Basic of Capacitor and type of capacitor (Spherical, Cylindrical) Questions in English

(C) For a cylindrical capacitor with inner radius $a$ and outer radius $b$,the electric field $E$ at a distance $r$ (where $a < r < b$) from the axis is determined using Gauss's Law.
Consider a Gaussian surface in the form of a cylinder of radius $r$ and length $L$.
The total charge enclosed is $q = \lambda L$,where $\lambda$ is the linear charge density.
According to Gauss's Law,$\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.
$E(2\pi r L) = \frac{\lambda L}{\varepsilon_0}$.
Thus,$E = \frac{\lambda}{2\pi \varepsilon_0 r}$.
This shows that the magnitude of the electric field $E$ varies inversely with the distance $r$ from the axis,i.e.,$E \propto 1/r$.

(D) Let the charge on the inner sphere be $+Q$.
Since the outer sphere is earthed,its potential is $0$.
The potential $V$ of the inner sphere is the sum of the potential due to its own charge and the potential due to the induced charge on the outer sphere.
$V = \frac{Q}{r_1} - \frac{Q}{r_2}$
Given $r_1 = 4\, cm$,$r_2 = 6\, cm$,and $V = 3\, e.s.u.$
$3 = \frac{Q}{4} - \frac{Q}{6}$
$3 = Q \left( \frac{3 - 2}{12} \right)$
$3 = \frac{Q}{12}$
$Q = 36\, e.s.u.$

(B) The electric field $E$ at the surface of a spherical conductor of radius $R$ carrying charge $Q$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} = k \frac{Q}{R^2}$.
Given $E_{max} = 3 \times 10^6 \, V/m$,$R = 3 \, m$,and $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
Substituting the values into the formula: $3 \times 10^6 = 9 \times 10^9 \times \frac{Q}{3^2}$.
$3 \times 10^6 = 9 \times 10^9 \times \frac{Q}{9}$.
$3 \times 10^6 = 10^9 \times Q$.
$Q = \frac{3 \times 10^6}{10^9} = 3 \times 10^{-3} \, C$.

(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.
Since the volume remains conserved when $8$ small drops combine to form one big drop:
$8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$8r^3 = R^3$
$R = 2r$
The capacitance of a spherical drop of radius $r$ is given by $C = 4 \pi \epsilon_0 r$.
Therefore,the capacitance of the small drop is $C_{small} = 4 \pi \epsilon_0 r$.
The capacitance of the big drop is $C_{big} = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (2r) = 2 \times (4 \pi \epsilon_0 r) = 2 C_{small}$.
Thus,the capacitance of the bigger drop is $2$ times that of each individual small drop.

(D) Given: The spacing between the two spheres is $(b - a) = 1\,mm = 1 \times 10^{-3}\,m$ ..... $(i)$
The capacitance of a spherical capacitor is given by $C = 4\pi \varepsilon_0 \left( \frac{ab}{b - a} \right)$.
Given $C = 1\,\mu F = 1 \times 10^{-6}\,F$ and $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$.
Substituting the values:
$1 \times 10^{-6} = \frac{1}{9 \times 10^9} \left( \frac{ab}{10^{-3}} \right)$
$ab = 9 \times 10^9 \times 10^{-6} \times 10^{-3} = 9$ ..... $(ii)$
From equations $(i)$ and $(ii)$,we have $a = b - 10^{-3}$.
Substituting $a$ in $(ii)$:
$(b - 10^{-3})b = 9$
$b^2 - 10^{-3}b - 9 = 0$
Using the quadratic formula $b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:
$b = \frac{10^{-3} + \sqrt{(10^{-3})^2 - 4(1)(-9)}}{2} = \frac{10^{-3} + \sqrt{10^{-6} + 36}}{2}$
Since $10^{-6}$ is negligible compared to $36$,$\sqrt{36.000001} \approx 6$.
$b \approx \frac{0.001 + 6}{2} \approx 3.0005\,m \approx 3\,m$.
Thus,the radius of the outer sphere is $3\,m$.

(C) The potential $V$ at the surface of a spherical conductor of radius $R$ carrying charge $q$ is given by:
$V = \frac{q}{4 \pi \varepsilon_{0} R}$
The capacitance $C$ is defined as the ratio of charge to potential:
$C = \frac{q}{V}$
Substituting the expression for $V$:
$C = \frac{q}{\frac{q}{4 \pi \varepsilon_{0} R}}$
Simplifying the expression:
$C = 4 \pi \varepsilon_{0} R$
Thus,the capacity of a spherical conductor is $4 \pi \varepsilon_{0} R$.

