ at each of its $8$ vertices.
The distance from the centre of the cube to any vertex is $r = \frac{1}{2} \times \text{diagonal of the cube}$.
The diagonal of the cube is $\sqrt{b^2 + b^2 + b^2} = b\sqrt{3}$.
Thus,$r = \frac{b\sqrt{3}}{2}$.
The electric potential $V$ at the centre due to one charge is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
Since there are $8$ charges,the total potential $V = 8 \times V_1 = 8 \times \frac{1}{4\pi\epsilon_0} \frac{q}{(b\sqrt{3}/2)}$.
$V = \frac{8q}{4\pi\epsilon_0} \times \frac{2}{b\sqrt{3}} = \frac{4q}{\sqrt{3}\pi\epsilon_0 b}$.
The electric field at the centre due to each charge is directed away from the charge (if $q>0$) towards the opposite vertex. Since the charges are placed symmetrically,the electric field vectors from opposite vertices cancel each other out. Therefore,the net electric field at the centre is $0$.

(N/A) The system consists of four charges: $+q$ at $x = -a$,$-q$ at $x = 0$,$-q$ at $x = 0$,and $+q$ at $x = a$. This is equivalent to a charge $+q$ at $x = -a$,a charge $-2q$ at $x = 0$,and a charge $+q$ at $x = a$.
Let point $P$ be at a distance $r$ from the origin (where the $-2q$ charge is located) along the axis.
The distances of the charges from $P$ are:
For $+q$ at $x = -a$: $d_1 = r + a$
For $-2q$ at $x = 0$: $d_2 = r$
For $+q$ at $x = a$: $d_3 = r - a$
The total electrostatic potential $V$ at point $P$ is:
$V = \frac{1}{4 \pi \epsilon_{0}} \left[ \frac{q}{r + a} - \frac{2q}{r} + \frac{q}{r - a} \right]$
$V = \frac{q}{4 \pi \epsilon_{0}} \left[ \frac{r(r - a) - 2(r^2 - a^2) + r(r + a)}{r(r^2 - a^2)} \right]$
$V = \frac{q}{4 \pi \epsilon_{0}} \left[ \frac{r^2 - ra - 2r^2 + 2a^2 + r^2 + ra}{r(r^2 - a^2)} \right]$
$V = \frac{q}{4 \pi \epsilon_{0}} \left[ \frac{2a^2}{r(r^2 - a^2)} \right]$
For $r >> a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{2qa^2}{4 \pi \epsilon_{0} r^3}$
Thus,for an electric quadrupole,$V \propto \frac{1}{r^3}$.
Comparing this with other systems:
For an electric monopole (single charge),$V \propto \frac{1}{r}$.
For an electric dipole,$V \propto \frac{1}{r^2}$.

(A) The potential $V_{1}$ of the inner sphere of radius $r_{1}$ is given by $V_{1} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}} \right)$.
The potential $V_{2}$ of the outer shell of radius $r_{2}$ is given by $V_{2} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{r_{2}} + \frac{q_{2}}{r_{2}} \right)$.
The potential difference between the sphere and the shell is $V = V_{1} - V_{2} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}} - \frac{q_{1}}{r_{2}} - \frac{q_{2}}{r_{2}} \right)$.
Simplifying this,we get $V = \frac{q_{1}}{4\pi\epsilon_{0}} \left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right)$.
Since $r_{2} > r_{1}$,the term $\left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right)$ is always positive.
If $q_{1} > 0$,then $V > 0$,which means $V_{1} > V_{2}$.
Therefore,charge will always flow from the inner sphere to the outer shell when connected by a wire,regardless of the value of $q_{2}$.

(N/A) force is conservative if the work done by it in moving a particle between two points is independent of the path taken. For an electrostatic force,the work done in moving a charge $q$ from point $A$ to point $B$ is $W = -(U_B - U_A)$,which depends only on the initial and final positions. Since the work done in a closed loop is zero,electrostatic forces are conservative.
Electrostatic potential energy is defined as: The work done by an external agent in bringing a unit positive charge from infinity to a specific point in an electric field at a constant speed (without acceleration) is defined as the electrostatic potential energy at that point.

