Question

A parallel plate capacitor consists of two circular metal plates of radius, R, separated by d -150 um of air. The capacitor is connected to a Vo- 10 V potential source, as shown in the figure below. This results in 167 nC of charge being stored on each plate. If the radius of the plate is doubled while leaving the capacitor connected to the same potential source, what is the new amount of charge stored on each plate?

Answer 1

Plate area (A_1) = pi R^2_1 theta = C_1 v_0 C-1 = epsilon_0 A/d So delta_1 = epsilon_0 A/d v-0 Capacitor shill connected to battery so voltage is constant theta, alpha A [d = same for both] theta_1/theta_2 = A_1/A_2 [Area = pi R^2] theta_1/theta_2 = pi r^2_1/pi r^2_2 [r_2 = 2 r_1 Radius is Doubled] 167 nC/theta_2 = r_1^2/(2r_1)^2 = r^2_1/4 r^2_1 theta_2 = (4) (167 nC) theta_2 = 668 nC