An isolated capacitor has a capacitance of 50 x 109F The par

An isolated capacitor has a capacitance of 5.0 x 10^-9F. The parallel capacitor plates are separated by 2-mm air carry uC of charge

With the capacitor still isolated and ethanol still between the plates, the capacitor plates are pulled apart so that the final plate separation is twice the initial plate separation

A) What is the change in the voltage between the plates

B) How much work did this require

Solution

Capcitance = 5.0 nF

the capacitor is isolated and hence charge Q on the capcitor remains consatnt,

voltage across the capacitor

V = Q/C

with paralle plate capacitor the area of the plate remained same, when the plates are pulled apart the capcitace C dcrease inversely with d, the plate seperation

new capcitance C = 5.0/2   = 2.5 nF, as the seepration has doubled.

New voltage V_n = Q/(C/2) = 2.0V_o voltage across the plates doubles

energy stored in the capacitor E = Q²/2C   

new energy stored = Q²/2(C/2) = Q²/C

increase in energy = Q²/5, this energy comes from the work done in pulling the plates apart.

The chareg Q is not given, put the value of Q to get the actaul workdone in Joules.