Answer to Question #289414 in Optics for Shortjohn

Question

A parallel plate capacitor is charged to 100v.its plate separation is 2mm and the area of separation of each plate is 120cm^2.calculate and account for the increase and decrease in energy when separation is reduced to 1mm at constant voltage and charge

Answer 1

Given:

$V=100\:\rm V$

$d_1=0.002\:\rm m$

$d_2=0.001\:\rm m$

$A=0.012\:\rm m^2$

The initial and final capacities of capacitor

$$C_1=\frac{\epsilon_0 A}{d_1}=\frac{8.85*10^{-12}*0.012}{0.002}=5.31*10^{-11}\:\rm F$$

$$C_2=\frac{\epsilon_0 A}{d_2}=\frac{8.85*10^{-12}*0.012}{0.001}=1.06*10^{-10}\:\rm F$$

The change in energy of capacitor

$$\Delta W=\frac{C_2-C_1}{2}V^2$$

$$=\frac{5.31*10^{-11}-1.06*10^{-10}}{2}100^2=-2.66*10^{-7}\rm\: J$$