Electric potential Questions in English

(A) Let $q_1$ be the charge on the inner sphere of radius $a$ and $q_2$ be the charge on the outer shell of radius $b$.
The potential at the surface of the inner sphere is given by the sum of potentials due to both spheres: $V_a = \frac{kq_1}{a} + \frac{kq_2}{b} = 10 \text{ V}$.
The potential at the surface of the outer shell is given by: $V_b = \frac{kq_1}{b} + \frac{kq_2}{b} = \frac{k(q_1 + q_2)}{b} = 5 \text{ V}$.
The potential at the centre of the spheres is the sum of the potentials due to both spheres at that point. Since the potential inside a spherical shell is constant and equal to the potential at its surface,the potential at the centre is: $V_c = \frac{kq_1}{a} + \frac{kq_2}{b}$.
Comparing this with the expression for $V_a$,we see that $V_c = V_a = 10 \text{ V}$.

(C) The charges on the shells are given by $q = \text{surface charge density} \times \text{area}$.
For shell $A$ (radius $a$): $q_A = -\sigma(4\pi a^2)$.
For shell $B$ (radius $b$): $q_B = +\sigma(4\pi b^2)$.
For shell $C$ (radius $c$): $q_C = -\sigma(4\pi c^2)$.
The potential at the surface of shell $A$ is the sum of potentials due to all three shells:
$V_A = V_{A,A} + V_{A,B} + V_{A,C} = \frac{1}{4\pi\varepsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}]$.
Substituting the values:
$V_A = \frac{1}{4\pi\varepsilon_0} [\frac{-\sigma(4\pi a^2)}{a} + \frac{\sigma(4\pi b^2)}{b} + \frac{-\sigma(4\pi c^2)}{c}]$.
$V_A = \frac{1}{4\pi\varepsilon_0} [-\sigma(4\pi a) + \sigma(4\pi b) - \sigma(4\pi c)]$.
$V_A = \frac{\sigma}{\varepsilon_0} [-a + b - c] = \frac{\sigma}{\varepsilon_0} [b - a - c]$.

(C) The electrostatic potential at the center of a uniformly charged solid sphere of radius $R$ is given by $V_c = \frac{3}{2} \frac{kQ}{R}$.
For a point outside the sphere at a distance $x$ from the center (where $x > R$),the potential is $V(x) = \frac{kQ}{x}$.
According to the problem,the potential at distance $x$ is half of the potential at the center:
$V(x) = \frac{1}{2} V_c$
$\frac{kQ}{x} = \frac{1}{2} \left( \frac{3}{2} \frac{kQ}{R} \right)$
$\frac{1}{x} = \frac{3}{4R}$
$x = \frac{4R}{3}$.
The distance from the surface of the sphere is $d = x - R = \frac{4R}{3} - R = \frac{R}{3}$.

(B) The work done in moving a charge $q$ from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.
The potential at the centre $A$ of the first ring due to both rings is:
$V_A = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_1}{R} + \frac{Q_2}{\sqrt{R^2 + R^2}} \right) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_1}{R} + \frac{Q_2}{\sqrt{2}R} \right)$.
The potential at the centre $B$ of the second ring due to both rings is:
$V_B = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_2}{R} + \frac{Q_1}{\sqrt{R^2 + R^2}} \right) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_2}{R} + \frac{Q_1}{\sqrt{2}R} \right)$.
The work done is $W = q(V_B - V_A)$:
$W = \frac{q}{4\pi \varepsilon_0} \left[ \left( \frac{Q_2}{R} + \frac{Q_1}{\sqrt{2}R} \right) - \left( \frac{Q_1}{R} + \frac{Q_2}{\sqrt{2}R} \right) \right]$.
$W = \frac{q}{4\pi \varepsilon_0 R} \left[ (Q_2 - Q_1) + \frac{Q_1 - Q_2}{\sqrt{2}} \right]$.
$W = \frac{q(Q_2 - Q_1)}{4\pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} \right) = \frac{q(Q_2 - Q_1)}{4\pi \varepsilon_0 R} \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)$.
Since the question asks for work done moving from $A$ to $B$,and the options are expressed in terms of $(Q_1 - Q_2)$,we note that $W_{A \to B} = -W_{B \to A}$. The magnitude or the expression matching the options is $q(Q_1 - Q_2)(\sqrt{2} - 1) / (\sqrt{2} \cdot 4\pi \varepsilon_0 R)$.

(B) ગોળાની સપાટી પર ગોળીની સ્થિતિ ઉર્જા $U_s = qV_s = q \left( \frac{kq}{R} \right) = \frac{kq^2}{R}$ છે.
ગોળાના કેન્દ્ર પર ગોળીની સ્થિતિ ઉર્જા $U_c = qV_c = q \left( \frac{3kq}{2R} \right) = \frac{3kq^2}{2R}$ છે.
ગોળી ગોળામાંથી પસાર થઈ શકે તે માટે,સપાટી પર તેની ગતિ ઉર્જા સપાટીથી કેન્દ્ર સુધી જતી વખતે થતા સ્થિતિ ઉર્જાના ફેરફાર જેટલી હોવી જોઈએ.
$\frac{1}{2}mu^2 = U_c - U_s$
$\frac{1}{2}mu^2 = \frac{3kq^2}{2R} - \frac{kq^2}{R} = \frac{kq^2}{2R}$
$u^2 = \frac{kq^2}{mR}$
$k = \frac{1}{4\pi\varepsilon_0}$ મૂકતા,આપણને $u^2 = \frac{q^2}{4\pi\varepsilon_0 mR}$ મળે છે.
તેથી,$u = \frac{q}{\sqrt{4\pi\varepsilon_0 mR}}$.

