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)1,330 V E)8,670 V

Question

Quizplus · Accepted Answer

The potential difference is reduced by 1.33 V. The charge added by the cosmic ray burst is 1.6 x 10^-13 C (since 1 electron charge is 1.6 x 10^-19 C and there are 10^6 electrons). The initial charge on the capacitor is 1.2 x 10^-12 C (C = Q/V, Q = 0.120 pF * 10 V). The new charge is the initial charge minus the charge from the cosmic rays, and the new potential difference is calculated using V = Q/C.

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