(N/A) At every point in an electric field,a particle with charge $q$ possesses a certain electrostatic potential energy. The work done by an external force in moving this charge increases its potential energy by an amount equal to the potential energy difference between points $R$ and $P$.
Thus,the potential energy difference is:
$\Delta U = U_{P} - U_{R} = W_{RP}$
Therefore,we can define the electric potential energy difference between two points as the work required to be done by an external force in moving (without acceleration) a charge $q$ from one point to another within the electric field of any arbitrary charge configuration.
Following comments may be made:
$(i)$ The work done depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another is independent of the path taken. This is the fundamental characteristic of a conservative force.
$(ii)$ The absolute value of potential energy is not significant; only the difference in potential energy is physically significant. If we add an arbitrary constant $\alpha$ to the potential energy at every point,the difference remains unchanged: $(U_{P} + \alpha) - (U_{R} + \alpha) = U_{P} - U_{R}$.
If we define the potential energy to be zero at infinity,then the work done in bringing a charge from infinity to a point $P$ is $W_{\infty P} = U_{P}$. Thus,the potential energy of a charge $q$ at a point is defined as the work done by an external force (equal and opposite to the electric field force) in bringing the charge $q$ from infinity to that point.

(N/A) Voltage,also known as electric potential difference,is the measure of the energy per unit charge that is required to move a charge between two points in an electric circuit.
It represents the 'pressure' or 'force' that drives electric current through a conductor.
Mathematically,it is defined as $V = \frac{W}{Q}$,where $W$ is the work done and $Q$ is the charge.
The $SI$ unit of voltage is the volt $(V)$.

(N/A) $1 \ eV$ (electron-volt) is defined as the amount of kinetic energy gained by an electron when it is accelerated through a potential difference of $1 \ V$.
Mathematically,$1 \ eV = 1.602 \times 10^{-19} \ J$.

(N/A) The electrostatic potential energy of a system of point charges is defined as the total amount of work done by an external agent in bringing the charges from infinity to their respective positions in the configuration,without any acceleration.
Mathematically,for a system of two charges $q_1$ and $q_2$ separated by a distance $r$,the electrostatic potential energy $U$ is given by:
$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$
where $\epsilon_0$ is the permittivity of free space.

(N/A) The work done by an external force in bringing a unit positive charge from infinity to a given point in an electric field,without acceleration,is called the electrostatic or electric potential at that point. It is denoted by $V$.
Consider a source charge $Q$ at the origin $O$. Let $P$ be a point at a certain distance from $O$,and $R$ be a point at infinity.
The work done in bringing a unit positive charge from infinity to point $P$ is the potential energy per unit charge.
If $U_P$ is the potential energy at point $P$ and $U_R$ is the potential energy at point $R$,then the electric potential difference between these points is given by $\frac{U_P - U_R}{q}$.
Thus,$V_P - V_R = \frac{U_P - U_R}{q}$,where $V_P$ and $V_R$ are potentials at points $P$ and $R$ respectively.
Since the absolute value of electric potential is not physically significant,only the potential difference is important. By convention,we take the potential at infinity to be zero $(V_R = 0)$.
Therefore,$V_P = \frac{U_P}{q}$.
The $SI$ unit of electric potential is the volt $(V)$,where $1 \ V = 1 \ J/C$. Other units include statvolt $(statV)$ in the $CGS$ system,where $1 \ V = \frac{1}{300} \ statV$.

(A) Electrostatic potential at a point is defined as the amount of work done in bringing a unit positive test charge from infinity to that point in an electric field.
Since work done is a scalar quantity,electrostatic potential is also a scalar quantity.
It has only magnitude and no specific direction associated with it.

(N/A) Consider a point charge $Q$ at the origin $O$. The electric potential at a point $P$ at a distance $r$ from $Q$ is defined as the work done by an external agent in bringing a unit positive test charge from infinity to point $P$.
Since the electrostatic force is conservative,the work done is independent of the path. We choose a radial path from infinity to $P$.
At an intermediate point $P^{\prime}$ at a distance $r^{\prime}$ from $O$,the electrostatic force on a unit positive test charge is:
$F = \frac{1}{4 \pi \epsilon_{0}} \frac{Q \times 1}{(r^{\prime})^{2}} = \frac{kQ}{(r^{\prime})^{2}}$
The work done by an external force $F_{ext}$ against the electrostatic force $F$ for a small displacement $dr^{\prime}$ towards the charge is:
$dW = F_{ext} \cdot dr^{\prime} = -F \cdot dr^{\prime} = -\frac{kQ}{(r^{\prime})^{2}} dr^{\prime}$
To find the total work done $W$ in bringing the charge from infinity to $r$,we integrate the expression:
$W = \int_{\infty}^{r} -\frac{kQ}{(r^{\prime})^{2}} dr^{\prime}$
$W = -kQ \int_{\infty}^{r} (r^{\prime})^{-2} dr^{\prime}$
$W = -kQ \left[ -\frac{1}{r^{\prime}} \right]_{\infty}^{r}$
$W = kQ \left[ \frac{1}{r} - \frac{1}{\infty} \right] = \frac{kQ}{r}$
Since electric potential $V = \frac{W}{q_{0}}$ and $q_{0} = 1 \text{ C}$,we have:
$V(r) = \frac{kQ}{r} = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r}$