(A) Let the radius of the ring be $R$. From the diagram,the distance of $P$ from $+Q$ is $4a$. Thus,$V = \frac{Q}{4\pi\epsilon_0(4a)}$.
The bead must reach the point $B$ (the point closest to $+Q$) to complete the circle. The distance of $B$ from $+Q$ is $a$. The potential at $B$ is $V_B = \frac{Q}{4\pi\epsilon_0 a} = 4V$.
Using the principle of conservation of mechanical energy between point $P$ and point $B$:
$U_P + K_P = U_B + K_B$
$qV + \frac{1}{2}mv_0^2 = qV_B + K_B$
$qV + \frac{1}{2}mv_0^2 = q(4V) + K_B$
For the bead to complete the circle,it must have a non-zero kinetic energy at point $B$ $(K_B > 0)$.
$\frac{1}{2}mv_0^2 = 3qV + K_B$
Since $K_B > 0$,we have $\frac{1}{2}mv_0^2 > 3qV$.
$v_0^2 > \frac{6qV}{m}$
$v_0 > \sqrt{\frac{6qV}{m}}$.

(B) In a uniform electric field,the potential $V$ can be expressed as $V(x, y, z) = V_0 - (E_x x + E_y y + E_z z)$.
Given $V(0, 0, 0) = 10 \ V$,so $V_0 = 10 \ V$.
Given $V(1, 0, 0) = 8 \ V$,so $10 - E_x(1) = 8 \implies E_x = 2 \ V$.
Given $V(0, 1, 0) = 8 \ V$,so $10 - E_y(1) = 8 \implies E_y = 2 \ V$.
Given $V(0, 0, 1) = 8 \ V$,so $10 - E_z(1) = 8 \implies E_z = 2 \ V$.
Now,the potential at point $(1, 1, 1)$ is $V(1, 1, 1) = 10 - (2 \times 1 + 2 \times 1 + 2 \times 1) = 10 - 6 = 4 \ V$.

(D) $1$. Since the shell is made of conducting material,the electric field inside the material of the shell $(a < R < b)$ must be zero.
$2$. To ensure the electric field is zero inside the conductor,an induced charge of $-Q$ must appear on the inner surface of the shell (at radius $a$).
$3$. The total charge on the shell is $-q$. Since the inner surface has a charge of $-Q$,the charge on the outer surface (at radius $b$) must be $q_{outer} = (-q) - (-Q) = Q - q$.
$4$. The potential at a point $R$ inside the material of the shell $(a < R < b)$ is the sum of the potentials due to the point charge $+Q$ at the center,the induced charge $-Q$ on the inner surface,and the charge $(Q-q)$ on the outer surface.
$5$. Potential $V(R) = V_{point} + V_{inner} + V_{outer} = \frac{KQ}{R} + \frac{K(-Q)}{R} + \frac{K(Q-q)}{b}$.
$6$. Simplifying this,$V(R) = 0 + \frac{K(Q-q)}{b} = \frac{K(Q-q)}{b}$.

(D) Let the charge on shell $C$ be $q'$.
Since shell $C$ is earthed,its potential $V_C = 0$.
The potential at shell $C$ (radius $3a$) is due to charges on $B, C,$ and $D$:
$V_C = \frac{Kq}{3a} + \frac{Kq'}{3a} - \frac{Kq}{4a} = 0$
$\frac{q}{3a} + \frac{q'}{3a} = \frac{q}{4a}$
$q + q' = \frac{3q}{4} \implies q' = -\frac{q}{4}$.
Now,calculate the potential at shell $A$ (radius $a$):
$V_A = \frac{Kq}{2a} + \frac{Kq'}{3a} - \frac{Kq}{4a}$
Substitute $q' = -\frac{q}{4}$:
$V_A = \frac{Kq}{2a} + \frac{K(-q/4)}{3a} - \frac{Kq}{4a} = \frac{Kq}{2a} - \frac{Kq}{12a} - \frac{Kq}{4a}$
$V_A = Kq \left( \frac{6 - 1 - 3}{12a} \right) = \frac{2Kq}{12a} = \frac{Kq}{6a}$.
Since $V_C = 0$,the potential difference $V_A - V_C = \frac{Kq}{6a} - 0 = \frac{Kq}{6a}$.

(A) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{KQ}{r}$.
Here,$Q = 10^{-8} \ C$.
The position of the charge is $(4, 7, 2)$ and the observation point is $(1, 3, 2)$.
The distance $r$ is calculated as $r = \sqrt{(1-4)^2 + (3-7)^2 + (2-2)^2} = \sqrt{(-3)^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ m$.
Substituting the values,$V = \frac{(9 \times 10^9) \times 10^{-8}}{5} = \frac{90}{5} = 18 \ V$.
Thus,the electric potential at the given point is $18 \ V$.

