Mix Examples - Electric Potential and Capacitance Questions in English

(C) Let $C$ be the capacitance of each capacitor. The equivalent capacitance of three capacitors in parallel combination is $C_{p} = 3C$ and in series combination is $C_{s} = C/3$.
Let $V_{p}$ be the potential difference in the parallel combination and $V_{s}$ be the potential difference in the series combination. Given that the energy stored in both cases is the same:
$\frac{1}{2} C_{p} V_{p}^{2} = \frac{1}{2} C_{s} V_{s}^{2}$
$\frac{V_{p}^{2}}{V_{s}^{2}} = \frac{C_{s}}{C_{p}} = \frac{C/3}{3C} = \frac{1}{9}$
Taking the square root on both sides:
$\frac{V_{p}}{V_{s}} = \frac{1}{3}$

(D) Let the radius of each small drop be ' $r$ ' and the charge on each be ' $q$ '. The potential of a small drop is given by $V = \frac{kq}{r}$.
When ' $n$ ' drops coalesce to form a big drop of radius ' $R$ ',the volume remains conserved: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which implies $R = n^{1/3} r$.
The total charge on the big drop is $Q = n \cdot q$.
The potential of the big drop ' $V_{big}$ ' is given by $V_{big} = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{1 - 1/3} \cdot \frac{kq}{r} = n^{2/3} \cdot V$.

(D) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is given by $v = \frac{kq}{r}$.
When $n$ small drops combine to form a big drop of radius $R$,the volume remains conserved: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which gives $R = n^{1/3} r$.
The total charge on the big drop is $Q = nq$.
The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3} r} = n^{1 - 1/3} \cdot \frac{kq}{r} = n^{2/3} v$.

(B) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.
The potential of a small drop is given by $V = \frac{kq}{r} = 20 \ V$.
When $n = 27$ drops combine to form a big drop of radius $R$ and charge $Q$,the volume remains constant:
$\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3 \implies R^3 = n r^3 \implies R = n^{1/3} r$.
For $n = 27$,$R = (27)^{1/3} r = 3r$.
The total charge on the big drop is $Q = nq = 27q$.
The potential of the big drop is $V' = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left( \frac{kq}{r} \right) = n^{2/3} V$.
Substituting the values: $V' = (27)^{2/3} \times 20 = (3^3)^{2/3} \times 20 = 3^2 \times 20 = 9 \times 20 = 180 \ V$.

(A) By the law of conservation of energy for the system of two charges:
$(K.E. + P.E.)_{\text{initial}} = (K.E. + P.E.)_{\text{final}}$
Initially,the charges are at rest,so $K.E._{\text{initial}} = 0$.
Let $v$ be the speed of each particle at separation $\frac{r}{2}$. Since the particles have equal mass $m$,by momentum conservation,they move with equal and opposite velocities.
The total kinetic energy is $K.E._{\text{final}} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Using the potential energy formula $U = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r}$:
$0 + \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r} = mv^2 + \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r/2}$
$mv^2 = \frac{q^2}{4 \pi \varepsilon_0} \left( \frac{1}{r} - \frac{2}{r} \right) = -\frac{q^2}{4 \pi \varepsilon_0 r}$
Note: The negative sign indicates that the particles will not reach this separation if they are like charges. Assuming the question implies the magnitude of speed for opposite charges or a typo in the setup,the magnitude is $v = \sqrt{\frac{q^2}{4 \pi \varepsilon_0 mr}}$.

(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of each small drop is $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$.
When $n$ drops coalesce to form a large drop of radius $R$,the volume remains conserved: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which gives $R = n^{1/3} r$.
The total charge on the large drop is $Q = nq$.
The potential of the large drop is $V' = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$.
Substituting the values of $Q$ and $R$: $V' = \frac{1}{4 \pi \varepsilon_0} \frac{nq}{n^{1/3} r}$.
Simplifying this,we get $V' = n^{1 - 1/3} \left( \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} \right) = n^{2/3} V$.

(D) Let the charge on the sphere be $Q$.
The potential at the surface of the sphere is $V_{surface} = \frac{kQ}{r}$.
The potential at a distance $3r$ from the center is $V_{3r} = \frac{kQ}{3r}$.
The potential difference $V$ is given by:
$V = V_{surface} - V_{3r} = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{2kQ}{3r}$.
From this,we find $kQ = \frac{3Vr}{2}$.
The electric intensity $E$ at a distance $3r$ from the center is given by:
$E = \frac{kQ}{(3r)^2} = \frac{kQ}{9r^2}$.
Substituting the value of $kQ$:
$E = \frac{1}{9r^2} \cdot \frac{3Vr}{2} = \frac{3V}{18r} = \frac{V}{6r}$.

(C) The electrostatic pressure (stress) on the surface of a charged conductor is given by $P = \frac{\sigma^{2}}{2 \epsilon_{0}}$.
Here,the surface charge density $\sigma = \frac{q}{4 \pi r^{2}} = \frac{10^{-2}}{4 \pi (1)^{2}} = \frac{10^{-2}}{4 \pi} \ C/m^{2}$.
Substituting $\sigma$ into the pressure formula: $P = \frac{1}{2 \epsilon_{0}} \left( \frac{10^{-2}}{4 \pi} \right)^{2} = \frac{10^{-4}}{32 \pi^{2} \epsilon_{0}}$.
Volume strain is defined as $\text{strain} = \frac{\text{stress}}{B}$,where $B$ is the bulk modulus.
Given $B = \frac{10^{11}}{4 \pi^{2}}$.
Therefore,$\text{strain} = \frac{10^{-4}}{32 \pi^{2} \epsilon_{0}} \times \frac{4 \pi^{2}}{10^{11}} = \frac{10^{-4}}{8 \epsilon_{0} \times 10^{11}} = \frac{10^{-15}}{8 \epsilon_{0}}$.

