Question

(a) Suppose you charge a 2.5 F capacitor with two 1.5 volt batteries. How much charge was on each plate? 7.5 Correct: Your answer is correct. C (b) How many excess electrons were on the negative plate? electrons

Answer 1

C = Q / V

Q = C * V = 2.5 F * 3 V = 7.5 Coulombs (assuming the batteries are in series)

N e = Q where N is the number of electrons and e the electronic charge

N = 7.5 / (1.60E-19) = 4.7E19 electrons

or 4.7 * 10^19 electrons