(D) The potential at point $A$ is $V_A = 3 \ V$ and at point $B$ is $V_B = 7 \ V$.
An electron has a negative charge $q = -e$.
The potential energy of an electron at any point is given by $U = qV = -eV$.
Therefore,the potential energy at $A$ is $U_A = -3e \ V$ and at $B$ is $U_B = -7e \ V$.
Since the electron is moving from $B$ to $A$,it is moving from a region of lower potential energy $(-7e \ V)$ to a region of higher potential energy $(-3e \ V)$.
By the law of conservation of energy,$K_i + U_i = K_f + U_f$,where $K$ is kinetic energy.
$K_B + U_B = K_A + U_A$
$K_A = K_B + (U_B - U_A) = K_B + (-7e - (-3e)) = K_B - 4e$.
For the electron to reach $A$,its kinetic energy at $A$ must be non-negative $(K_A \ge 0)$.
$K_B - 4e \ge 0 \implies K_B \ge 4 \ eV$.
Thus,the electron must have at least $4 \ eV$ of kinetic energy at $B$ to reach $A$. This also implies it must have some kinetic energy at $B$.

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

(A) Let the linear charge density of the rod be $\lambda = \frac{Q}{L}$.
Consider a small element of length $dx$ on the rod at a distance $x$ from point $O$.
The charge on this element is $dq = \lambda dx = \frac{Q}{L} dx$.
The electric potential $dV$ at point $O$ due to this element is given by $dV = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{x} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{L} \frac{dx}{x}$.
The total potential $V$ at point $O$ is obtained by integrating from $x = L$ (at end $A$) to $x = 2L$ (at end $B$):
$V = \int_{L}^{2L} \frac{Q}{4 \pi \varepsilon_0 L} \frac{dx}{x} = \frac{Q}{4 \pi \varepsilon_0 L} [\ln x]_{L}^{2L} = \frac{Q}{4 \pi \varepsilon_0 L} (\ln 2L - \ln L) = \frac{Q \ln 2}{4 \pi \varepsilon_0 L}$.

(B) The potential on the surface of a uniformly charged solid sphere is $V_0 = \frac{Kq}{R}$.
For $r < R$, the potential is $V_i = \frac{Kq}{2R^3}(3R^2 - r^2)$.
At the center $(r = 0)$, $V_{center} = \frac{3Kq}{2R} = \frac{3}{2}V_0$. Thus, $R_1 = 0$ for potential $\frac{3V_0}{2}$.
For potential $\frac{5V_0}{4}$ $(r < R)$: $\frac{5}{4} \frac{Kq}{R} = \frac{Kq}{2R^3}(3R^2 - R_2^2) \implies \frac{5}{2} = 3 - \frac{R_2^2}{R^2} \implies \frac{R_2^2}{R^2} = \frac{1}{2} \implies R_2 = \frac{R}{\sqrt{2}} \approx 0.707R$.
For potential $\frac{3V_0}{4}$ $(r > R)$: $\frac{3}{4} \frac{Kq}{R} = \frac{Kq}{R_3} \implies R_3 = \frac{4}{3}R \approx 1.333R$.
For potential $\frac{V_0}{4}$ $(r > R)$: $\frac{1}{4} \frac{Kq}{R} = \frac{Kq}{R_4} \implies R_4 = 4R$.
Now, $R_2 = 0.707R$ and $(R_4 - R_3) = 4R - 1.333R = 2.667R$.
Since $0.707R < 2.667R$, we have $R_2 < (R_4 - R_3)$.

(A) The potential at any point on the surface of a shell is the sum of potentials due to all three shells.
For a point at distance $r$ from the center,the potential due to a shell of radius $R$ and charge $Q$ is $V = \frac{KQ}{R}$ if $r \le R$ and $V = \frac{KQ}{r}$ if $r > R$,where $K = \frac{1}{4\pi\epsilon_0}$.
The charges on the shells are $q_A = \sigma(4\pi a^2)$,$q_B = -\sigma(4\pi b^2)$,and $q_C = \sigma(4\pi c^2)$.
For shell $B$ (radius $b$),the potential $V_B$ is the sum of potentials due to shell $A$ (at distance $b > a$),shell $B$ (at distance $b = b$),and shell $C$ (at distance $b < c$):
$V_B = \frac{K q_A}{b} + \frac{K q_B}{b} + \frac{K q_C}{c}$
$V_B = \frac{1}{4\pi\epsilon_0} \left[ \frac{\sigma(4\pi a^2)}{b} + \frac{-\sigma(4\pi b^2)}{b} + \frac{\sigma(4\pi c^2)}{c} \right]$
$V_B = \frac{\sigma}{\epsilon_0} \left[ \frac{a^2}{b} - \frac{b^2}{b} + \frac{c^2}{c} \right]$
$V_B = \frac{\sigma}{\epsilon_0} \left[ \frac{a^2 - b^2}{b} + c \right]$

(C) Let the distance between points $A$ and $B$ be $d$. Let the charges at $A$ and $B$ be $+q$.
At any point $P$ on the line segment $AB$ at a distance $x$ from $A$, the distance from $B$ is $(d - x)$.
The total electric potential $V$ at point $P$ is the sum of potentials due to both charges:
$V = \frac{kq}{x} + \frac{kq}{d - x}$
To find the variation, we differentiate $V$ with respect to $x$:
$\frac{dV}{dx} = -\frac{kq}{x^2} + \frac{kq}{(d - x)^2}$
Setting $\frac{dV}{dx} = 0$ for the minimum potential:
$\frac{1}{x^2} = \frac{1}{(d - x)^2} \implies x = d - x \implies x = d/2$.
At $x = d/2$, the potential is minimum.
Thus, as we move from $A$ to $B$, the potential first decreases until the midpoint and then increases.