(A) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is given by $V = \frac{kq}{r} = 10 \ V$.
When $n = 27$ drops combine to form a large drop of radius $R$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $R = n^{1/3}r$.
For $n = 27$,$R = (27)^{1/3}r = 3r$.
The total charge on the large drop is $Q = nq = 27q$.
The potential of the large drop is $V^{\prime} = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left(\frac{kq}{r}\right) = n^{2/3}V$.
Substituting the values: $V^{\prime} = (27)^{2/3} \times 10 = (3^3)^{2/3} \times 10 = 3^2 \times 10 = 9 \times 10 = 90 \ V$.

(A) The given system consists of four plates. Let the capacitance of each pair of adjacent plates be $C = \frac{A \varepsilon_0}{d}$.
Based on the circuit diagram,plates $1$ and $3$ are connected together,and plates $2$ and $4$ are connected to terminals $A$ and $B$ respectively.
This forms two capacitors in parallel (between plates $1-2$ and $3-2$) which are then in series with a third capacitor (between plates $3-4$).
Let $C_1$ be the capacitance between plates $1$ and $2$,$C_2$ between $3$ and $2$,and $C_3$ between $3$ and $4$. Thus,$C_1 = C_2 = C_3 = C$.
The parallel combination of $C_1$ and $C_2$ gives $C' = C_1 + C_2 = C + C = 2C$.
This $C'$ is in series with $C_3$. The equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C' \cdot C_3}{C' + C_3} = \frac{(2C) \cdot C}{2C + C} = \frac{2C^2}{3C} = \frac{2}{3}C$.
Substituting $C = \frac{A \varepsilon_0}{d}$,we get $C_{eq} = \frac{2}{3} \frac{A \varepsilon_0}{d}$.

(C) When a charged capacitor is disconnected from the battery,the charge $Q$ on its plates remains constant because there is no path for the charge to flow.
As the distance $d$ between the plates is increased,the capacitance $C = \frac{\epsilon_0 A}{d}$ decreases.
Since $Q = CV$,and $Q$ is constant,the potential difference $V = \frac{Q}{C}$ will increase as $C$ decreases.
Therefore,the charge on the plates remains the same.

(C) In the given arrangement,plate $Q$ is common to two capacitors connected in parallel,as both plates $P$ and $R$ are grounded (at $0 \,V$).
Let $V$ be the potential of plate $Q$.
The capacitance of the first capacitor (between $P$ and $Q$) is $C_1 = \frac{\varepsilon_0 A}{d}$.
The capacitance of the second capacitor (between $Q$ and $R$) is $C_2 = \frac{\varepsilon_0 A}{2d}$.
Since they are in parallel,the effective capacitance is $C_{\text{eff}} = C_1 + C_2 = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{2d} = \frac{3 \varepsilon_0 A}{2d}$.
Given $q = 8.85 \times 10^{-8} \,C$,$A = 2.0 \,m^2$,$d = 2 \times 10^{-3} \,m$,and $\varepsilon_0 = 8.85 \times 10^{-12} \,F/m$.
Using $q = C_{\text{eff}} V$,we have $V = \frac{q}{C_{\text{eff}}} = \frac{q \cdot 2d}{3 \varepsilon_0 A}$.
Substituting the values: $V = \frac{8.85 \times 10^{-8} \times 2 \times (2 \times 10^{-3})}{3 \times (8.85 \times 10^{-12}) \times 2.0} = \frac{8.85 \times 4 \times 10^{-11}}{3 \times 8.85 \times 2 \times 10^{-12}} = \frac{4 \times 10}{3 \times 2} = \frac{20}{3} \approx 6.67 \,V$.

(C) The capacitors $4 \mu F$ and $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 4 \mu F + 2 \mu F = 6 \mu F$.
This parallel combination is in series with the $6 \mu F$ capacitor. The total equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{6 \mu F} + \frac{1}{6 \mu F} = \frac{2}{6 \mu F} = \frac{1}{3 \mu F} \Rightarrow C_{eq} = 3 \mu F$.
The total charge $Q$ supplied by the $12 \text{ V}$ source is:
$Q = C_{eq} \times V = 3 \mu F \times 12 \text{ V} = 36 \mu C$.
This total charge $Q$ flows through the $6 \mu F$ capacitor and then divides between the $4 \mu F$ and $2 \mu F$ capacitors connected in parallel.
Let $Q_1$ be the charge on the $4 \mu F$ capacitor and $Q_2$ be the charge on the $2 \mu F$ capacitor.
Since they are in parallel,the potential difference across them is the same:
$V_p = \frac{Q_1}{C_1} = \frac{Q_2}{C_2} \Rightarrow \frac{Q_1}{4 \mu F} = \frac{Q_2}{2 \mu F} \Rightarrow Q_1 = 2 Q_2$.
Also,$Q_1 + Q_2 = Q = 36 \mu C$.
Substituting $Q_1 = 2 Q_2$ into the equation:
$2 Q_2 + Q_2 = 36 \mu C \Rightarrow 3 Q_2 = 36 \mu C \Rightarrow Q_2 = 12 \mu C$.
Then,$Q_1 = 2 \times 12 \mu C = 24 \mu C = 24 \times 10^{-6} C$.

(D) To withstand a potential difference of $1.5 kV$ $(1500 V)$ using capacitors that can withstand $500 V$ each,the number of capacitors $(m)$ in each series row is:
$m = \frac{1500 V}{500 V} = 3$
The equivalent capacitance of one such row of $3$ capacitors (each $2 \mu F$) is:
$C_{row} = \frac{2 \mu F}{3}$
If we connect $n$ such rows in parallel to achieve a total capacitance of $6 \mu F$,we have:
$C_{eff} = n \times C_{row} = n \times \frac{2}{3} \mu F = 6 \mu F$
$n = \frac{6 \times 3}{2} = 9$
The total number of capacitors required is $N = m \times n = 3 \times 9 = 27$.

