If a 0.20 pF capacitor has a voltage of 40 mV,how many more electrons are on the negative plate than on the positive plate?

Question

Quizplus · Accepted Answer

The capacitance of a capacitor is given by the formula C = Q/V, where Q is the charge stored on the plates and V is the voltage across the plates. Rearranging this formula, we get Q = CV.

Substituting the given values, we have:

Q = (0.20 x 10^-12 F) x (40 x 10^-3 V)
Q = 8 x 10^-15 C

Since the capacitor is electrically neutral, the total charge on the positive plate must be equal in magnitude but opposite in sign to the total charge on the negative plate. Therefore, each plate has a charge of 4 x 10^-15 C.

The number of electrons on each plate can be found by dividing the total charge by the charge on a single electron (1.6 x 10^-19 C):

n = (4 x 10^-15 C) / (1.6 x 10^-19 C)
n = 2.5 x 10^4

Therefore, the negative plate has 2.5 x 10^4 more electrons than the positive plate.

The best choice is C.

If a 020 PF Capacitor Has a Voltage of 40 MV,how Many