Question

How many electrons should be removed from an initially uncharged spherical conductor of radius 0.300 m to product a potential of 7.50 KV at the surface?

Answer 1

The potential at the surface of a conductor is given by the following equation:

V=Kq/r. Where K = 9x10⁹ Nm²/C²

For V= 7.5kV and r= 0.3m, we can solve for q:

q = Vr/K

q = 7500 x 0.3 / 9x10⁹ = 2.5x10^-7 C

This total charge has Ne electrons of 1.6x10^-19 C each, so the number of electrons in 2.5x10-7 Coulombs is given by the following equation:

Ne=Q/e So solving for Ne:

Ne = 2.5x10^-7 / 1.6x10^-19 = 1.5625x10¹² electrons