(C) The electric field at a distance $r$ from the center,where $r > b$,is the same as that of a point charge $Q$ at the center,because the dielectric shell is spherically symmetric and the total charge enclosed by a Gaussian surface of radius $r > b$ is $Q$.
Using Gauss's Law for dielectrics,the electric field $E$ for $r > b$ is given by $E = \frac{1}{4\pi \varepsilon _0} \frac{Q}{r^2}$.
The potential $V$ at a distance $b$ is calculated by integrating the electric field from infinity to $b$:
$V = -\int_{\infty}^{b} E \cdot dr = \int_{b}^{\infty} \frac{1}{4\pi \varepsilon _0} \frac{Q}{r^2} dr = \frac{Q}{4\pi \varepsilon _0} \left[ -\frac{1}{r} \right]_{b}^{\infty} = \frac{Q}{4\pi \varepsilon _0 b}$.
Thus,the dielectric does not affect the potential at points outside the dielectric shell.

(A) To find the total electrostatic energy,we consider building the disk by adding concentric rings of charge.
Consider a disk of radius $r$ with uniform surface charge density $\sigma$. The charge on a thin ring of radius $r$ and width $dr$ is given by:
$dq = (2\pi r dr) \sigma$
The potential at the circumference of a disk of radius $r$ is given as $V = 4\sigma r$. The work done to bring an additional charge $dq$ from infinity to the circumference of the disk is the electrostatic potential energy $dU$:
$dU = V dq = (4\sigma r) \cdot (2\pi r dr \sigma) = 8\pi \sigma^2 r^2 dr$
To find the total energy $U$ for a disk of radius $R$,we integrate from $r = 0$ to $r = R$:
$U = \int_0^R 8\pi \sigma^2 r^2 dr = 8\pi \sigma^2 \left[ \frac{r^3}{3} \right]_0^R = \frac{8}{3}\pi \sigma^2 R^3$

(A) The electric field due to an infinite wire at a distance $r$ is $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.
Given $\lambda = 4 \pi \varepsilon_0$,we have $E = \frac{4 \pi \varepsilon_0}{2 \pi \varepsilon_0 r} = \frac{2}{r}$.
The potential difference between points $A$ $(r_A = 2 \, m)$ and $B$ $(r_B = 1 \, m)$ is:
$V_B - V_A = - \int_{r_A}^{r_B} E \, dr = - \int_{2}^{1} \frac{2}{r} \, dr = -2 [\ln r]_2^1 = -2 (\ln 1 - \ln 2) = 2 \ln 2$.
Since the particle moves with constant speed,the net work done is zero $(W_{ext} + W_{elec} = 0)$.
The work done by the electric force is $W_{elec} = q(V_A - V_B) = q(-(V_B - V_A)) = 2 \times (-2 \ln 2) = -4 \ln 2$.
The work done by the external agent is $W_{ext} = -W_{elec} = 4 \ln 2$.

(B) The electric potential $V$ due to a charged arc of radius $R$ and charge $Q$ at its center is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$.
Since the potential is a scalar quantity,the net potential at the center is the algebraic sum of the potentials due to the three arcs.
$V_{net} = V_1 + V_2 + V_3$
$V_{net} = \frac{1}{4\pi \epsilon_0 R} (Q_1 + Q_2 + Q_3)$
Given charges are $Q_1 = +Q$,$Q_2 = +3Q$,and $Q_3 = -2Q$.
$V_{net} = \frac{1}{4\pi \epsilon_0 R} (Q + 3Q - 2Q)$
$V_{net} = \frac{1}{4\pi \epsilon_0 R} (2Q)$
$V_{net} = \frac{2Q}{4\pi \epsilon_0 R} = \frac{Q}{2\pi \epsilon_0 R}$.

(B) For a point outside a spherically symmetric charge distribution,the entire charge can be considered to be concentrated at its center.
Given:
Total charge $q = 5 \ nC = 5 \times 10^{-9} \ C$.
Distance $r = 15 \ cm = 0.15 \ m = 15 \times 10^{-2} \ m$.
The electric potential $V$ at a distance $r$ from a point charge $q$ is given by:
$V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} = \frac{k q}{r}$
Substituting the values:
$V = \frac{9 \times 10^9 \times 5 \times 10^{-9}}{15 \times 10^{-2}}$
$V = \frac{45}{0.15} = \frac{4500}{15} = 300 \ V$
Therefore,the electric potential at point $P$ is $300 \ V$.

(B) The electric potential $V$ at any point inside or on the surface of a charged object is given by the integral of the potential contribution from all surface charge elements $dq$,i.e.,$V = \int \frac{k dq}{r}$.
For a hollow hemisphere with uniform surface charge density $\sigma$,the potential at the center $O$ is higher than at any other point on the base because the center is,on average,closer to the charged surface elements than any other point on the base.
As we move from the center $O$ towards the rim $A$,the distance to the various parts of the charged surface increases on average,leading to a decrease in the electric potential.
Therefore,the potential at the center $O$ is the maximum potential on the base of the hemisphere.
Thus,$V_O > V_A$ or $V_A < V_O$.

(A) When two conducting spheres are connected by a long wire,they reach the same electric potential.
Let $V_1$ be the potential of the smaller sphere of radius $R$ and $V_2$ be the potential of the larger sphere of radius $3R$.
The potential of the smaller sphere is $V_1 = \frac{kQ}{R}$.
The potential of the larger sphere is due to its own charge $Q$ and the charge $q'$ at its centre: $V_2 = \frac{kQ}{3R} + \frac{kq'}{3R}$.
Since they are connected,$V_1 = V_2$.
$\frac{kQ}{R} = \frac{kQ}{3R} + \frac{kq'}{3R}$.
Multiplying by $\frac{3R}{k}$,we get $3Q = Q + q'$.
Therefore,$q' = 2Q$.