(A) The capacitance $C$ of a spherical capacitor with radii $a$ and $b$ $(b > a)$ and dielectric constant $K$ is given by the formula: $C = 4\pi \varepsilon_0 K \left( \frac{ab}{b - a} \right)$.
Given: $a = 9 \; cm = 9 \times 10^{-2} \; m$,$b = 12 \; cm = 12 \times 10^{-2} \; m$,$K = 6$,and $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \; N \cdot m^2/C^2$.
Substituting the values:
$C = \frac{6}{9 \times 10^9} \times \left( \frac{12 \times 10^{-2} \times 9 \times 10^{-2}}{12 \times 10^{-2} - 9 \times 10^{-2}} \right)$
$C = \frac{6}{9 \times 10^9} \times \left( \frac{108 \times 10^{-4}}{3 \times 10^{-2}} \right)$
$C = \frac{6}{9 \times 10^9} \times (36 \times 10^{-2})$
$C = \frac{216 \times 10^{-2}}{9 \times 10^9} = 24 \times 10^{-11} \; F$
$C = 240 \times 10^{-12} \; F = 240 \; pF$.

(A) The capacitance $C$ of an isolated spherical conductor of radius $r$ is given by the formula: $C = 4\pi \varepsilon_0 r$.
We know that $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$,therefore $4\pi \varepsilon_0 = \frac{1}{9 \times 10^9}$.
Given $C = 1\,\mu F = 1 \times 10^{-6}\,F$.
Substituting the values into the formula: $1 \times 10^{-6} = \frac{r}{9 \times 10^9}$.
Solving for $r$: $r = 1 \times 10^{-6} \times 9 \times 10^9 = 9 \times 10^3\,m$.
Converting to kilometers: $r = 9\,km$.

(D) For a charged spherical capacitor with inner radius $a$ and outer radius $b$,the electric field $E$ at a distance $r$ (where $a < r < b$) from the centre is given by Gauss's Law as $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.
Since $E \propto \frac{1}{r^2}$,the electric field decreases as the distance $r$ from the centre increases.
Therefore,the correct option is $D$.

(C) When the outer sphere is earthed,the capacitance $C_1$ of the spherical capacitor is given by:
$C_1 = 4\pi \varepsilon_0 \frac{ab}{b - a}$
When the inner sphere is earthed,the outer sphere acts as an isolated conductor with capacitance $4\pi \varepsilon_0 b$ in parallel with the spherical capacitor. Thus,the total capacitance $C_2$ is:
$C_2 = 4\pi \varepsilon_0 b + \frac{4\pi \varepsilon_0 ab}{b - a} = 4\pi \varepsilon_0 \left( \frac{b(b - a) + ab}{b - a} \right) = 4\pi \varepsilon_0 \left( \frac{b^2 - ab + ab}{b - a} \right) = 4\pi \varepsilon_0 \frac{b^2}{b - a}$
The difference in capacities is:
$C_2 - C_1 = 4\pi \varepsilon_0 \left( \frac{b^2}{b - a} - \frac{ab}{b - a} \right) = 4\pi \varepsilon_0 \left( \frac{b^2 - ab}{b - a} \right) = 4\pi \varepsilon_0 \frac{b(b - a)}{b - a} = 4\pi \varepsilon_0 b$

(C) The capacitance $C$ of a spherical capacitor with inner radius $a$ and outer radius $b$ is given by $C = 4\pi\epsilon_0 \frac{ab}{b - a}$.
Given, the radius of the outer sphere $b = R$.
The difference in radii is $b - a = x$, so the inner radius $a = R - x$.
Substituting these values into the formula, we get $C = 4\pi\epsilon_0 \frac{(R - x)R}{x}$.
Since $4\pi\epsilon_0$ is a constant, the capacity $C$ is proportional to $\frac{R(R - x)}{x}$.

(C) The capacitance of a spherical drop of radius $r$ is given by $C = 4\pi\epsilon_0 r$.
When $n$ identical drops of radius $r$ merge to form a single drop of radius $R$, the volume remains conserved.
So, $n \times (\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$, which implies $R = n^{1/3}r$.
The capacitance of the new drop is $C' = 4\pi\epsilon_0 R = 4\pi\epsilon_0 (n^{1/3}r) = n^{1/3}C$.
For $n = 2$, the new capacitance is $C' = 2^{1/3}C$.
Since $1 < 2^{1/3} < 2$, the resultant capacitance $C'$ is less than $2C$ but greater than $C$.

