Question

An isolated solid sphere of aluminum having radius 7.0 cm is at a potential of 500 V. Calculate the number of electrons which have been removed from the sphere to raise it to this potential. 

Answer 1

The potential and charge are

$$V=\frac{kq}{r},\\\space\\ q=\frac{Vr}{k}.$$

The number of electrons:

$$N=\frac qe=\frac{Vr}{ke}=2.43·10^{10}.$$