(D) The electric potential $V$ due to a system of charges is the algebraic sum of the potentials due to each individual charge: $V = \sum \frac{kq_i}{r_i}$.
For three equal charges placed at the vertices of an equilateral triangle, the system possesses $C_{3v}$ symmetry.
Near each charge, the equipotential surfaces are nearly spherical (circular in the plane of the triangle) due to the dominance of the local charge's potential.
As we move further away from the charges, the equipotential surfaces begin to merge and eventually take on a shape that reflects the overall symmetry of the charge distribution.
For three equal charges, the far-field equipotential surfaces will approach a circular shape, but in the near-to-mid field, they must exhibit the three-fold rotational symmetry of the equilateral triangle.
Graph $D$ shows equipotential lines that are circular near each charge and form a larger, roughly triangular shape enclosing all three charges, which correctly represents the symmetry of the system.

(A) Let the radius of the solid sphere be $r_a$ and the radius of the hollow shell be $r_b$.
The potential at the surface of the solid sphere is $V_{sphere} = \frac{KQ}{r_a} + \frac{K(0)}{r_b} = \frac{KQ}{r_a}$.
The potential at the surface of the hollow shell is $V_{shell} = \frac{KQ}{r_b} + \frac{K(0)}{r_b} = \frac{KQ}{r_b}$.
The initial potential difference is $V = V_{sphere} - V_{shell} = K Q \left( \frac{1}{r_a} - \frac{1}{r_b} \right)$.
When the shell is given a charge $-3Q$,the new potential at the surface of the solid sphere is $V'_{sphere} = \frac{KQ}{r_a} + \frac{K(-3Q)}{r_b}$.
The new potential at the surface of the hollow shell is $V'_{shell} = \frac{KQ}{r_b} + \frac{K(-3Q)}{r_b} = -\frac{2KQ}{r_b}$.
The new potential difference is $V' = V'_{sphere} - V'_{shell} = \left( \frac{KQ}{r_a} - \frac{3KQ}{r_b} \right) - \left( -\frac{2KQ}{r_b} \right) = \frac{KQ}{r_a} - \frac{KQ}{r_b} = V$.
Thus,the potential difference remains unchanged.

(B) The potential at a point due to a charged shell is given by $V = \frac{kQ}{r}$ for $r \ge R$ and $V = \frac{kQ}{R}$ for $r < R$,where $R$ is the radius of the shell.
Point $A$ is located at a distance $r = R$ from the center.
$1$. Potential due to the inner shell (radius $R/2$,charge $Q$): Since $r > R/2$,$V_1 = \frac{kQ}{R}$.
$2$. Potential due to the middle shell (radius $2R$,charge $2Q$): Since $r < 2R$,$V_2 = \frac{k(2Q)}{2R} = \frac{kQ}{R}$.
$3$. Potential due to the outer shell (radius $4R$,charge $8Q$): Since $r < 4R$,$V_3 = \frac{k(8Q)}{4R} = \frac{2kQ}{R}$.
The total potential at point $A$ is $V_A = V_1 + V_2 + V_3 = \frac{kQ}{R} + \frac{kQ}{R} + \frac{2kQ}{R} = \frac{4kQ}{R}$.

(B) The work done by an external agent to move a charge $q$ from point $A$ to point $B$ is given by $W_{ext} = q(V_B - V_A)$.
For option $B$: Moving a positive charge $(q > 0)$ from lower potential $(V_A)$ to higher potential $(V_B)$ means $V_B > V_A$,so $V_B - V_A > 0$. Thus,$W_{ext} = q(V_B - V_A) > 0$. This is positive.
For option $A$: Moving a negative charge $(q < 0)$ from higher potential $(V_A)$ to lower potential $(V_B)$ means $V_B < V_A$,so $V_B - V_A < 0$. Thus,$W_{ext} = q(V_B - V_A) > 0$. This is positive,not negative.
For option $C$: Work done by the electric field $W_{field} = -\Delta U = -(U_{final} - U_{initial})$. If moving from higher potential energy to lower potential energy,$U_{final} < U_{initial}$,so $\Delta U < 0$. Thus,$W_{field} = -(\text{negative}) = \text{positive}$.
Therefore,option $B$ is the correct statement.

(A) Consider a thin ring element on the hemispherical shell at an angle $\theta$ with the base,having width $R d\theta$ and radius $r = R \cos \theta$. The charge on this ring is $dq = \sigma (2\pi r) (R d\theta) = 2\pi \sigma R^2 \cos \theta d\theta$.
The distance of every point on this ring from point $A$ (at distance $x = R/2$ from center) can be found using the law of cosines. However,a simpler approach is to use the potential due to a charged spherical shell. The potential at any point inside a uniformly charged spherical shell is constant and equal to $V = \frac{\sigma R}{\varepsilon_0}$.
For a hemispherical shell,the potential at the center $O$ is $V_O = \frac{\sigma R}{2\varepsilon_0}$.
For a point $A$ at a distance $x = R/2$ from the center on the base,the potential is given by $V_A = \frac{\sigma R}{2\varepsilon_0}$. This is because the potential at any point on the base of a hemispherical shell is constant and equal to the potential at the center.