(C) The capacitance $C$ of an isolated spherical conductor of radius $R$ is given by the formula: $C = 4\pi \varepsilon_0 R$.
We know that $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$,so $4\pi \varepsilon_0 = \frac{1}{9 \times 10^9}$.
Given $C = 1/9\,F$.
Substituting the values into the formula: $R = \frac{C}{4\pi \varepsilon_0} = C \times (9 \times 10^9)$.
$R = (1/9) \times (9 \times 10^9) = 10^9\,m$.
Therefore,the radius of the sphere is $10^9\,m$.

(A) The capacitance of a body is defined as its ability to store electric charge.
The mathematical relationship is given by:
$C = \frac{Q}{V}$
where $Q$ is the charge on the body and $V$ is the electric potential.
Therefore,the ratio of charge to potential is known as capacitance.

(B) The capacitance $C$ of a spherical conductor of radius $R$ is given by the formula $C = 4\pi \varepsilon_0 R$.
Given $C = 1 \text{ pF} = 1 \times 10^{-12} \text{ F}$.
We know that $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.
Therefore,$R = C \times (9 \times 10^9) = 10^{-12} \times 9 \times 10^9 = 9 \times 10^{-3} \text{ m}$.
The diameter $D$ is $2R = 2 \times 9 \times 10^{-3} \text{ m} = 18 \times 10^{-3} \text{ m}$.

(B) Let the radius of each small drop be $r$ and the radius of the bigger drop be $R$.
Since the volume remains constant,the volume of the bigger drop is equal to the sum of the volumes of $64$ small drops:
$\frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3$
$R^3 = 64 r^3$
$R = 4r$
The capacitance of a spherical drop is given by $C = 4 \pi \epsilon_0 r$.
Therefore,the capacitance of the bigger drop $C'$ is:
$C' = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (4r) = 4(4 \pi \epsilon_0 r) = 4C$.
Thus,the capacitance of the bigger drop is $4C$.

(D) The capacitance $C$ of a conductor is defined by the relation $Q = CV$,where $Q$ is the charge and $V$ is the potential.
However,the capacitance $C$ is a geometric property of the conductor.
It depends on the shape,size,and the surrounding medium (dielectric constant) of the conductor.
It does not depend on the charge $Q$ given to the conductor or the potential $V$ developed on it.
Furthermore,for a conductor,the capacitance is independent of the material of the conductor itself,as the charge resides on the surface.
Therefore,the capacity does not depend on charge,voltage,or the nature of the material.
Thus,the correct option is $D$.

(D) The capacitance $C$ of a spherical capacitor consisting of two concentric conducting spheres of radii $R_1$ and $R_2$ $(R_2 > R_1)$ is given by the formula:
$C = 4\pi \varepsilon_0 \frac{R_1 R_2}{R_2 - R_1}$
Since $4\pi \varepsilon_0$ is a constant,the capacitance is proportional to the term $\frac{R_1 R_2}{R_2 - R_1}$.
Therefore,the correct option is $D$.

(D) When the inner sphere $A$ (radius $a$) is grounded,its potential $V_A$ becomes $0$.
Let the charge on sphere $A$ be $q$ and the charge on sphere $B$ be $Q$.
The potential of sphere $A$ is given by $V_A = \frac{1}{4\pi \varepsilon_0} \frac{q}{a} + \frac{1}{4\pi \varepsilon_0} \frac{Q}{b} = 0$.
From this,we get $q = -Q \frac{a}{b}$.
The potential of sphere $B$ is $V_B = \frac{1}{4\pi \varepsilon_0} \frac{q}{b} + \frac{1}{4\pi \varepsilon_0} \frac{Q}{b}$.
Substituting $q = -Q \frac{a}{b}$ into the equation for $V_B$:
$V_B = \frac{1}{4\pi \varepsilon_0} \left( -Q \frac{a}{b^2} + \frac{Q}{b} \right) = \frac{Q}{4\pi \varepsilon_0 b} \left( 1 - \frac{a}{b} \right) = \frac{Q}{4\pi \varepsilon_0 b} \left( \frac{b - a}{b} \right) = \frac{Q(b - a)}{4\pi \varepsilon_0 b^2}$.
The capacitance $C$ is defined as $C = \frac{Q}{V_B}$.
Therefore,$C = \frac{Q}{\frac{Q(b - a)}{4\pi \varepsilon_0 b^2}} = \frac{4\pi \varepsilon_0 b^2}{b - a}$.

