Question

there is an electric potential of 120V at a point that is 0.25m away from a charge. find the magnitude and the sign of charge

Answer 1

By the definition of the electric potential, we have:

$$V=\dfrac{kQ}{r},$$ $$Q=\dfrac{Vr}{k}=\dfrac{120\ V\cdot0.25\ m}{9\cdot10^9\ \dfrac{Nm^2}{C^2}}=3.3\cdot10^{-9}\ C=3.3\ nC.$$