Electric potential Questions in English

(B) The radius of the hollow metallic sphere is $R = 15 \,cm$. The potential on the surface is $V_{surface} = 20 \,V$.
For a hollow metallic sphere, the electric field inside the sphere is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -dV/dr)$, if $E = 0$, then the potential $V$ must be constant throughout the interior of the sphere.
Therefore, the potential at the centre of the sphere is equal to the potential on its surface.
Thus, the potential at the centre is $20 \,V$.

(C) The given situation is shown in the figure. The distance $AC = CB = L$. The point $D$ is at a distance $L$ from $B$,so $AD = 3L$ and $BD = L$.
Work done $W$ in moving a charge $Q$ along the semi-circle $CRD$ is equal to the change in potential energy of the system.
$W = U_{\text{final}} - U_{\text{initial}}$
$U_{\text{initial}}$ is the potential energy when $Q$ is at $C$:
$U_{\text{initial}} = \frac{kqQ}{AC} + \frac{k(-q)Q}{BC} + \frac{kq(-q)}{AB} = \frac{kqQ}{L} - \frac{kqQ}{L} - \frac{kq^2}{2L} = -\frac{kq^2}{2L}$
$U_{\text{final}}$ is the potential energy when $Q$ is at $D$:
$U_{\text{final}} = \frac{kqQ}{AD} + \frac{k(-q)Q}{BD} + \frac{kq(-q)}{AB} = \frac{kqQ}{3L} - \frac{kqQ}{L} - \frac{kq^2}{2L} = -\frac{2kqQ}{3L} - \frac{kq^2}{2L}$
$W = (-\frac{2kqQ}{3L} - \frac{kq^2}{2L}) - (-\frac{kq^2}{2L}) = -\frac{2kqQ}{3L}$
Substituting $k = \frac{1}{4\pi\varepsilon_0}$:
$W = -\frac{2}{3} \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{L} = -\frac{qQ}{6\pi\varepsilon_0 L}$

(B) Given,charge of each small spherical drop $= Q$ and potential at its surface $= V_0$. Let $r$ be the radius of each small drop. The potential is given by $V_0 = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.
When two such drops combine to form a single large drop of radius $R$,the volume is conserved: $\frac{4}{3} \pi R^3 = 2 \times \frac{4}{3} \pi r^3$,which gives $R^3 = 2r^3$ or $R = 2^{1/3} r$.
The total charge on the new large drop is $Q' = Q + Q = 2Q$.
The potential at the surface of the new drop is $V' = \frac{1}{4 \pi \varepsilon_0} \frac{Q'}{R} = \frac{1}{4 \pi \varepsilon_0} \frac{2Q}{2^{1/3} r}$.
Substituting $V_0 = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$,we get $V' = V_0 \times \frac{2}{2^{1/3}} = V_0 \times 2^{1 - 1/3} = V_0 \times 2^{2/3} = V_0 \times (2^2)^{1/3} = 4^{1/3} V_0$.

(A) The potential $V$ at a distance $r$ from the center of a conducting spherical shell of radius $R$ with charge $Q$ is given by:
$V = \frac{kQ}{R}$ for $r \le R$ and $V = \frac{kQ}{r}$ for $r > R$,where $k = \frac{1}{4\pi\epsilon_0}$.
For the point at distance $r = \frac{R}{2}$ (inside both shells):
The potential $V_2$ is the sum of potentials due to both shells:
$V_2 = \frac{kq}{R} + \frac{k(2q)}{2R} = \frac{kq}{R} + \frac{kq}{R} = \frac{2kq}{R}$.
For the point at distance $r = 3R$ (outside both shells):
The potential $V_1$ is the sum of potentials due to both shells treated as point charges at the center:
$V_1 = \frac{kq}{3R} + \frac{k(2q)}{3R} = \frac{3kq}{3R} = \frac{kq}{R}$.
Therefore,the ratio $\frac{V_2}{V_1} = \frac{2kq/R}{kq/R} = 2$.