(A) The capacitance $C$ of an isolated spherical conductor of radius $R$ is given by the formula: $C = 4\pi \epsilon_0 R$.
We know that the value of the Coulomb constant $k = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2$.
Therefore,$4\pi \epsilon_0 = \frac{1}{9 \times 10^9} \, F/m$.
Given $R = 1 \, m$,we substitute the values:
$C = \frac{1}{9 \times 10^9} \times 1 = 0.111 \times 10^{-9} \, F = 1.11 \times 10^{-10} \, F$.
Thus,the correct option is $A$.

(C) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the electric permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.
Rearranging the formula for $\varepsilon_0$,we get $\varepsilon_0 = \frac{C \cdot d}{A}$.
Substituting the units: The unit of capacitance $C$ is $Farad$ $(F)$,the unit of distance $d$ is $m$,and the unit of area $A$ is $m^2$.
Therefore,the unit of $\varepsilon_0 = \frac{F \cdot m}{m^2} = F/m$.

(C) Let the radius of the large drop be $R$ and its capacitance be $C = 4\pi\epsilon_0 R = 1\,\mu F$.
Let the radius of each small drop be $r$. Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $8$ small drops:
$\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$
$R^3 = 8r^3 \implies R = 2r$ or $r = R/2$.
The capacitance of each small drop is $c = 4\pi\epsilon_0 r$.
Substituting $r = R/2$,we get $c = 4\pi\epsilon_0 (R/2) = C/2$.
Given $C = 1\,\mu F$,the capacitance of each small drop is $c = 1/2 = 0.5\,\mu F$.

(A) The potential difference $(V)$ across the plates of the capacitor is calculated as the difference between the potentials of the two plates.
$V = V_1 - V_2 = 10\,V - (-10\,V) = 20\,V$.
The charge $(Q)$ on the plate is given as $40\,C$.
The capacitance $(C)$ of a capacitor is defined by the formula $C = \frac{Q}{V}$.
Substituting the values,we get $C = \frac{40\,C}{20\,V} = 2\,F$.

(A) The capacitance $C$ of a spherical capacitor with inner radius $a$ and outer radius $b$ is given by the formula $C = 4\pi \varepsilon_0 \frac{ab}{b - a}$.
Since shell $A$ is grounded,its potential $V_A = 0$.
The potential of shell $B$ is $V_B = \frac{1}{4\pi \varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi \varepsilon_0} \frac{q_A}{b}$,where $q_A$ is the induced charge on shell $A$.
However,for a grounded inner shell,the capacitance is defined as the ratio of the charge on the outer shell to the potential difference between the shells.
The potential difference is $V = V_B - V_A = \frac{Q}{4\pi \varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right)$.
Thus,$C = \frac{Q}{V} = \frac{4\pi \varepsilon_0}{\frac{1}{a} - \frac{1}{b}} = 4\pi \varepsilon_0 \frac{ab}{b - a}$.

(D) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $A$ is the area of the plates (related to the size),$d$ is the distance between the plates (related to the thickness/separation),and $\epsilon_0$ (or $\epsilon$) depends on the dielectric material between the plates. Since all these factors influence the capacitance,the correct answer is that it depends on all of the above.

(D) The capacitance $C$ of a spherical capacitor consisting of two concentric conducting spheres of radii $R_1$ and $R_2$ (where $R_2 > R_1$) is given by the formula:
$C = 4 \pi \varepsilon_0 \frac{R_1 R_2}{R_2 - R_1}$
Here,$4 \pi \varepsilon_0$ is a constant.
Therefore,the capacitance $C$ is proportional to the term $\frac{R_1 R_2}{R_2 - R_1}$.

(D) Given,circumference of the sphere $2\pi R = 2 \ m$.
So,$R = \frac{1}{\pi} \ m$.
The capacitance of a spherical conductor in a medium is given by $C = 4\pi \varepsilon_0 K R$.
We know that $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$,so $4\pi \varepsilon_0 = \frac{1}{9 \times 10^9}$.
Substituting the values: $C = \frac{1}{9 \times 10^9} \times 80 \times \frac{1}{\pi}$.
$C = \frac{80}{9 \times 3.14159 \times 10^9} \approx 2.828 \times 10^{-9} \ F$.
Converting to $pF$ $(1 \ F = 10^{12} \ pF)$: $C = 2.828 \times 10^{-9} \times 10^{12} \ pF = 2828 \ pF$.