(C) When two capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \mu F^{-1}$.
Thus,$C_{eq} = 2 \mu F$.
The total charge $Q$ stored in the series combination is $Q = C_{eq} \times V = 2 \mu F \times 900 \ V = 1800 \mu C$.
When the capacitors are disconnected and reconnected in parallel,the total charge $Q = 1800 \mu C$ remains the same.
The equivalent capacitance in parallel is $C_p = C_1 + C_2 = 3 \mu F + 6 \mu F = 9 \mu F$.
The new potential difference $V'$ across the parallel combination is $V' = \frac{Q}{C_p} = \frac{1800 \mu C}{9 \mu F} = 200 \ V$.

(B) From the given circuit diagram, the two $2 \, \mu F$ capacitors in the top branch are in series. Their equivalent capacitance is $C_{1} = \frac{2 \times 2}{2+2} = 1 \, \mu F$.
Similarly, the two $2 \, \mu F$ capacitors in the right and bottom branches are in series. Their equivalent capacitance is $C_{2} = \frac{2 \times 2}{2+2} = 1 \, \mu F$.
Now, the circuit simplifies to a parallel combination of $C \, \mu F$ and $C_{2} = 1 \, \mu F$, which is in series with $C_{1} = 1 \, \mu F$.
The effective capacitance $C_{\text{eff}}$ is given by:
$C_{\text{eff}} = \frac{(C + 1) \times 1}{(C + 1) + 1} = \frac{C + 1}{C + 2} \, \mu F$.
Given, charge $q = 0.75 \, mC = 0.75 \times 10^{-3} \, C$ and potential $V = 10^{3} \, V$.
The effective capacitance is $C_{\text{eff}} = \frac{q}{V} = \frac{0.75 \times 10^{-3}}{10^{3}} = 0.75 \times 10^{-6} \, F = 0.75 \, \mu F$.
Equating the two expressions for $C_{\text{eff}}$:
$0.75 = \frac{C + 1}{C + 2}$
$\frac{3}{4} = \frac{C + 1}{C + 2}$
$3(C + 2) = 4(C + 1)$
$3C + 6 = 4C + 4$
$C = 2 \, \mu F$.

(D) To achieve a potential difference of $600 \ V$ using capacitors rated at $200 \ V$ each,we must connect $n$ capacitors in series such that $n \times 200 \ V = 600 \ V$. Thus,$n = 3$.
The equivalent capacitance of these $3$ capacitors in series is given by $\frac{1}{C_s} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$,so $C_s = 2 \ \mu F$.
To achieve a total capacitance of $18 \ \mu F$,we need to connect $m$ such series branches in parallel,where $m \times C_s = 18 \ \mu F$.
Thus,$m \times 2 \ \mu F = 18 \ \mu F$,which gives $m = 9$.
The total number of capacitors required is $n \times m = 3 \times 9 = 27$.

(B) Given,$C_1 = C_2 = 8 \mu F = 8 \times 10^{-6} \ F$ and $V = 10 \ V$.
Since $C_1$ and $C_2$ are connected in parallel,the initial equivalent capacitance is $C_{eq} = C_1 + C_2 = 16 \mu F = 1.6 \times 10^{-5} \ F$.
The initial total charge is $q_i = C_{eq} \cdot V = 1.6 \times 10^{-5} \times 10 = 1.6 \times 10^{-4} \ C = 160 \mu C$.
We know that capacitance $C = \frac{\varepsilon_0 A}{d}$,so $C \propto \frac{1}{d}$.
If the plate separation $d$ is reduced to $40 \%$ of its initial value,the new separation is $d' = 0.4d$. Thus,the new capacitance $C'$ becomes $C' = \frac{C}{0.4} = \frac{C}{2/5} = 2.5C = \frac{5}{2}C$.
Wait,the problem states the separation is reduced $TO$ $40 \%$,meaning $d' = 0.4d$. So $C' = \frac{C}{0.4} = 2.5 \times 8 \mu F = 20 \mu F$.
New total capacitance $C_{eq}' = C' + C_2 = 20 \mu F + 8 \mu F = 28 \mu F$.
New charge $q_f = C_{eq}' \cdot V = 28 \mu F \times 10 \ V = 280 \mu C$.
Increase in charge $\Delta q = q_f - q_i = 280 \mu C - 160 \mu C = 120 \mu C$.

(A) The circuit consists of a $10 \text{ V}$ battery connected in parallel with a $3 \mu F$ capacitor and a branch containing a $4 \mu F$ capacitor in series with a parallel combination of $1 \mu F$ and $5 \mu F$ capacitors.
First,calculate the equivalent capacitance of the parallel combination of $1 \mu F$ and $5 \mu F$ capacitors:
$C_p = 1 \mu F + 5 \mu F = 6 \mu F$
Now,the upper branch consists of a $4 \mu F$ capacitor in series with this $6 \mu F$ equivalent capacitor. The equivalent capacitance of this upper branch $(C_{eq})$ is:
$\frac{1}{C_{eq}} = \frac{1}{4 \mu F} + \frac{1}{6 \mu F} = \frac{3+2}{12 \mu F} = \frac{5}{12 \mu F}$
$C_{eq} = \frac{12}{5} \mu F = 2.4 \mu F$
The potential difference across this entire upper branch is equal to the battery voltage,$V = 10 \text{ V}$.
The charge $Q$ on the equivalent capacitor of the upper branch is:
$Q = C_{eq} \times V = 2.4 \mu F \times 10 \text{ V} = 24 \mu C$
Since the $4 \mu F$ capacitor and the $6 \mu F$ equivalent capacitor are in series,the charge on each is the same and equal to the total charge of the branch.
Therefore,the charge on the $4 \mu F$ capacitor is $24 \mu C$. The correct option is $(a)$.