(A) Let the surface charge density be $\sigma$. The total charge $Q$ is distributed on the two shells such that $\sigma_1 = \sigma_2 = \sigma$.
Total charge $Q = \sigma(4 \pi r^2) + \sigma(4 \pi R^2) = 4 \pi \sigma(r^2 + R^2)$.
Thus,$\sigma = \frac{Q}{4 \pi (r^2 + R^2)}$.
The electric potential at the centre of a spherical shell of radius $a$ with charge $q$ is $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{a}$.
For the two shells,the potential at the centre is the sum of the potentials due to each shell:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{\sigma 4 \pi r^2}{r} + \frac{\sigma 4 \pi R^2}{R} \right)$.
$V = \frac{\sigma}{\varepsilon_0} (r + R)$.
Substituting the value of $\sigma$:
$V = \frac{Q}{4 \pi \varepsilon_0 (r^2 + R^2)} (r + R) = \frac{Q(R+r)}{4 \pi \varepsilon_0 (R^2+r^2)}$.

(C) The potential at point $P(0,0,z)$ due to the charge $+q$ located at $(0,0,a)$ is:
$V_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}$
The potential at point $P(0,0,z)$ due to the charge $-q$ located at $(0,0,-a)$ is:
$V_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z-(-a))} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z+a)}$
The total electric potential $V$ at point $P$ is the algebraic sum of the potentials due to individual charges:
$V = V_1 + V_2$
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{z-a} - \frac{q}{z+a} \right]$
$V = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{(z+a) - (z-a)}{(z-a)(z+a)} \right]$
$V = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{z+a-z+a}{z^2-a^2} \right]$
$V = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a}{z^2-a^2} \right]$
$V = \frac{2qa}{4 \pi \varepsilon_0(z^2-a^2)}$

(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$. Point $P$ is at distance $r$ from the charge $-q$ at $x=a$. Since $P$ is on the $x$-axis at distance $r$ from $x=a$,its coordinate is $x = a + r$.
The potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
Factoring out $q/2$:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$
Combining the terms:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$
Simplifying the numerator:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right] = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$
Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$

(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
Let the point $P$ be at $x = a + r$. The distances of the charges from $P$ are:
For charge $\frac{q}{2}$ at $x=0$: distance $d_1 = (a+r) - 0 = r+a$
For charge $-q$ at $x=a$: distance $d_2 = (a+r) - a = r$
For charge $\frac{q}{2}$ at $x=2a$: distance $d_3 = |(a+r) - 2a| = |r-a| = r-a$ (since $r >> a$)
The total potential $V_P$ is:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$
Since $r >> a$,we have $r^2 - a^2 \approx r^2$.
Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.

(B) Let $q_1 = 9 \mu C$ and $q_2 = -3 \mu C$. The distance between them is $d = 0.16 \ m$.
Let the point $P$ be at a distance $x$ from $q_1$. Then the distance of $P$ from $q_2$ is $(0.16 - x)$.
The electric potential $V$ at point $P$ due to both charges is the algebraic sum of the potentials:
$V = V_1 + V_2 = 0$
$\frac{1}{4 \pi \epsilon_0} \frac{q_1}{x} + \frac{1}{4 \pi \epsilon_0} \frac{q_2}{0.16 - x} = 0$
$\frac{9 \times 10^{-6}}{x} = - \frac{-3 \times 10^{-6}}{0.16 - x}$
$\frac{9}{x} = \frac{3}{0.16 - x}$
$3(0.16 - x) = x$
$0.48 - 3x = x$
$4x = 0.48$
$x = 0.12 \ m$
Thus,the distance of $P$ from the $9 \mu C$ charge is $0.12 \ m$.

