Question

2 Calculate the following quantities in the circuit below, assuming the switch is open. С 1-6 F Switch a) Equivalent capacitance Coq. b) Charge stored in the capacitor C2 c) Potential difference in the capacitor Ci d) Potential energy stored in the capacitor C Now the switch is closed. Find the following quantities. e) Equivalent capacitance Ceq f) Charge stored in the capacitor C2. g) Potential difference in the capacitor C2 h) Potential energy stored in the capacitor C2

Answer 1

Given circuit switch open (a) Equivalent capacitance C_eq. c_1, c_2 are in series. C_eq = c_1 times c_2/c_1 + c_2 = 6 times 6/6 + 6 = 36/12 = 3 mu F (b) charge stored in the capacitor c_2. Q = CV (because circuit is in series) Q_net = c_eq delta v Q_net = (3 mu F)(12 V) Q_net = 36 mu c delta v = 12/2 = 6 V (c) Potential difference in the capacitor c_2 delta v_c_2 = Q/C = 36 mu c/6 mu F = 6V. (d) Potential energy stored in the capacitor c_2. delta v_c_2 = Q^2/2C = 36 times 36/2 times 6 = 108 mu J.