Question

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V.

Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
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.

Answer 1

U_eq = 1.99 * 10^(-10) J

Step-by-step explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12 C^2 / N .m

Equations used:

U = 0.5 C*V^2 .... Eq 1

C = e_o * k*A /d .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d

U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J