A 0.120 pF parallel-plate capacitor is charged to a potential difference of 10.0 V and then disconnected from the battery.A cosmic ray burst creates 1.00 * 106 electrons and 1.00 * 106 positive charges between the plates.If the charges do not recombine, but reach the oppositely charged plates, by how much is the potential difference between the capacitor plates reduced?
A)1.33 V
B)7.34 V
C)8.67 V
D)1330 V
E)8670 V

Question

Quizplus · Accepted Answer

The charge added by the cosmic ray burst is 1.6 * 10^-13 C (since 1 electron charge is 1.6 * 10^-19 C). The initial charge on the capacitor is 1.2 * 10^-12 C (C = Q/V, Q = 0.120 pF * 10 V). The new charge is the initial charge minus the charge added by the cosmic rays. The new potential difference is calculated using V = Q/C, resulting in a reduction of 1.33 V.

A 0120 PF Parallel-Plate Capacitor Is Charged to a Potential Difference