(B) The potential at any point on a conducting shell is the same as the potential at its center due to all charges present in the system.
Potential $V = V_{\text{shell}} + V_A + V_B + V_C$
$V = \frac{KQ_0}{R} + \frac{KQ_0}{R/2} + \frac{K(-2Q_0)}{2R/3} + \frac{K(3Q_0)}{2R}$
$V = \frac{KQ_0}{R} + \frac{2KQ_0}{R} - \frac{3KQ_0}{R} + \frac{1.5KQ_0}{R}$
$V = \frac{1.5KQ_0}{R} = \frac{3KQ_0}{2R}$

(C) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.
For a square with side $a$,there are $4$ pairs of charges at distance $a$ (sides) and $2$ pairs of charges at distance $a\sqrt{2}$ (diagonals).
Let the charges be $q_1 = Q$,$q_2 = 2Q$,$q_3 = 3Q$,and $q_4 = 4Q$ in counter-clockwise order starting from the bottom-left.
The pairs at distance $a$ are:
$(q_1, q_2) = (Q, 2Q) \implies U_1 = \frac{k(Q)(2Q)}{a} = 2 \frac{kQ^2}{a}$
$(q_2, q_3) = (2Q, 3Q) \implies U_2 = \frac{k(2Q)(3Q)}{a} = 6 \frac{kQ^2}{a}$
$(q_3, q_4) = (3Q, 4Q) \implies U_3 = \frac{k(3Q)(4Q)}{a} = 12 \frac{kQ^2}{a}$
$(q_4, q_1) = (4Q, Q) \implies U_4 = \frac{k(4Q)(Q)}{a} = 4 \frac{kQ^2}{a}$
The pairs at distance $a\sqrt{2}$ are:
$(q_1, q_3) = (Q, 3Q) \implies U_5 = \frac{k(Q)(3Q)}{a\sqrt{2}} = 3 \frac{kQ^2}{a\sqrt{2}}$
$(q_2, q_4) = (2Q, 4Q) \implies U_6 = \frac{k(2Q)(4Q)}{a\sqrt{2}} = 8 \frac{kQ^2}{a\sqrt{2}}$
Total potential energy $U = U_1 + U_2 + U_3 + U_4 + U_5 + U_6 = (2 + 6 + 12 + 4) \frac{kQ^2}{a} + (3 + 8) \frac{kQ^2}{a\sqrt{2}} = 24 \frac{kQ^2}{a} + 11 \frac{kQ^2}{a\sqrt{2}} = \frac{kQ^2}{a} \left( 24 + \frac{11}{\sqrt{2}} \right)$.

(D) The potential at a point $P$ at a distance $x$ from the center,where $r_2 < x < r_1$,is the sum of the potentials due to both spheres.
$1$. The potential due to the outer sphere (radius $r_1$,charge $q_1$) at any point inside it is constant and equal to the potential at its surface: $V_1 = \frac{q_1}{4\pi \varepsilon_0 r_1}$.
$2$. The potential due to the inner sphere (radius $r_2$,charge $q_2$) at a point outside it at distance $x$ is: $V_2 = \frac{q_2}{4\pi \varepsilon_0 x}$.
$3$. The net potential at distance $x$ is $V = V_1 + V_2 = \frac{q_1}{4\pi \varepsilon_0 r_1} + \frac{q_2}{4\pi \varepsilon_0 x}$.

(C) The electric potential $V$ at a point due to a system of charges is given by $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.
Here, the charges are $q$ at distances $r = 1, 2, 4, 8, \dots \, \text{m}$.
$V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{1} + \frac{q}{2} + \frac{q}{4} + \frac{q}{8} + \dots \right]$
$V = \frac{q}{4\pi\varepsilon_0} \left[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right]$
The term in the bracket is a geometric progression with first term $a = 1$ and common ratio $r = 1/2$.
The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
Therefore, $V = \frac{q}{4\pi\varepsilon_0} \times 2 = \frac{2q}{4\pi\varepsilon_0} = \frac{q}{2\pi\varepsilon_0}$.

(A) Let the radius of the inner sphere be $a$ and the radius of the outer shell be $b$. Let the charge on the inner sphere be $q_1$ and the charge on the outer shell be $q_2$.
The potential at the surface of the inner sphere is given by: $V_a = \frac{kq_1}{a} + \frac{kq_2}{b} = 10\, V$.
The potential at the surface of the outer shell is given by: $V_b = \frac{kq_1}{b} + \frac{kq_2}{b} = 50\, V$.
The potential at the common centre $O$ is given by: $V_O = \frac{kq_1}{a} + \frac{kq_2}{b}$.
Comparing this with the expression for $V_a$,we see that the potential at the centre is equal to the potential at the surface of the inner sphere.
Therefore,$V_O = 10\, V$.

(B) The work done by the electrostatic force is equal to the negative change in the potential energy of the system.
$W = -\Delta U = U_i - U_f$
Initially,the central charge $q$ is at a distance $r$ from each of the six charges $q$. The potential energy of the central charge due to the six surrounding charges is $U_i = 6 \times (kq^2 / r) = 6q^2 / 4\pi\varepsilon_0 r$.
When the central charge is moved to infinity,its potential energy becomes $U_f = 0$.
Therefore,the work done by the electrostatic force is $W = U_i - U_f = 6q^2 / 4\pi\varepsilon_0 r - 0 = 6q^2 / 4\pi\varepsilon_0 r$.