(C) Let the radius of each small drop be $r$ and the charge be $q$. The capacitance of a small drop is $C_s = 4 \pi \epsilon_0 r$.
When $n = 8$ drops coalesce, the volume remains constant. Thus, $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$, which gives $R = n^{1/3} r$.
For $n = 8$, $R = 8^{1/3} r = 2r$.
The capacitance of the larger drop is $C_L = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (2r) = 2 C_s$.
Therefore, the capacitance of the larger drop is $2$ times that of the smaller drop.

(A) The capacitance $C$ of an isolated spherical conductor of radius $R$ is given by the formula: $C = 4\pi \epsilon_0 R$.
We know that $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$.
Therefore,$4\pi \epsilon_0 = \frac{1}{9 \times 10^9} \ F/m$.
Given $R = 1 \ m$,substituting the values:
$C = \frac{1}{9 \times 10^9} \times 1 \ F$.
$C = 0.111 \times 10^{-9} \ F = 1.11 \times 10^{-10} \ F$.

(B) The capacitance $C$ of a conductor is defined by the formula $C = \frac{Q}{V}$.
Given:
Charge $Q = 50\ \mu C = 50 \times 10^{-6}\ C$
Potential difference $V = 5\, V$
Substituting the values into the formula:
$C = \frac{50\ \mu C}{5\, V} = 10\ \mu F$.

(C) Let the radius of the large drop be $R$ and the radius of each small drop be $r$. The capacitance of a spherical drop is given by $C = 4 \pi \epsilon_0 R$.
Since the volume is conserved,the volume of the large drop equals the sum of the volumes of the $8$ small drops: $\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3$.
This simplifies to $R^3 = 8 r^3$,which means $R = 2r$.
The capacitance of the large drop is $C = 4 \pi \epsilon_0 R = 1 \ \mu F$.
The capacitance of each small drop is $c = 4 \pi \epsilon_0 r$.
Substituting $r = R/2$,we get $c = 4 \pi \epsilon_0 (R/2) = C/2$.
Therefore,$c = 1 \ \mu F / 2 = 0.5 \ \mu F$.

(B) The charge $Q$ transferred is given by $Q = ne$,where $n = 6.25 \times 10^{15}$ and $e = 1.6 \times 10^{-19} \, C$.
$Q = (6.25 \times 10^{15}) \times (1.6 \times 10^{-19}) = 10 \times 10^{-4} = 10^{-3} \, C$.
The capacitance $C$ is given by $C = \frac{Q}{V}$.
Given $V = 100 \, V$,we have $C = \frac{10^{-3}}{100} = 10^{-5} \, F$.
To convert to $\mu F$,$C = 10^{-5} \times 10^6 \, \mu F = 10 \, \mu F$.

(B) The capacitance $C$ of a spherical capacitor with radii $a$ and $b$ (where $b > a$) filled with a dielectric of constant $k$ is given by the formula:
$C = \frac{4\pi \epsilon_0 k ab}{b - a}$
Given: $a = 0.5 \ m$,$b = 0.6 \ m$,$k = 6$,and $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \ N \ m^2/C^2$.
Substituting the values:
$C = \frac{6 \times 0.5 \times 0.6}{9 \times 10^9 \times (0.6 - 0.5)}$
$C = \frac{1.8}{9 \times 10^9 \times 0.1}$
$C = \frac{1.8}{0.9 \times 10^9}$
$C = 2 \times 10^{-9} \ F$.

(D) The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$,where $L$ is the length,$b$ is the outer radius,and $a$ is the inner radius.
Since the capacitance $C$ remains constant,we have $L \propto \ln(b/a)$.
Let the initial radii be $a_1$ and $b_1$,where $b_1/a_1 = 10$.
For the second case,the new inner radius $a_2 = a_1/2$. The outer radius $b_2$ remains the same as $b_1$ (i.e.,$b_2 = b_1 = 10a_1$).
Thus,the new ratio is $b_2/a_2 = (10a_1) / (a_1/2) = 20$.
Setting the capacitances equal: $L_1 \ln(b_1/a_1) = L_2 \ln(b_2/a_2)$.
Therefore,the ratio of the lengths is $L_2/L_1 = \frac{\ln(b_1/a_1)}{\ln(b_2/a_2)} = \frac{\ln(10)}{\ln(20)}$.
Using $\ln(10) \approx 2.302$ and $\ln(20) = \ln(10) + \ln(2) \approx 2.302 + 0.693 = 2.995$.
$L_2/L_1 = 2.302 / 2.995 \approx 0.768$.
Wait,re-evaluating the ratio: $L_2/L_1 = \frac{\ln(20)}{\ln(10)} = \frac{2.995}{2.302} \approx 1.301$.
Thus,the length must be increased by a factor of approximately $1.3$.