(A) The work done in moving a charge $Q$ in an electrostatic field is path-independent and depends only on the initial and final positions.
Work done $W = Q(V_{final} - V_{initial})$.
Here, the initial point is $C$ and the final point is $D$.
Distance $AC = L$, $CB = L$, $BD = L$.
Potential at $C$ $(V_C)$: $V_C = \frac{1}{4 \pi \epsilon_0} [\frac{q}{AC} + \frac{-q}{CB}] = \frac{1}{4 \pi \epsilon_0} [\frac{q}{L} - \frac{q}{L}] = 0$.
Potential at $D$ $(V_D)$: $V_D = \frac{1}{4 \pi \epsilon_0} [\frac{q}{AD} + \frac{-q}{BD}] = \frac{1}{4 \pi \epsilon_0} [\frac{q}{3L} - \frac{q}{L}] = \frac{1}{4 \pi \epsilon_0} [\frac{q - 3q}{3L}] = \frac{-2q}{12 \pi \epsilon_0 L} = \frac{-q}{6 \pi \epsilon_0 L}$.
Since the electrostatic force is conservative, the work done along any path from $C$ to $D$ is the same:
$W_1 = W_2 = Q(V_D - V_C) = Q(\frac{-q}{6 \pi \epsilon_0 L} - 0) = \frac{-Qq}{6 \pi \epsilon_0 L}$.
Thus, $W_1 = W_2 = \frac{-Qq}{6 \pi \epsilon_0 L}$.

(A) Let $q$ be the charge appearing on the outer shell of radius $2R$ after earthing.
Since the outer shell is grounded,its potential must be zero.
The potential at the outer shell is the sum of potentials due to the inner charge $Q$ and the outer charge $q$:
$V_{\text{outer}} = \frac{KQ}{2R} + \frac{Kq}{2R} = 0$
Solving for $q$,we get $q = -Q$.
Now,we calculate the potential at a distance $r = \frac{3R}{2}$ from the center.
Since $R < r < 2R$,the point lies outside the inner sphere and inside the outer sphere.
The potential at this point is the sum of the potential due to the inner sphere (as a point charge) and the potential due to the outer shell (constant inside):
$V(r) = \frac{KQ}{r} + \frac{Kq}{2R}$
Substituting $r = \frac{3R}{2}$ and $q = -Q$:
$V = \frac{KQ}{3R/2} + \frac{K(-Q)}{2R} = \frac{2KQ}{3R} - \frac{KQ}{2R} = \frac{4KQ - 3KQ}{6R} = \frac{KQ}{6R} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{6R}$

(D) The potential at the surface of shell $B$ is the sum of potentials due to all three shells $A, B$ and $C$.
$V_B = V_{A,B} + V_{B,B} + V_{C,B}$
Since the potential inside a shell is constant and equal to the potential at its surface,we have:
$V_{A,B} = \frac{k Q_A}{b} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma(4\pi a^2)}{b} = \frac{\sigma a^2}{\varepsilon_0 b}$
$V_{B,B} = \frac{k Q_B}{b} = \frac{1}{4\pi\varepsilon_0} \frac{-\sigma(4\pi b^2)}{b} = -\frac{\sigma b}{\varepsilon_0}$
$V_{C,B} = \frac{k Q_C}{c} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma(4\pi c^2)}{c} = \frac{\sigma c}{\varepsilon_0}$
Adding these,we get:
$V_B = \frac{\sigma a^2}{\varepsilon_0 b} - \frac{\sigma b}{\varepsilon_0} + \frac{\sigma c}{\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{b} - b + c \right)$

(C) $1$. The inner sphere has charge $q$. Due to induction, a charge $-q$ is induced on the inner surface of the shell (radius $a$).
$2$. Since the shell is grounded, its potential is zero. Let the charge on the outer surface (radius $b$) be $q'$. The total charge on the shell is $Q_{shell} = -q + q' = 0$, which implies $q' = q$. However, because the shell is grounded, the charge $q'$ flows to the ground, leaving the outer surface with zero charge. Thus, the inner surface of the shell has charge $-q$ and the outer surface has charge $0$.
$3$. The potential at the centre of the sphere is the sum of potentials due to the sphere (radius $R$, charge $q$), the inner surface of the shell (radius $a$, charge $-q$), and the outer surface of the shell (radius $b$, charge $0$).
$4$. $V_{centre} = V_{sphere} + V_{inner_shell} + V_{outer_shell} = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{R} + \frac{1}{4 \pi \epsilon_{0}} \frac{-q}{a} + \frac{1}{4 \pi \epsilon_{0}} \frac{0}{b} = \frac{q}{4 \pi \epsilon_{0}} \left( \frac{1}{R} - \frac{1}{a} \right)$.