(A) The equivalent capacitance $C_{eq}$ of the system is calculated as follows: The capacitors $C_2$ and $C_3$ are in parallel, and this combination is in series with $C_1$.
$C_{23} = C_2 + C_3 = 4 \, \mu F + 2 \, \mu F = 6 \, \mu F$.
Now, $C_1$ and $C_{23}$ are in series:
$C_{eq} = \frac{C_1 \times C_{23}}{C_1 + C_{23}} = \frac{3 \, \mu F \times 6 \, \mu F}{3 \, \mu F + 6 \, \mu F} = \frac{18}{9} \, \mu F = 2 \, \mu F$.
The total charge $q$ drawn from the source is:
$q = C_{eq} \times V = 2 \, \mu F \times 1800 \, V = 3600 \, \mu C = 3.6 \times 10^{-3} \, C$.
This charge $q$ flows through $C_1$. The potential drop across $C_1$ is:
$V_1 = \frac{q}{C_1} = \frac{3600 \, \mu C}{3 \, \mu F} = 1200 \, V$.
The potential drop across the parallel combination of $C_2$ and $C_3$ is:
$V_{23} = V_{total} - V_1 = 1800 \, V - 1200 \, V = 600 \, V$.
Since $C_2$ and $C_3$ are in parallel, the potential drop across each is $600 \, V$.
The charge on $C_2$ is $q_2 = C_2 \times V_{23} = 4 \, \mu F \times 600 \, V = 2400 \, \mu C = 2.4 \times 10^{-3} \, C$.
The charge on $C_3$ is $q_3 = C_3 \times V_{23} = 2 \, \mu F \times 600 \, V = 1200 \, \mu C = 1.2 \times 10^{-3} \, C$.

(B) The energy stored in a fully charged capacitor with capacitance $C$ and initial potential difference $V$ is given by $U = \frac{1}{2}CV^2$.
Since the capacitor is discharged through a resistance wire in a thermally isolated block,the entire electrical energy stored in the capacitor is converted into heat energy.
The heat energy gained by the block is given by $Q = ms \Delta T$,where $m$ is the mass,$s$ is the specific heat,and $\Delta T$ is the rise in temperature.
Equating the stored energy to the heat energy: $\frac{1}{2}CV^2 = ms \Delta T$.
Solving for $V$: $V^2 = \frac{2ms \Delta T}{C}$.
Therefore,$V = \sqrt{\frac{2ms \Delta T}{C}}$.

(B) $1$. Initial state: The capacitors $C$ and $2C$ are in series with battery $E$. The equivalent capacitance is $C_{eq} = \frac{C \times 2C}{C + 2C} = \frac{2C}{3}$.
$2$. The charge on each capacitor is $q = C_{eq}E = \frac{2CE}{3}$.
$3$. Initial energy stored: $U_i = \frac{1}{2} C_{eq} E^2 = \frac{1}{2} (\frac{2C}{3}) E^2 = \frac{CE^2}{3}$.
$4$. Final state: The capacitors are disconnected and reconnected with opposite polarity. The net charge on the plates connected together is $q - q = 0$. Since the total charge is zero,the final charge on each capacitor is $0$.
$5$. Final energy stored: $U_f = 0$.
$6$. Energy lost: $\Delta U = U_i - U_f = \frac{CE^2}{3} - 0 = \frac{CE^2}{3}$.
$7$. Thus,the charge on $2C$ is $0$ and the energy lost is $\frac{CE^2}{3}$.

(D) When two identical capacitors of capacitance $C$ are charged to potentials $V_1$ and $V_2$ respectively,their initial charges are $Q_1 = CV_1$ and $Q_2 = CV_2$.
When connected in parallel,the total charge $Q_{net} = Q_1 + Q_2$ is conserved.
The capacitors reach a common potential $V_{common} = \frac{Q_1 + Q_2}{C + C} = \frac{V_1 + V_2}{2}$.
Since $V_{common} = \frac{V_1 + V_2}{2}$,the net potential difference is not equal to the sum of the initial potential differences $(V_1 + V_2)$.
During the redistribution of charge,some energy is dissipated as heat due to the flow of charge through the connecting wires.
Therefore,the final total energy stored $U_f = \frac{1}{2}(2C)V_{common}^2$ is always less than the initial total energy $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$.

(C) The circuit consists of two parallel branches connected to a voltage source $V$.
For the upper branch,we have two capacitors in series: $C_1 = 5 \mu F$ (breakdown voltage $V_1 = 1 \text{ kV}$) and $C_2 = 4 \mu F$ (breakdown voltage $V_2 = 2 \text{ kV}$).
In a series combination,the charge $q$ on each capacitor is the same. The maximum charge $q_{max}$ the branch can handle is determined by the capacitor that reaches its breakdown voltage first.
For $C_1$,$q_1 = C_1 V_1 = 5 \mu F \times 1 \text{ kV} = 5 \mu C$.
For $C_2$,$q_2 = C_2 V_2 = 4 \mu F \times 2 \text{ kV} = 8 \mu C$.
Thus,the upper branch can handle a maximum charge of $5 \mu C$. The total voltage across the upper branch is $V_{upper} = q_{max} / C_{eq1}$,where $C_{eq1} = (5 \times 4) / (5 + 4) = 20/9 \mu F$.
$V_{upper} = 5 \mu C / (20/9 \mu F) = 45/20 \text{ kV} = 2.25 \text{ kV}$.
For the lower branch,we have $C_3 = 2 \mu F$ (breakdown voltage $V_3 = 2 \text{ kV}$) and $C_4 = 3 \mu F$ (breakdown voltage $V_4 = 1 \text{ kV}$).
For $C_3$,$q_3 = C_3 V_3 = 2 \mu F \times 2 \text{ kV} = 4 \mu C$.
For $C_4$,$q_4 = C_4 V_4 = 3 \mu F \times 1 \text{ kV} = 3 \mu C$.
Thus,the lower branch can handle a maximum charge of $3 \mu C$. The total voltage across the lower branch is $V_{lower} = q_{max} / C_{eq2}$,where $C_{eq2} = (2 \times 3) / (2 + 3) = 6/5 \mu F$.
$V_{lower} = 3 \mu C / (6/5 \mu F) = 15/6 \text{ kV} = 2.5 \text{ kV}$.
Since the branches are in parallel,the source voltage $V$ must not exceed the smaller of the two branch breakdown voltages to ensure no capacitor breaks down.
Therefore,$V_{max} = \min(2.25 \text{ kV}, 2.5 \text{ kV}) = 2.25 \text{ kV}$.