(A) Given: Length $l = 15\,cm = 0.15\,m$,inner radius $a = 1.4\,cm = 0.014\,m$,outer radius $b = 1.5\,cm = 0.015\,m$,charge $q = 3.5\,\mu C = 3.5 \times 10^{-6}\,C$.
The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \epsilon_0 l}{\ln(b/a)} = \frac{2\pi \epsilon_0 l}{2.303 \log_{10}(b/a)}$.
Substituting the values: $C = \frac{2 \times 3.14159 \times 8.854 \times 10^{-12} \times 0.15}{2.303 \times \log_{10}(1.5/1.4)} \approx \frac{8.345 \times 10^{-12}}{2.303 \times 0.0299} \approx 1.21 \times 10^{-10}\,F$.
The potential of the inner cylinder relative to the earthed outer cylinder is $V = \frac{q}{C}$.
$V = \frac{3.5 \times 10^{-6}}{1.21 \times 10^{-10}} \approx 2.89 \times 10^4\,V$.

(B) The capacitance of a spherical capacitor is given by $C = 4\pi \varepsilon_0 \frac{ab}{b - a}$.
Here,$a$ is the radius of the Earth $= 6400 \ km = 6.4 \times 10^6 \ m$.
The outer radius $b$ is the distance to the stratosphere $= 6400 \ km + 50 \ km = 6450 \ km = 6.45 \times 10^6 \ m$.
Given $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$.
Substituting the values:
$C = \frac{1}{9 \times 10^9} \times \frac{(6.4 \times 10^6) \times (6.45 \times 10^6)}{6.45 \times 10^6 - 6.4 \times 10^6}$
$C = \frac{1}{9 \times 10^9} \times \frac{41.28 \times 10^{12}}{0.05 \times 10^6}$
$C = \frac{41.28 \times 10^{12}}{9 \times 10^9 \times 0.05 \times 10^6} = \frac{41.28 \times 10^{12}}{0.45 \times 10^{15}} \approx 0.0917 \ F \approx 0.09 \ F$.

(B) The capacitance $C$ of an isolated spherical conductor of radius $R$ is given by the formula $C = 4\pi \varepsilon_0 R$.
From this,the permittivity of free space $\varepsilon_0$ can be expressed as $\varepsilon_0 = \frac{C}{4\pi R}$.
Since the unit of capacitance $C$ is $Farad$ $(F)$ and the unit of radius $R$ is $meter$ $(m)$,the unit of $\varepsilon_0$ is $Farad / meter$ $(F/m)$.

(C) The capacity of a conductor to hold charge depends on its capacitance. For a spherical conductor of radius $R$, the capacitance is given by $C = 4 \pi \epsilon_0 R$.
Since both the solid sphere and the hollow sphere have the same radius $R$, their capacitance is identical.
Furthermore, for a charged conductor, the charge resides entirely on the outer surface of the conductor.
Therefore, both spheres have the same capacity to hold electric charge.

(D) Let the radius of each small drop be $r$ and the radius of the large drop be $R$.
Since the volume remains constant,the volume of $8$ small drops equals the volume of the large drop:
$8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 8r^3 \implies R = 2r$.
The capacitance of a spherical drop is given by $C = 4 \pi \epsilon_0 r$.
For the large drop,the capacitance $C' = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (2r) = 2 \times (4 \pi \epsilon_0 r) = 2C$.
Therefore,the capacitance of the large drop is $2C$.

(B) Let $Q_1$ be the charge on the inner shell and $Q_2$ be the charge on the outer shell.
The potential at the inner shell $(R_1)$ is given by $V_1 = \frac{k Q_1}{R_1} + \frac{k Q_2}{R_2}$.
The potential at the outer shell $(R_2)$ is given by $V_2 = \frac{k Q_1}{R_2} + \frac{k Q_2}{R_2}$.
Subtracting the two equations: $V_1 - V_2 = k Q_1 (\frac{1}{R_1} - \frac{1}{R_2}) = k Q_1 \frac{R_2 - R_1}{R_1 R_2}$.
Thus,$k Q_1 = \frac{(V_1 - V_2) R_1 R_2}{R_2 - R_1}$.
The potential at a distance $x$ $(R_1 < x < R_2)$ is $V(x) = \frac{k Q_1}{x} + \frac{k Q_2}{R_2}$.
Since $V_2 = \frac{k Q_1}{R_2} + \frac{k Q_2}{R_2}$,we have $\frac{k Q_2}{R_2} = V_2 - \frac{k Q_1}{R_2}$.
Substituting this into the expression for $V(x)$: $V(x) = k Q_1 (\frac{1}{x} - \frac{1}{R_2}) + V_2 = k Q_1 \frac{R_2 - x}{x R_2} + V_2$.
Substituting the value of $k Q_1$: $V(x) = \frac{(V_1 - V_2) R_1 R_2}{R_2 - R_1} \cdot \frac{R_2 - x}{x R_2} + V_2$.
Simplifying: $V(x) = \frac{(V_1 - V_2) R_1 (R_2 - x) + V_2 x (R_2 - R_1)}{x (R_2 - R_1)} = \frac{V_1 R_1 (R_2 - x) + V_2 R_2 (x - R_1)}{x (R_2 - R_1)}$.