(B) The potential $V$ at a distance $x$ from the center of a ring of radius $r$ carrying charge $q$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2} + r^{2}}}$.
For point $A$ at distance $x_A = \frac{4}{3}r$,the distance from the ring circumference is $d_A = \sqrt{(\frac{4}{3}r)^2 + r^2} = \sqrt{\frac{16}{9}r^2 + r^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r$.
Thus,$V_A = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r/3} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{5r}$.
For point $B$ at distance $x_B = \frac{3}{4}r$ on the opposite side,the distance from the ring circumference is $d_B = \sqrt{(\frac{3}{4}r)^2 + r^2} = \sqrt{\frac{9}{16}r^2 + r^2} = \sqrt{\frac{25}{16}r^2} = \frac{5}{4}r$.
Thus,$V_B = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r/4} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4q}{5r}$.
The work done $W$ in moving a charge $-q$ from $A$ to $B$ is $W = (-q)(V_B - V_A)$.
$W = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4q}{5r} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{5r} \right) = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r} \right) = -\frac{1}{5} \cdot \frac{q^2}{4 \pi \varepsilon_{0} r}$.

(A) The total electric potential $V$ at a point $P$ between the charges is given by $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{r} + \frac{4Q}{d-r} \right)$,where $r$ is the distance of $P$ from $A$.
For the potential to be minimum,we set $\frac{dV}{dr} = 0$.
$\frac{dV}{dr} = \frac{1}{4\pi\varepsilon_0} \left( -\frac{Q}{r^2} + \frac{4Q}{(d-r)^2} \right) = 0$.
This implies $\frac{Q}{r^2} = \frac{4Q}{(d-r)^2}$,which is equivalent to the condition where the net electric field $E = 0$.
Taking the square root on both sides: $\frac{1}{r} = \frac{2}{d-r}$.
Solving for $r$: $d - r = 2r \Rightarrow 3r = d \Rightarrow r = \frac{d}{3}$.
Thus,the potential is minimum at a distance of $\frac{d}{3}$ units to the right of $A$.

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{q}{4 \pi \varepsilon_{0} r}$.
Since the electric field is conservative,the work done in moving a charge $Q$ from point $A$ to point $B$ is independent of the path and is given by $W = Q(V_B - V_A)$.
The potential at point $A(a, 0)$ is $V_A = \frac{q}{4 \pi \varepsilon_{0} a}$.
The potential at point $B(0, b)$ is $V_B = \frac{q}{4 \pi \varepsilon_{0} b}$.
Therefore,the work done is $W = Q \left( \frac{q}{4 \pi \varepsilon_{0} b} - \frac{q}{4 \pi \varepsilon_{0} a} \right)$.
$W = \frac{q Q}{4 \pi \varepsilon_{0}} \left( \frac{1}{b} - \frac{1}{a} \right) = \frac{q Q}{4 \pi \varepsilon_{0}} \left( \frac{a - b}{a b} \right)$.

(D) The electrostatic force is a conservative force.
By definition,the work done by a conservative force in moving a charge along any closed path is always zero.
Since the charge $Q$ is moved once around a circle,which is a closed path,the total work done is $0$.

(C) In physics,the absolute electric potential at a single point is not a uniquely measurable quantity because it depends on the choice of the reference point (where potential is assumed to be zero).
However,the potential difference (voltage) between two points is a well-defined and measurable quantity.
Resistance and displacement current are both physical quantities that can be measured directly using appropriate instruments.
Therefore,'Voltage' (when referring to absolute potential at a point) is not considered a measurable quantity in an absolute sense without a defined reference.

