Question
A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates. Please add explanation!
Joshi sir comment
due to induction the charges on the four faces will be
so V = q/C = 1.5 * 10-8/1.2 * 10-9
solve now