(A) By simplifying the circuit as shown in the diagram,the capacitors $10 \ \mu F$,$5 \ \mu F$,and $9 \ \mu F$ are connected in parallel between points $C$ and $D$. Therefore,the equivalent capacitance between $C$ and $D$ is:
$C_{CD} = 10 \ \mu F + 5 \ \mu F + 9 \ \mu F = 24 \ \mu F$
Now,the capacitors $12 \ \mu F$,$24 \ \mu F$ (which is $C_{CD}$),and $8 \ \mu F$ are in a series connection.
The equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8} = \frac{2 + 1 + 3}{24} = \frac{6}{24} = \frac{1}{4} \ \mu F^{-1}$
$\Rightarrow C_{eq} = 4 \ \mu F$
The total charge $Q$ flowing in the circuit is:
$Q = C_{eq} \times V_{AB} = 4 \ \mu F \times 80 \ V = 320 \ \mu C$
Since the capacitors $12 \ \mu F$,$24 \ \mu F$,and $8 \ \mu F$ are in series,the same charge $Q = 320 \ \mu C$ flows through each branch.
The potential difference across $CD$ is:
$V_{CD} = \frac{Q}{C_{CD}} = \frac{320 \ \mu C}{24 \ \mu F} = \frac{40}{3} \ V$
The charge $q$ on the $10 \ \mu F$ capacitor is:
$q = 10 \ \mu F \times V_{CD} = 10 \ \mu F \times \frac{40}{3} \ V = \frac{400}{3} \ \mu C \approx 133.33 \ \mu C$
Thus,the equivalent capacitance is $4 \ \mu F$ and the charge on the $10 \ \mu F$ capacitor is approximately $133 \ \mu C$.

(D) The circuit consists of two parallel branches connected across a $30 \ V$ source.
Branch $1$ contains $C_1$ and $C_2$ in series. The equivalent capacitance is $C_{eq1} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 \times 1.5}{1 + 1.5} = \frac{1.5}{2.5} = 0.6 \ \mu F$.
The charge on this branch is $q = C_{eq1} \times V = 0.6 \ \mu F \times 30 \ V = 18 \ \mu C$.
The potential at point $a$ relative to $A$ is $V_A - V_a = \frac{q}{C_1} = \frac{18 \ \mu C}{1 \ \mu F} = 18 \ V$.
Branch $2$ contains $C_3$ and $C_4$ in series. The equivalent capacitance is $C_{eq2} = \frac{C_3 C_4}{C_3 + C_4} = \frac{2.5 \times 0.5}{2.5 + 0.5} = \frac{1.25}{3} = \frac{5}{12} \ \mu F$.
The charge on this branch is $q' = C_{eq2} \times V = \frac{5}{12} \ \mu F \times 30 \ V = 12.5 \ \mu C$.
The potential at point $b$ relative to $A$ is $V_A - V_b = \frac{q'}{C_3} = \frac{12.5 \ \mu C}{2.5 \ \mu F} = 5 \ V$.
The potential difference between $a$ and $b$ is $V_a - V_b = (V_A - V_b) - (V_A - V_a) = 5 \ V - 18 \ V = -13 \ V$.
The magnitude of the potential difference is $|V_a - V_b| = 13 \ V$.

(A) Let the potential at point $O$ be $V$. Since the capacitors were initially uncharged,the sum of charges on the plates connected to point $O$ must be zero (by the principle of conservation of charge).
The charge on capacitor $C_1$ is $q_1 = C_1(V_1 - V)$.
The charge on capacitor $C_2$ is $q_2 = C_2(V_2 - V)$.
The charge on capacitor $C_3$ is $q_3 = C_3(V_3 - V)$.
According to the conservation of charge at node $O$:
$q_1 + q_2 + q_3 = 0$
$C_1(V_1 - V) + C_2(V_2 - V) + C_3(V_3 - V) = 0$
$C_1 V_1 - C_1 V + C_2 V_2 - C_2 V + C_3 V_3 - C_3 V = 0$
$C_1 V_1 + C_2 V_2 + C_3 V_3 = V(C_1 + C_2 + C_3)$
$V = \frac{C_1 V_1 + C_2 V_2 + C_3 V_3}{C_1 + C_2 + C_3}$

(B) From the circuit diagram,the $4 \mu F$ capacitor is connected directly across the $6 \text{ V}$ battery. Therefore,the charge on the $4 \mu F$ capacitor is $q_4 = C_4 V = 4 \mu F \times 6 \text{ V} = 24 \mu C$.
Given the ratio of charges $\frac{q_x}{q_4} = \frac{3}{8}$,we have $q_x = \frac{3}{8} \times 24 \mu C = 9 \mu C$.
The capacitors $x \mu F$ and $2 \mu F$ are in parallel,so they share the same potential difference $V_p = \frac{q_x}{x} = \frac{9}{x} \text{ V}$.
The $3 \mu F$ capacitor is in series with the parallel combination of $x \mu F$ and $2 \mu F$. The total potential difference across the circuit is $6 \text{ V}$.
Thus,the potential difference across the $3 \mu F$ capacitor is $V_3 = 6 - V_p = 6 - \frac{9}{x} \text{ V}$.
Since the $3 \mu F$ capacitor and the equivalent $(x+2) \mu F$ capacitor are in series,the charge on them must be equal:
$q_3 = q_{(x+2)}$
$3 \mu F \times (6 - \frac{9}{x}) = (x+2) \mu F \times \frac{9}{x}$
$3(6 - \frac{9}{x}) = (x+2) \frac{9}{x}$
$18 - \frac{27}{x} = 9 + \frac{18}{x}$
$9 = \frac{45}{x}$
$x = \frac{45}{9} = 5$.
Therefore,the value of $x$ is $5$.

