A capacitor has a capacitance of 2.5*10^-8 F. The capacitor is charged by removing electrons from one plate and placing them on the other plate. the process continues until the potential difference between the plates is 450V. How many electrons have been transferred in order to accomplish this?

Question

anonymous · Answer

Q=CV.
Q= 2.5x10^-8 x 450
= 1125 x 10^-8
= 1.125 x 10^-5 C
1 electron = 1.6 x 10^-19 Coulombs
 so no. of electrons = (1.125x 10^-5)/(1.6x 10^-19)
= 7.03125 x 10^13 electrons