Question

A charge of 1 $\mu C$ is given to one plate of a parallel-plate capacitor of capacitance 0.1 $\mu F$ and a charge of 2 $\mu C$ given to the other plate, Find the potential difference developed between the plates

Answer 1

The charge on the first plate of capacitor is $q_1=1\mu$

The capacitance of te capacitor is $C=0.1\mu F=1\times {10}^{-7}F$

The chare on the second plate of the capacitor is $q_2=2\mu C=2\times {10}^{-6}C$

The net charge on the plates is given as:

$q=\frac{q_1-q_2}{2}$

$=\frac{(1-2)\times {10}^{-6}}{2}=-0.5\times {10}^{-6}C$

The Potential of the capacitor can be obtained as:

${}^{\prime }{V}^{\prime }=\frac{q}{C}=\frac{1\times {10}^{-7}}{-5\times {10}^{-7}}=-5V$

But potential can never be negative. So, $V=5V$