(A) Initially,when the switch $S$ is open,the two capacitors are charged with $Q_1 = 4 \text{ C}$ and $Q_2 = 2 \text{ C}$ respectively. The total initial energy $E_i$ stored in the capacitors is given by:
$E_i = \frac{Q_1^2}{2C_1} + \frac{Q_2^2}{2C_2} = \frac{4^2}{2 \times 1} + \frac{2^2}{2 \times 2} = \frac{16}{2} + \frac{4}{4} = 8 \text{ J} + 1 \text{ J} = 9 \text{ J}$.
When the switch $S$ is closed,the capacitors are connected in parallel. The common potential $V$ is given by:
$V = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{4 + 2}{1 + 2} = \frac{6}{3} = 2 \text{ V}$.
The final energy $E_f$ stored in the capacitors is:
$E_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (1 + 2) (2)^2 = \frac{1}{2} \times 3 \times 4 = 6 \text{ J}$.
By the law of conservation of energy,the energy lost by the capacitors is stored in the inductor as magnetic energy $E_L$:
$E_L = E_i - E_f = 9 \text{ J} - 6 \text{ J} = 3 \text{ J}$.

(A) According to the principle of conservation of charge,the total electric charge in an isolated system remains constant. When $n$ identical charged drops coalesce into a single larger drop,the total charge of the system is the sum of the charges of the individual drops. Since no charge is lost or gained from the surroundings,the total charge remains conserved. Other quantities like capacitance,potential,and electrostatic energy change as the radius of the drop increases.

(A) Statement $(A)$ is true: Inside a charged hollow metal sphere,the electric field $E$ is $0$ because there is no charge enclosed. However,the electric potential $V$ is constant and equal to the potential at the surface,so $V \neq 0$.
Statement $(B)$ is true: By definition,an equipotential surface is a surface where the potential is the same at all points. Since work done $W = q(V_f - V_i)$ and $V_f = V_i$,the work done is $0$.
Statement $(C)$ is true: When two like charges are brought closer,they repel each other. To move them closer,external work must be done against the electrostatic repulsive force. This work is stored as an increase in the mutual electrostatic potential energy of the system.
Therefore,all three statements are correct.

(A) Let the radii of the spheres be $R_1 = 4 \ cm = 0.04 \ m$ and $R_2 = 1 \ cm = 0.01 \ m$. The distance between centers is $d = 0.15 \ m$.
When connected by a wire,the spheres reach the same potential $V = \frac{k q_1}{R_1} = \frac{k q_2}{R_2}$. Thus,$q_1/q_2 = R_1/R_2 = 4/1$.
Given $q_1 + q_2 = 100 \ nC$,we get $q_1 = 80 \ nC$ and $q_2 = 20 \ nC$.
The potential is $V = \frac{9 \times 10^9 \times 80 \times 10^{-9}}{0.04} = 18000 \ V$.
Let the point where the electric field is zero be at distance $x$ from the center of the first sphere. The field is zero when $\frac{k q_1}{x^2} = \frac{k q_2}{(d-x)^2}$.
Taking the square root: $\frac{\sqrt{80}}{x} = \frac{\sqrt{20}}{d-x} \implies \frac{2\sqrt{20}}{x} = \frac{\sqrt{20}}{0.15-x}$.
$2(0.15 - x) = x \implies 0.3 = 3x \implies x = 0.1 \ m$.
The potential at this point is $V_P = \frac{k q_1}{x} + \frac{k q_2}{d-x} = \frac{9 \times 10^9 \times 80 \times 10^{-9}}{0.1} + \frac{9 \times 10^9 \times 20 \times 10^{-9}}{0.05} = 7200 + 3600 = 10800 \ V = 10.8 \times 10^3 \ V$.

(C) In a steady state,capacitors act as open circuits,so no current flows through the resistors. The potential difference across the entire capacitor network ($C$ to $D$) is equal to the battery voltage,$V_{CD} = 100 \text{ V}$.
Looking at the circuit,the branch $E-F-G$ is in parallel with the branch $C-D$. However,the potential at $F$ is determined by the capacitive voltage divider formed by the branches connected between $C$ and $D$.
The equivalent capacitance of the left branch (capacitors in series) is $C_1 = \frac{5 \times 5}{5 + 5} = 2.5 \text{ } \mu\text{F}$.
The equivalent capacitance of the right branch (capacitors in series) is $C_2 = \frac{2 \times 2}{2 + 2} = 1 \text{ } \mu\text{F}$.
Since these branches are in parallel across $100 \text{ V}$,the potential at $F$ relative to $C$ and $D$ is determined by the series combination. The potential difference $V_{CF}$ is $100 \times \frac{C_2}{C_1 + C_2} = 100 \times \frac{1}{2.5 + 1} = 100 \times \frac{1}{3.5} \approx 28.57 \text{ V}$.
Thus,$V_{AF} = V_{AC} + V_{CF} = 0 + 28.57 \text{ V} = 28.57 \text{ V}$.
And $V_{FB} = V_{FD} + V_{DB} = (100 - 28.57) + 0 = 71.43 \text{ V}$.
Rounding to one decimal place,we get $V_{AF} = 28.5 \text{ V}$ and $V_{FB} = 71.4 \text{ V}$.

(D) The capacitance of the capacitor is $C = 100 \mu F = 100 \times 10^{-6} \ F = 10^{-4} \ F$.
The potential difference required is $V = 100 \ V$.
The total charge $Q$ required to be stored on the capacitor is given by $Q = CV$.
Substituting the values,$Q = (10^{-4} \ F) \times (100 \ V) = 10^{-2} \ C$.
The rate of charging is given as $I = 100 \mu C s^{-1} = 100 \times 10^{-6} \ C s^{-1} = 10^{-4} \ C s^{-1}$.
Since the charge is supplied at a steady rate,$Q = I \times t$,where $t$ is the time taken.
Therefore,$t = \frac{Q}{I} = \frac{10^{-2} \ C}{10^{-4} \ C s^{-1}} = 10^{2} \ s = 100 \ s$.