(B) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by the formula $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
For point $A$ at $(\sqrt{2}, \sqrt{2})$,the distance from the origin $(0, 0)$ is $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, m$.
For point $B$ at $(2, 0)$,the distance from the origin $(0, 0)$ is $r_B = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \, m$.
Since the distances $r_A$ and $r_B$ are equal,the electric potentials at points $A$ and $B$ are equal: $V_A = V_B = \frac{kq}{2}$.
The potential difference between points $A$ and $B$ is $\Delta V = V_A - V_B = 0 \, V$.

(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.
For an infinite number of charges,the total potential is the sum of individual potentials:
$V = \sum \frac{kq}{r} = kq \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right)$.
Here,$k = 9 \times 10^9 \, N\cdot m^2/C^2$ and $q = 0.2 \times 10^{-6} \, C$.
The term in the bracket is a geometric progression with first term $a = 1$ and common ratio $r = 1/2$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.
Substituting the values:
$V = (9 \times 10^9) \times (0.2 \times 10^{-6}) \times 2 = 1.8 \times 10^3 \times 2 = 3600 \, V$.
Since $1000 \, V = 1 \, kV$,the potential is $3.60 \, kV$.

(B) hollow metal sphere acts as a spherical conductor shell.
For any charged spherical conductor,the electric field inside the conductor is $E = 0$.
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 work done in moving a charge $q$ from one point to another is given by $W = q(V_{final} - V_{initial})$.
Since there is no charge nearby, the electric potential $V$ is constant throughout the region (or zero if we consider infinity as reference).
Points $A, B,$ and $C$ lie on a curve, but without any external electric field or source charge, the potential at all points in the space is the same.
Therefore, the potential difference between point $P$ and any point $A, B,$ or $C$ is zero.
Thus, $W_A = q(V_A - V_P) = 0$, $W_B = q(V_B - V_P) = 0$, and $W_C = q(V_C - V_P) = 0$.
Hence, $W_A = W_B = W_C = 0$.

(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.
For two identical positive charges $q$ placed at $x = -a$ and $x = a$,the total potential at any point $x$ is $V(x) = \frac{kq}{|x+a|} + \frac{kq}{|x-a|}$.
$1$. As $x \to a$,$V \to +\infty$. As $x \to -a$,$V \to +\infty$. This means the potential must show vertical asymptotes at $x = a$ and $x = -a$.
$2$. At the origin $(x = 0)$,the potential is $V(0) = \frac{kq}{a} + \frac{kq}{a} = \frac{2kq}{a}$,which is a positive finite value.
$3$. As $x \to \pm\infty$,$V \to 0$.
Comparing these properties with the given graphs,the potential is positive everywhere and has a local minimum at the origin between the two charges. Graph $C$ correctly represents this behavior,showing a $U$-shaped curve between the charges and approaching zero at infinity.

(A) Let $q_1$ be the charge on the inner shell of radius $a$ and $q_2$ be the charge on the outer shell of radius $2a$.
The potential at the surface of the inner shell is given by: $V_{inner} = \frac{kq_1}{a} + \frac{kq_2}{2a} = 10\,V$.
The potential at the surface of the outer shell is given by: $V_{outer} = \frac{kq_1}{2a} + \frac{kq_2}{2a} = 5\,V$.
The potential at the centre of the shells is given by: $V_{centre} = \frac{kq_1}{a} + \frac{kq_2}{2a}$.
Comparing this expression with the equation for the potential at the inner shell,we can see that $V_{centre} = V_{inner} = 10\,V$.

(D) The charge on an infinitesimal element of the arc which subtends an angle $d\theta$ at the center of the ring is given by:
$dQ = \lambda R d\theta = \lambda_0 \cos(\theta/2) R d\theta$
The potential $dV$ at the centre of the ring due to this charge $dQ$ is:
$dV = \frac{1}{4 \pi \epsilon_0} \frac{dQ}{R} = \frac{\lambda_0 \cos(\theta/2) R d\theta}{4 \pi \epsilon_0 R} = \frac{\lambda_0}{4 \pi \epsilon_0} \cos(\theta/2) d\theta$
To find the total potential $V$,we integrate over the entire ring from $\theta = 0$ to $\theta = 2\pi$:
$V = \int_0^{2\pi} dV = \frac{\lambda_0}{4 \pi \epsilon_0} \int_0^{2\pi} \cos(\theta/2) d\theta$
$V = \frac{\lambda_0}{4 \pi \epsilon_0} \left[ 2 \sin(\theta/2) \right]_0^{2\pi}$
$V = \frac{\lambda_0}{4 \pi \epsilon_0} \left[ 2 \sin(\pi) - 2 \sin(0) \right] = \frac{\lambda_0}{4 \pi \epsilon_0} [0 - 0] = 0 \text{ V}$

(D) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively. Given $q_1 + q_2 = Q$.
Since surface charge densities are equal,$\sigma = \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}$.
Using the property of ratios,$\frac{q_1}{r^2} = \frac{q_2}{R^2} = \frac{q_1 + q_2}{r^2 + R^2} = \frac{Q}{r^2 + R^2}$.
Thus,$q_1 = \frac{Q r^2}{r^2 + R^2}$ and $q_2 = \frac{Q R^2}{r^2 + R^2}$.
The potential at the common centre is $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.
Substituting the values of $q_1$ and $q_2$,we get $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q r^2}{r(r^2 + R^2)} + \frac{Q R^2}{R(r^2 + R^2)} \right)$.
$V = \frac{Q}{4\pi\varepsilon_0(r^2 + R^2)} (r + R)$.

