Question

Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the radius of the outer plate of capacitor A is four times the radius of the outer plate of capacitor B. Both inner plates have a radius that is half the outer radius of capacitor B.
(a) What is the relative capacitance of each capacitor? Express your answer as a ratio CA/CB.

(b) What is the relative energy stored by each capacitor? Express your answer as a ratio UA/UB.

(c) What is the relative charge stored by each capacitor? Express your answer as a ratio QA/QB.

Answer 1

Answer:

Part(a):  The relative capacitance is $\dfrac{C_{A}}{C_{B}} = 0.33$

Part(b): The relative energy stored is $\dfrac{U_{A}}{U_{B}} = 0.33$

Part(c): The relative charge stored is $\dfrac{Q_{A}}{Q_{B}} = 0.33$

Explanation:

We know the capacitance ($C$) of a capacitor having charge ($Q$) and subjected to a potential difference of ($V$) is given by

$C = \dfrac{Q}{V}$

Also, the energy ($U$) stored by a capacitor can be written as

$U = \dfrac{1}{2}C~V^{2}$

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be $\textbf{r_{i}^{B}}$$\bf{r_{i}^{B}}$, the outer radius be $\bf{r_{o}^{B}}$, the inner radius of Capacitor A be $\bf{r_{i}^{A}}$ and the outer radius be $\bf{r_{o}^{B}}$.

Given in the problem,

$&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}$

Now, the capacitance ($C$) of a cylindrical capacitor is given by,

$\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}$

where $\epsilon_{o}$ is the permittivity of the free space, $L$ is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

$C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}$

and the capacitance of capacitor B,

$C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}$

giving the relative capacitance of each capacitor to be

$\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33$

Part(b):

Energy stored by capacitor A,

$U_{A} = \dfrac{1}{2}~C_{A}~V^{2}$

Energy stored by capacitor B,

$U_{B} = \dfrac{1}{2}~C_{B}~V^{2}$

giving the relative energy stored by each capacitor to be

$\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

Part(c):

The charge stored by capacitor A,

$Q_{A} = C_{A}~V$

The charge stored by capacitor B,

$Q_{B} = C_{B}~V$

giving the relative charge stored by each capacitor to be

$\dfrac{Q_{A}}{Q_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$