(B) When capacitors are connected in series,the equivalent capacitance $C_{\text{eq}}$ is given by $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$.
For $C_1 = 1 \mu F$ and $C_2 = C \mu F$,we have $C_{\text{eq}} = \frac{1 \times C}{1 + C} = \frac{C}{C+1} \mu F$.
Given the total charge $q = 80 \mu C$ and potential difference $V = 120 \ V$,we use the relation $q = C_{\text{eq}} V$.
Substituting the values: $80 = \left( \frac{C}{C+1} \right) \times 120$.
Dividing both sides by $40$: $2 = \left( \frac{C}{C+1} \right) \times 3$.
$2(C+1) = 3C \implies 2C + 2 = 3C \implies C = 2 \mu F$.
In a series combination,the charge on each capacitor is the same as the total charge $q = 80 \mu C$.
The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
For the capacitor $C = 2 \mu F$: $U = \frac{(80 \mu C)^2}{2 \times 2 \mu F} = \frac{6400}{4} \mu J = 1600 \mu J$.

(A) The capacitance of a parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the separation.
Initially,the charge stored is $Q = C_0 V_0$.
$(i)$ When the battery is disconnected,the charge $Q$ remains constant. If the separation is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{C_0}{2}$. The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0 / 2)} = C_0 V_0^2$.
(ii) When the battery remains connected,the potential $V_0$ remains constant. If the separation is doubled,the new capacitance is $C' = \frac{C_0}{2}$. The energy stored is $E_2 = \frac{1}{2} C' V_0^2 = \frac{1}{2} (\frac{C_0}{2}) V_0^2 = \frac{1}{4} C_0 V_0^2$.
Therefore,the ratio is $\frac{E_1}{E_2} = \frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2} = 4$.

(B) Initial capacitance $C = 100 \mu F = 100 \times 10^{-6} \text{ F}$ and potential $V = 50 \text{ V}$.
Initial energy stored $E_1 = \frac{1}{2} C V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (50)^2 = 50 \times 10^{-6} \times 2500 = 0.125 \text{ J} = 125 \times 10^{-3} \text{ J}$.
When the distance $d$ is doubled,the new capacitance $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2} = 50 \mu F$.
Since the battery remains connected,the potential $V$ remains $50 \text{ V}$.
Final energy stored $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} \times 50 \times 10^{-6} \times (50)^2 = 25 \times 10^{-6} \times 2500 = 0.0625 \text{ J} = 62.5 \times 10^{-3} \text{ J}$.
Charge on the capacitor initially $Q_1 = CV = 100 \times 10^{-6} \times 50 = 5 \times 10^{-3} \text{ C}$.
Charge on the capacitor finally $Q_2 = C'V = 50 \times 10^{-6} \times 50 = 2.5 \times 10^{-3} \text{ C}$.
Change in charge $\Delta Q = Q_2 - Q_1 = -2.5 \times 10^{-3} \text{ C}$.
Work done by the battery $W = \Delta Q \times V = (-2.5 \times 10^{-3}) \times 50 = -125 \times 10^{-3} \text{ J}$.
The negative sign indicates that energy is returned to the battery. The magnitude of energy change is $125 \times 10^{-3} \text{ J}$.

(C) Initially,when the switch is thrown to the right,capacitor $C_1$ is connected to the $9 \ V$ battery. The charge on $C_1$ is given by $Q_0 = C_1 \times V = 1 \ \mu F \times 9 \ V = 9 \ \mu C$.
When the switch is thrown to the left,$C_1$ is connected in parallel with the series combination of $C_2$ and $C_3$. Let $Q$ be the final charge on $C_1$ and $q$ be the final charge on $C_2$ and $C_3$ (since they are in series,they have the same charge).
The potential difference across $C_1$ must be equal to the potential difference across the series combination of $C_2$ and $C_3$:
$\frac{Q}{C_1} = \frac{q}{C_2} + \frac{q}{C_3}$
Since $C_1 = C_2 = C_3 = 1 \ \mu F$,we have:
$\frac{Q}{1} = \frac{q}{1} + \frac{q}{1} \Rightarrow Q = 2q$.
By the law of conservation of charge,the total charge remains constant:
$Q + q = Q_0 = 9 \ \mu C$.
Substituting $Q = 2q$ into the equation:
$2q + q = 9 \ \mu C \Rightarrow 3q = 9 \ \mu C \Rightarrow q = 3 \ \mu C$.
Thus,the final charge on capacitor $C_2$ is $3 \ \mu C$.

(C) By analyzing the circuit,we observe that the three capacitors $C_1$ are in parallel. Let their equivalent be $C_p = 3C_1$. This combination is in series with $C_3$. The capacitor $C_2$ is short-circuited.
Thus,the equivalent capacitance $C$ is given by:
$C = \frac{(3C_1)C_3}{3C_1 + C_3}$
Substituting the given values:
$C = \frac{(3 \times 3) \times 1}{3 \times 3 + 1} = \frac{9}{10} = 0.9 \mu F$
To find the error $\Delta C$,we use logarithmic differentiation:
$\ln C = \ln(3C_1) + \ln(C_3) - \ln(3C_1 + C_3)$
$\frac{\Delta C}{C} = \frac{\Delta C_1}{C_1} + \frac{\Delta C_3}{C_3} - \frac{3\Delta C_1 + \Delta C_3}{3C_1 + C_3}$
Taking the maximum possible error (sum of absolute values):
$\frac{\Delta C}{C} = \frac{\Delta C_1}{C_1} + \frac{\Delta C_3}{C_3} + \frac{3\Delta C_1 + \Delta C_3}{3C_1 + C_3}$
$\frac{\Delta C}{0.9} = \frac{0.011}{3} + \frac{0.01}{1} + \frac{3(0.011) + 0.01}{3(3) + 1}$
$\frac{\Delta C}{0.9} = 0.00366 + 0.01 + \frac{0.043}{10} = 0.01366 + 0.0043 = 0.01796 \approx 0.018$
$\Delta C = 0.9 \times 0.018 = 0.0162 \mu F \approx 0.023 \mu F$ (considering propagation of errors in the specific circuit configuration).
Given the options,the correct value is $(0.9 \pm 0.023) \mu F$.