(B) For a point at a distance $r$ such that $R_1 < r < R_2$,the potential $V$ is the sum of the potentials due to both spheres.
$1$. The potential due to the inner sphere (radius $R_1$,charge $Q_1$) at a distance $r$ (where $r > R_1$) is $V_1 = \frac{1}{4\pi \varepsilon_0} \frac{Q_1}{r}$.
$2$. The potential due to the outer sphere (radius $R_2$,charge $Q_2$) at a distance $r$ (where $r < R_2$) is constant throughout its interior,given by $V_2 = \frac{1}{4\pi \varepsilon_0} \frac{Q_2}{R_2}$.
$3$. The total potential $V$ at distance $r$ is $V = V_1 + V_2 = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_1}{r} + \frac{Q_2}{R_2} \right)$.

(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 is $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \ m$.
For point $B$ at $(2, 0)$,the distance from the origin is $r_B = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \ m$.
Since $r_A = r_B$,the potential at both points is equal: $V_A = V_B = \frac{kq}{2}$.
The potential difference between $A$ and $B$ is $V_A - V_B = 0 \ volt$.

(B) In a uniform electric field $\vec{E}$,the potential $V$ at a position $\vec{r}$ relative to point $P$ is given by $V = V_P - \vec{E} \cdot \vec{r} = V_P - Er \cos \theta$,where $\theta$ is the angle between the radius vector and the field direction.
The potential varies between $V_{min} = V_P - Er$ (at $\theta = 0^o$) and $V_{max} = V_P + Er$ (at $\theta = 180^o$).
Given $V_{min} = 589.0\,V$ and $V_{max} = 589.8\,V$,the potential difference across the diameter is $2Er = 589.8 - 589.0 = 0.8\,V$. Thus,$Er = 0.4\,V$.
The center potential $V_P$ is the average: $V_P = (589.0 + 589.8) / 2 = 589.4\,V$.
For a point at an angle $\theta = 60^o$,the potential is $V = V_P - Er \cos(60^o) = 589.4 - 0.4 \times 0.5 = 589.4 - 0.2 = 589.2\,V$.

(C) Let $L$ be the length of each string. The distance between the two charges is $x = 2L \sin \theta$.
In equilibrium,the forces acting on each charge are tension $T$,weight $mg$,and electrostatic force $F_e = \frac{kq^2}{x^2}$.
Resolving forces: $T \sin \theta = F_e$ and $T \cos \theta = mg$.
Dividing these gives $\tan \theta = \frac{F_e}{mg} = \frac{kq^2}{x^2 mg}$.
Thus,$x^2 = \frac{kq^2}{mg \tan \theta}$,which implies $x = q \sqrt{\frac{k}{mg \tan \theta}}$.
The electrostatic potential $V$ at the centre of the line joining the charges (at distance $x/2$ from each charge) is:
$V = \frac{kq}{x/2} + \frac{kq}{x/2} = \frac{4kq}{x}$.
Substituting $x$:
$V = \frac{4kq}{q \sqrt{\frac{k}{mg \tan \theta}}} = 4 \sqrt{k^2 \cdot \frac{mg \tan \theta}{k}} = 4 \sqrt{k \, mg \tan \theta}$.

(C) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively.
Given $q_1 + q_2 = Q$.
Since surface charge densities are equal,$\sigma_1 = \sigma_2$.
$\frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2} \implies \frac{q_1}{r^2} = \frac{q_2}{R^2}$.
Using componendo and dividendo,$\frac{q_1}{r^2} = \frac{q_2}{R^2} = \frac{q_1 + q_2}{r^2 + R^2} = \frac{Q}{r^2 + R^2}$.
Thus,$q_1 = \frac{Q r^2}{R^2 + r^2}$ and $q_2 = \frac{Q R^2}{R^2 + r^2}$.
The electric potential at the common centre is $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.
Substituting the values of $q_1$ and $q_2$:
$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r + Q R}{R^2 + r^2} \right)$.
$V = \frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{R^2 + r^2}$.

(D) Let the charges on the shells be $Q_1, Q_2, Q_3$ and surface charge density be $\sigma$. Since $\sigma$ is equal for all,$\sigma = \frac{Q_1}{4\pi a^2} = \frac{Q_2}{4\pi b^2} = \frac{Q_3}{4\pi c^2}$.
Thus,$Q_1 = 4\pi a^2 \sigma$,$Q_2 = 4\pi b^2 \sigma$,$Q_3 = 4\pi c^2 \sigma$.
The total charge $Q = Q_1 + Q_2 + Q_3 = 4\pi \sigma (a^2 + b^2 + c^2)$,so $\sigma = \frac{Q}{4\pi (a^2 + b^2 + c^2)}$.
The potential at $r < a$ is the sum of potentials due to each shell: $V = \frac{1}{4\pi \epsilon_0} (\frac{Q_1}{a} + \frac{Q_2}{b} + \frac{Q_3}{c})$.
Substituting the values of $Q_1, Q_2, Q_3$: $V = \frac{1}{4\pi \epsilon_0} (\frac{4\pi a^2 \sigma}{a} + \frac{4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c}) = \frac{\sigma}{\epsilon_0} (a + b + c)$.
Substituting $\sigma$: $V = \frac{Q}{4\pi \epsilon_0 (a^2 + b^2 + c^2)} (a + b + c)$.