Question

A charge of +2·0×10-8 C is placed on the positive plate and a charge of -1·0×10-8 C on the negative plate of a parallel-plate capacitor of capacitance 1·2×10-3 μF. Calculate the potential difference developed between the plates.

Answer 1

Explanation:

Given:

The charge on the positive plate is q1 and that on the negative plate is q2.

q1=2.0×10-8 Cq2=-1.0×10-8 C

Now,

Net charge on the capacitor =  q1-q22=1.5×10-8 C

The  potential difference developed between the plates is given by

q=VC⇒V=1.5×10-81.2×10-9=12.5 V