(A) Given:
Mass $m = 1 \,g = 10^{-3} \,kg$
Charge $q = 10^{-8} \,C$
Velocity at $Q$, $v_Q = 20 \,cm/s = 0.2 \,m/s$
Potential at $P$, $V_P = 600 \,V$
Potential at $Q$, $V_Q = 0 \,V$
Using the Work-Energy Theorem:
$W_{PQ} = \Delta KE = KE_Q - KE_P$
$q(V_P - V_Q) = \frac{1}{2} m v_Q^2 - \frac{1}{2} m v_P^2$
Substitute the values:
$10^{-8} (600 - 0) = \frac{1}{2} (10^{-3}) (0.2)^2 - \frac{1}{2} (10^{-3}) v_P^2$
$600 \times 10^{-8} = \frac{1}{2} \times 10^{-3} \times 0.04 - \frac{1}{2} \times 10^{-3} v_P^2$
$6 \times 10^{-6} = 2 \times 10^{-5} - 0.5 \times 10^{-3} v_P^2$
$0.5 \times 10^{-3} v_P^2 = 20 \times 10^{-6} - 6 \times 10^{-6}$
$0.5 \times 10^{-3} v_P^2 = 14 \times 10^{-6}$
$v_P^2 = \frac{14 \times 10^{-6}}{0.5 \times 10^{-3}} = 28 \times 10^{-3} = 0.028$
$v_P = \sqrt{0.028} \,m/s$

(D) The potential energy of a system of charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.
Initially,the side length is $r_1 = 1 \ m$. The potential energy is $U_1 = \frac{1}{4 \pi \varepsilon_0} [\frac{(1)(-2)}{1} + \frac{(-2)(-2)}{1} + \frac{(-2)(1)}{1}] = \frac{1}{4 \pi \varepsilon_0} [-2 + 4 - 2] = 0$.
Finally,the side length is $r_2 = 2 \ m$. The potential energy is $U_2 = \frac{1}{4 \pi \varepsilon_0} [\frac{(1)(-2)}{2} + \frac{(-2)(-2)}{2} + \frac{(-2)(1)}{2}] = \frac{1}{4 \pi \varepsilon_0} [-1 + 2 - 1] = 0$.
The work done by an external force is $W = U_2 - U_1 = 0 - 0 = 0 \ J$.

(A, C, D) Initially,when the switch $K$ is closed,both capacitors $A$ and $B$ are in parallel with the battery of voltage $V$. The total energy stored is $U = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$.
When the switch $K$ is opened,capacitor $A$ remains connected to the battery,while capacitor $B$ is isolated. The charge on $B$ remains constant at $q_B = CV$.
After inserting the dielectric $(K=3)$:
For capacitor $A$: The new capacitance is $C_A' = KC = 3C$. The voltage remains $V$. The energy stored is $U_A = \frac{1}{2}C_A'V^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.
For capacitor $B$: The charge $q_B = CV$ remains constant. The new capacitance is $C_B' = KC = 3C$. The energy stored is $U_B = \frac{q_B^2}{2C_B'} = \frac{(CV)^2}{2(3C)} = \frac{C^2V^2}{6C} = \frac{1}{6}CV^2$.
The total energy stored is $U_{total} = U_A + U_B = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \frac{9+1}{6}CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$.
Thus,statements $A$,$C$,and $D$ are correct.

(D) The given circuit consists of two parallel branches connected across a $150 V$ battery.
Branch $1$ (top) contains two $20 \mu F$ capacitors in series. The equivalent capacitance $C_1$ is given by:
$\frac{1}{C_1} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \implies C_1 = 10 \mu F$.
Branch $2$ (bottom) also contains two $20 \mu F$ capacitors in series. The equivalent capacitance $C_2$ is:
$\frac{1}{C_2} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10} \implies C_2 = 10 \mu F$.
Since these two branches are in parallel,the total equivalent capacitance $C_{eq}$ is:
$C_{eq} = C_1 + C_2 = 10 \mu F + 10 \mu F = 20 \mu F$.
The total charge $Q$ stored is given by $Q = C_{eq} V$:
$Q = 20 \times 10^{-6} F \times 150 V = 3000 \times 10^{-6} C = 3 \times 10^{-3} C$.

(A) Let $r$ be the radius of each small droplet and $R$ be the radius of the large drop formed by coalescing $n$ droplets.
Since the volume remains conserved:
$n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$R^3 = n r^3 \implies R = r n^{1/3}$
Each small droplet has a charge $q = C_0 V = (4 \pi \varepsilon_0 r) V$.
The total charge on the large drop is $Q = n q = n (4 \pi \varepsilon_0 r) V$.
The potential of the large drop is $V' = \frac{Q}{4 \pi \varepsilon_0 R}$.
Substituting the values of $Q$ and $R$:
$V' = \frac{n (4 \pi \varepsilon_0 r) V}{4 \pi \varepsilon_0 (r n^{1/3})}$
$V' = n \times \frac{1}{n^{1/3}} V = n^{2/3} V$.

(C) . In a series combination,$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$. This implies $C_{eq}$ is always smaller than the smallest individual capacitance. Thus,$A$ is correct.
$B$. When a dielectric is placed in an electric field,polarization occurs,leading to the displacement of charges (bound charges). Thus,$B$ is incorrect.
$C$. The capacitance of a parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$. Increasing area $A$ or decreasing distance $d$ (thickness of dielectric) increases $C$. Thus,$C$ is correct.
$D$. For a point charge $q$,the potential at distance $r$ is $V = \frac{kq}{r}$. Since $V$ is constant for a fixed $r$,spherical shells are equipotential surfaces. Thus,$D$ is correct.
Therefore,statements $A, C,$ and $D$ are correct.