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.

Part(a):  The relative capacitance is

Part(b): The relative energy stored is

Part(c): The relative charge stored is

Explanation:

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

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

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be , the outer radius be , the inner radius of Capacitor A be and the outer radius be .

Given in the problem,

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

where is the permittivity of the free space, is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

and the capacitance of capacitor B,

giving the relative capacitance of each capacitor to be

Part(b):

Energy stored by capacitor A,

Energy stored by capacitor B,

giving the relative energy stored by each capacitor to be

Part(c):

The charge stored by capacitor A,

The charge stored by capacitor B,

giving the relative charge stored by each capacitor to be

electrical power

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation