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?

Question

Quizplus · Accepted Answer

The charge on the capacitor before the burst is Q = CV = (0.120 pF)(10.0 V) = 1.20 × 10^-12 C. After the burst, the total charge on the capacitor is Q' = Q + 1.00 × 10^6e, where e is the magnitude of the charge on an electron. The electric field between the plates is then E = Q'/C = (Q + 1.00 × 10^6e)/(0.120 pF) = 8.33 × 10^7 V/m. The potential difference between the plates is then V' = Ed = (8.33 × 10^7 V/m)(d), where d is the distance between the plates. Since the charges do not recombine and reach the opposite plates, the potential difference is reduced by the work done by the charges moving through the electric field, which is W = QV' = (1.00 × 10^6)(-1.60 × 10^-19 C)(8.33 × 10^7 V/m)(0.001 m) = -1.33 V. Therefore, the new potential difference is 10.0 V - 1.33 V = 8.67 V.

A 0120-PF Parallel-Plate Capacitor Is Charged to a Potential Difference of of 10.0