Question

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 6.00 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

a) Find the energy U1 of the dielectric-filled capacitor. Express your answer numerically in joules.

b) The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. Express your answer numerically in joules.

c) 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, U3. Express your answer numerically in joules.

d) 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.

Answer 1

formula for capacitence is

C= k eo A/ d

= 4.00 * 8.85 * 10^-12 * 10 * 10^-4 / 6.00 * 10^-3

               =   5.9 * 10^-12   F

U1 = 1/2 * C V^2 = .0.5 *  5.9 * 10^-12   F ( 15)^2 = 663.75 * 10^-12 J

(2)

when half of th ecapacitor is filled with dielectric

C₂   =    ε₀ * (A/2) / d   +   k * ε₀ * (A/2) / d

= 8.85 * 10^-12 *( 10 * 10^-4/2) / 6.00 * 10^-3* ( 1+ 4)

               = 3.68* 10^-12 F

U2 = 1/2 * C2 V^2 = 0.5 * 3.68* 10^-12 F * ( 15)^2 = 4.14 * 10^-10 J

(c)

Q2 = C2V

= 3.68* 10^-12 F ( 15)

=55.31 * 10^-12 C

New capacitence is

Co = eo A/ d = 8.85 * 10^-12 * 10 * 10^-4 / 6.00 * 10^-3= 1.475 * 10^-12   C

the nergy stored in capaciotr is

U3 = Q2^2/ 2Co

= 0.5 ( 55.31 * 10^-12 C)^2/ 1.475 * 10^-12   C

=10.37 * 10^-10 J

(d)

wordone by external agent is

W = U3 - U2 = 10.37 * 10^-10 J-4.14 * 10^-10 J = 6.23 * 10^-10 J