Question

Can someone please help me solve PART D

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 Cm^2. plate separation d = 5.00 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V. Throughout the problem, use epsilon_0 = 8.85 times 10^-12C^2/N middot m^2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U_3. Express your answer numerically in joules. U_3 = 1.66 times 10^-9 J In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric? Express your answer numerically in joules. W = ____________

Answer 1

D) When the battery is connected energy stored in the capacitor,

U2 = (1/2)*C*V^2

= (1/2)*(k*A*epsilon/d)*V^2

= (1/2)*(3*30*10^-4*8.854*10^-12/(5*10^-3))*12.5^2

= 1.24*10^-9 J

when the battery is disconnetced,

U3 = 1.66*10^-9 J

Workdone, W = decrease in potential energy

= U3 - U2

= 1.66*10^-9 - 1.24*10^-9

= 4.2*10^-10 J