Question

A capacitor of 9 micro-farads is connected to a 16 volt battery. The distance between the plates of the capacitor is increased by a factor of 3.6 while the capacitor is still connected to the battery. The battery is then disconnected from the plates, and then the plates are moved closer by a factor of 4.2. In their final configuration, what is the charge on each sheet in micro-coulombs?

Answer 1

Initial capacitance, C = 9.0 micro F

So. C1 = 9.0 / 3.6 = 2.5 micro F.

So, Q1 = C1*V = 2.5 x 10^-6 x 16 = 40 x 10^-6 C = 40 micro C.

The expression for the capacitance C = k*A / d

Initially, C1 = k*A / (3.6*d)

and finally, C2 = k*A / (d/4.2)

therefore -

C1 / C2 = 1/ (3.6 x 4.2) = 1 / 15.12

Now in the whole process, the energy will be coserved.

means, U = (1/2)*Q1^2 / C1 = (1/2)*Q1^2 / C2

=> Q1^2 / Q2^2 = C1 / C2 = 1/ 15.12

=> Q2^2 = 15.12 x Q1^2 = 15.12 x (40 x 10^-6)^2

=> Q2 = 155.54 x 10^-6 C = 155.54 micro C.

So, the charge on each plate is 155.54 micro C.