(A) The work done $W$ in bringing a charge $q_{3}$ to a point in the presence of charges $q_{1}$ and $q_{2}$ is given by the potential energy of the configuration:
$W = V \times q_{3} = (\frac{kq_{1}}{\ell} + \frac{kq_{2}}{\ell}) q_{3}$
Given: $q_{1} = 1 \times 10^{-9} \text{ C}$,$q_{2} = 2 \times 10^{-9} \text{ C}$,$q_{3} = 3 \times 10^{-9} \text{ C}$,$\ell = 3 \times 10^{-2} \text{ m}$,$k = 9 \times 10^{9} \text{ N.m}^{2}/\text{C}^{2}$.
Substituting the values:
$W = \frac{9 \times 10^{9}}{3 \times 10^{-2}} (1 \times 10^{-9} + 2 \times 10^{-9}) \times 3 \times 10^{-9}$
$W = (3 \times 10^{11}) \times (3 \times 10^{-9}) \times (3 \times 10^{-9})$
$W = 27 \times 10^{-7} \text{ J} = 2.7 \times 10^{-6} \text{ J} = 2.7 \text{ } \mu\text{J}$.

(C) The potential at any point on a conducting spherical shell is the sum of the potentials due to all the charges present on the shells.
For a point at distance $r$ from the center,the potential due to a shell of radius $R$ and charge $q$ is $\frac{kq}{R}$ if $r \le R$ and $\frac{kq}{r}$ if $r > R$.
For sphere $A$ (radius $a$): It is inside $B$ and $C$,so the potential is $V_A = \frac{kq_1}{a} + \frac{kq_2}{b} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c} \right)$.
For sphere $B$ (radius $b$): It is on the surface of $B$,inside $C$,and outside $A$. So,$V_B = \frac{kq_1}{b} + \frac{kq_2}{b} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1+q_2}{b} + \frac{q_3}{c} \right)$.
For sphere $C$ (radius $c$): It is on the surface of $C$,and outside $A$ and $B$. So,$V_C = \frac{kq_1}{c} + \frac{kq_2}{c} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1+q_2+q_3}{c} \right)$.

(D) Point '$A$' is at a distance $l$ from the top $+q$ charge and the bottom $+q$ charge.
The distance from point '$A$' to both $-q$ charges is $\sqrt{(2l)^2 + l^2} = \sqrt{4l^2 + l^2} = \sqrt{5l^2} = l\sqrt{5}$.
The electric potential $V$ at point '$A$' due to all four charges is given by the sum of potentials from each charge:
$V_A = k(\frac{q}{l} + \frac{q}{l} - \frac{q}{l\sqrt{5}} - \frac{q}{l\sqrt{5}})$
$V_A = k(\frac{2q}{l} - \frac{2q}{l\sqrt{5}})$
$V_A = \frac{2kq}{l}(1 - \frac{1}{\sqrt{5}})$
Therefore,option $(D)$ is correct.

(B) The electric potential $V$ inside a spherical conductor is constant and equal to the potential on its surface,given by $V = \frac{kq}{R}$.
The electric field $E$ on the surface of the spherical conductor is given by $E = \frac{kq}{R^2}$.
To find the ratio of the electric potential inside to the electric field on the surface,we calculate:
$\frac{V}{E} = \frac{kq/R}{kq/R^2} = \frac{kq}{R} \times \frac{R^2}{kq} = R$.
Therefore,the ratio is $R$. Option $(B)$ is correct.

(A) For a uniformly charged spherical shell,the electric potential at any point inside the shell is constant and equal to the potential at its surface.
The potential at the surface of the shell is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$.
Since the distance $r = R/2$ is less than the radius $R$ (i.e.,the point is inside the shell),the electric potential at this point is the same as the potential at the surface.
Therefore,the potential at $r = R/2$ is $V = \frac{Q}{4\pi \epsilon_0 R}$.