A 0.160 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 x 106 electrons and 1.00 x 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

1.0 volts

Explanation:

Given that:

The potential difference between the plate V = 10 V

C = 0.160 pF = 0.160 × 10⁻¹² F

The charge on the capacitor

Q = CV

Q = 0.160 × 10⁻¹² × 10

Q = 1.6 × 10⁻¹² C

It implies that the positive plate of capacitor has +1.6 × 10⁻¹² C charge while the negative plate has -1.6 × 10⁻¹² C

The number of excess electrons on the negative plate is:

Thus, electron deficiency on the positive plate is

The net negative charge that moves towards the positive plate is :

q = number of electrons moved × e

Now, the net charge on the positive plate is:

The potential difference between the plate;

The reduction in potential difference

continuous conducting path for current, power supply

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