(N/A) The electrostatic potential $V$ due to a point charge $Q$ at a distance $r$ is given by the formula $V = \frac{kQ}{r}$,where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.
From this relation,we can see that $V \propto \frac{1}{r}$.
As the distance $r$ increases,the potential $V$ decreases hyperbolically.
If the charge $Q$ is positive,the potential $V$ is positive for all $r > 0$. If the charge $Q$ is negative,the potential $V$ is negative for all $r > 0$.
The graph of $V$ versus $r$ is a hyperbola that approaches the $r$-axis as $r \to \infty$ and approaches infinity as $r \to 0$.

(N/A)

Electric Potential $(V)$	Electric Potential Energy $(U)$
$(1)$ Work done against an electric field to bring a unit positive charge from infinity to a point is electric potential.	$(1)$ Work done against an electric field to bring a charge $q$ from infinity to a point is electric potential energy.
$(2)$ $SI$ Unit: $J \cdot C^{-1}$ or $\text{Volt} (V)$.	$(2)$ $SI$ Unit: $J$ (Joule).
$(3)$ Formula: $V = \frac{W}{q}$, where $q$ is a unit positive charge.	$(3)$ Formula: $U = qV$.
$(4)$ Dimensional Formula: $[M^1 L^2 T^{-3} A^{-1}]$.	$(4)$ Dimensional Formula: $[M^1 L^2 T^{-2}]$.

(N/A) The electrostatic potential $V$ at a distance $r$ from a point charge $q$ is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
For a negative point charge,we substitute $q = -|q|$.
Therefore,the equation for the electrostatic potential of a negative point charge is: $V = -\frac{1}{4\pi\epsilon_0} \frac{|q|}{r}$,where $\epsilon_0$ is the permittivity of free space.

(B) The electrostatic potential $V$ at a distance $r$ from a point charge $q$ is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
From this expression,it is clear that the potential $V$ is inversely proportional to the distance $r$,i.e.,$V \propto \frac{1}{r}$.
As $r$ increases,$V$ decreases following a rectangular hyperbola curve.
Therefore,the graph representing $V$ versus $r$ is a curve where $V$ decreases as $r$ increases,which corresponds to the relationship $V \propto \frac{1}{r}$.

(N/A) Consider a system of charges $q_{1}, q_{2}, q_{3}, \ldots, q_{N}$ located at distances $r_{1P}, r_{2P}, r_{3P}, \ldots, r_{NP}$ from a point $P$.
The electric potential at point $P$ due to a single charge $q_{i}$ is given by $V_{i} = \frac{1}{4 \pi \epsilon_{0}} \frac{q_{i}}{r_{iP}}$.
Since electric potential is a scalar quantity,the total electric potential $V$ at point $P$ due to the system of $N$ charges is the algebraic sum of the potentials due to individual charges:
$V = V_{1} + V_{2} + V_{3} + \ldots + V_{N}$
Substituting the expression for each potential:
$V = \frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}}{r_{1P}} + \frac{1}{4 \pi \epsilon_{0}} \frac{q_{2}}{r_{2P}} + \ldots + \frac{1}{4 \pi \epsilon_{0}} \frac{q_{N}}{r_{NP}}$
Taking the common factor $\frac{1}{4 \pi \epsilon_{0}}$ out:
$V = \frac{1}{4 \pi \epsilon_{0}} \sum_{i=1}^{N} \frac{q_{i}}{r_{iP}}$
Here,$\epsilon_{0}$ is the permittivity of free space and $r_{iP}$ is the distance of the $i$-th charge from point $P$.

(N/A) For a continuous charge distribution,the electric potential $V$ at a point $P$ with position vector $\vec{r}$ is given by the integral of the potential contributions from infinitesimal charge elements $dq$.
$1$. For linear charge distribution:
$V = k \int_{L} \frac{\lambda dl}{|\vec{r} - \vec{r}'|}$
where $\lambda$ is the linear charge density,$dl$ is the length element,and $\vec{r}'$ is the position vector of the charge element.
$2$. For surface charge distribution:
$V = k \int_{S} \frac{\sigma dS}{|\vec{r} - \vec{r}'|}$
where $\sigma$ is the surface charge density and $dS$ is the area element.
$3$. For volume charge distribution:
$V = k \int_{V} \frac{\rho dV}{|\vec{r} - \vec{r}'|}$
where $\rho$ is the volume charge density and $dV$ is the volume element.
In all expressions,$k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.