(C) Case $1$: When the outer shell of radius $b$ is grounded,the capacitance $C_1$ is given by the formula for a spherical capacitor: $C_1 = 4\pi \varepsilon_0 \frac{ab}{b-a}$.
Case $2$: When the inner shell of radius $a$ is grounded,the system acts as two capacitors in parallel: the inner shell (isolated) and the outer shell (acting as a spherical conductor of radius $b$). The total capacitance $C_2$ is the sum of the capacitance of the spherical shell $b$ $(4\pi \varepsilon_0 b)$ and the mutual capacitance between the shells $(4\pi \varepsilon_0 \frac{ab}{b-a})$. Thus,$C_2 = 4\pi \varepsilon_0 b + 4\pi \varepsilon_0 \frac{ab}{b-a} = 4\pi \varepsilon_0 \left( \frac{b^2 - ab + ab}{b-a} \right) = 4\pi \varepsilon_0 \frac{b^2}{b-a}$.
The difference is $C_2 - C_1 = 4\pi \varepsilon_0 \left( \frac{b^2}{b-a} - \frac{ab}{b-a} \right) = 4\pi \varepsilon_0 \left( \frac{b(b-a)}{b-a} \right) = 4\pi \varepsilon_0 b$.

(B) The capacitance $C$ of a spherical capacitor with a dielectric medium is given by $C = K \cdot 4\pi \varepsilon_0 \left( \frac{r_1 r_2}{r_2 - r_1} \right)$.
Given: $r_1 = 9 \times 10^{-2}\,m$,$r_2 = 10 \times 10^{-2}\,m$,$K = 6$,and $q = 18 \times 10^{-9}\,C$.
Substituting the values: $C = 6 \times \frac{1}{9 \times 10^9} \times \left( \frac{9 \times 10^{-2} \times 10 \times 10^{-2}}{10 \times 10^{-2} - 9 \times 10^{-2}} \right)$.
$C = \frac{6}{9 \times 10^9} \times \left( \frac{90 \times 10^{-4}}{1 \times 10^{-2}} \right) = \frac{6}{9 \times 10^9} \times 90 \times 10^{-2} = 6 \times 10^{-10}\,F$.
The potential $V$ of the inner sphere relative to the earthed outer sphere is $V = \frac{q}{C}$.
$V = \frac{18 \times 10^{-9}}{6 \times 10^{-10}} = 3 \times 10 = 30\,V$.

(B) The capacitance of an isolated conducting sphere of radius $a$ is $C_0 = 4\pi \varepsilon_0 a$.
When this sphere is enclosed by an earthed concentric sphere of radius $b$ $(b > a)$,the system forms a spherical capacitor.
The capacitance of a spherical capacitor is given by $C = 4\pi \varepsilon_0 \left( \frac{ab}{b-a} \right)$.
We can rewrite this as $C = 4\pi \varepsilon_0 a \left( \frac{b}{b-a} \right)$.
Dividing the numerator and denominator by $a$,we get $C = 4\pi \varepsilon_0 a \left( \frac{b/a}{b/a - 1} \right)$.
Given the ratio $\frac{b}{a} = n$,we substitute this into the expression:
$C = (4\pi \varepsilon_0 a) \left( \frac{n}{n-1} \right) = C_0 \left( \frac{n}{n-1} \right)$.
Thus,the capacitance is increased by a factor of $\frac{n}{n-1}$.

(D) In the steady state condition,the capacitors act as open circuits,meaning no current flows through them.
Since all branches are connected in parallel across the battery of $EMF$ $E$,the potential difference across each capacitor is equal to the $EMF$ of the battery,$E$.
The charge stored in the first capacitor $(C_1 = C)$ is $Q_1 = C \times E$.
The charge stored in the $n^{th}$ capacitor $(C_n = nC)$ is $Q_n = (nC) \times E$.
Therefore,the ratio of the charge stored in the first capacitor to the charge stored in the $n^{th}$ capacitor is $\frac{Q_1}{Q_n} = \frac{CE}{nCE} = \frac{1}{n}$.

(C) When two conducting spherical shells are connected by a conductor,they reach the same potential.
Since the inner shell is connected to the outer shell,the entire charge given to the system resides on the outer surface of the outer shell.
Thus,the system behaves like a single isolated spherical conductor of radius $b$.
The capacitance of an isolated spherical conductor of radius $R$ is given by $C = 4 \pi \varepsilon_0 R$.
Therefore,for the given system,the capacitance is $C = 4 \pi \varepsilon_0 b$.

(B) For a large separation $L$,a charged capacitor can be treated as an electric dipole.
The dipole moment $p$ of a parallel plate capacitor is given by $p = qd$,where $q$ is the charge on the plates and $d$ is the separation between them. Thus,$p \propto d$.
The force of interaction $F$ between two dipoles separated by a distance $L$ is given by $F \propto \frac{p_1 p_2}{L^4}$.
Substituting $p \propto d$,we get $F \propto \frac{d \cdot d}{L^4} = \frac{d^2}{L^4}$.
Therefore,the force of interaction is proportional to $d^2/L^4$.

(A) The capacitance formulas for different configurations are as follows:
$1$. For a cylindrical capacitor of length $\ell$ and radii $r_1, r_2$,the capacitance is $C = \frac{2\pi \epsilon_0 \ell}{\ln(r_2/r_1)}$. Thus,$A \rightarrow iii$.
$2$. For a spherical capacitor with inner radius $r_1$ and outer radius $r_2$,the capacitance is $C = \frac{4\pi \epsilon_0 r_1 r_2}{r_2 - r_1}$. Thus,$B \rightarrow iv$.
$3$. For a parallel plate capacitor with a dielectric of constant $K$,the capacitance is $C = \frac{K A \epsilon_0}{d}$. Thus,$C \rightarrow ii$.
$4$. For an isolated spherical conductor of radius $R$,the capacitance is $C = 4\pi \epsilon_0 R$. Thus,$D \rightarrow i$.
Therefore,the correct matching is $A-(iii), B-(iv), C-(ii), D-(i)$.

(D) The capacitance of a spherical conductor is given by $C = 4 \pi \epsilon_0 r$.
For $C = 1 \, F$,the required radius is $r = \frac{C}{4 \pi \epsilon_0} = 9 \times 10^9 \, m$.
This radius is approximately $1500$ times the radius of the Earth $(6.4 \times 10^6 \, m)$,making it physically impossible to construct such a sphere.
Therefore,Statement $1$ is true.
Statement $2$ claims it is possible for Earth,but the Earth's capacitance is only $C = 4 \pi \epsilon_0 R_e \approx 711 \, \mu F$,which is much less than $1 \, F$.
Thus,Statement $2$ is false.

(B) Let the radius of the isolated sphere be $a$ and the radius of the earthed outer sphere be $b$.
The capacitance of an isolated sphere is given by $C = 4 \pi \epsilon_0 a$.
When this sphere is enclosed by an earthed concentric sphere of radius $b$,the new capacitance $C'$ is given by $C' = \frac{4 \pi \epsilon_0 a b}{b - a}$.
According to the problem,the capacity is increased $n$ times,so $C' = nC$.
Substituting the expressions,we get $n(4 \pi \epsilon_0 a) = \frac{4 \pi \epsilon_0 a b}{b - a}$.
Dividing both sides by $4 \pi \epsilon_0 a$,we get $n = \frac{b}{b - a}$.
Rearranging the equation: $n(b - a) = b$,which implies $nb - na = b$.
Grouping the terms with $b$: $nb - b = na$,or $b(n - 1) = na$.
Therefore,the ratio of their radii is $\frac{b}{a} = \frac{n}{n - 1}$.

(B) The voltmeter is connected across the two capacitors on the left. Since these two capacitors are connected in parallel,the potential difference across each is $200\,V$.
Given,capacitance $C = 25\,\mu F = 25 \times 10^{-6}\,F$ and potential difference $V = 200\,V$.
The charge $Q$ on each plate of the capacitor is given by the formula:
$Q = \pm CV$
Substituting the values:
$Q = \pm (25 \times 10^{-6}\,F) \times (200\,V)$
$Q = \pm 5000 \times 10^{-6}\,C$
$Q = \pm 5 \times 10^{-3}\,C$