Q. Two identical metal plates are given positive charges $Q_{1}$ and $Q_{2}( < Q_{1})$ respectively. If they are now brought close together to form a parallel plates capacitor with capacitance $C$, the potential difference between them is
A) $\frac{Q_{1} + Q_{2}}{2C}$
B) $\frac{Q_{1} + Q_{2}}{C}$
C) $\frac{Q_{1} - Q_{2}}{C}$
D) $\frac{Q_{1} - Q_{2}}{2C}$
Explanation
(d) For charge $Q_{1}$, electronic field is
$E_{1} = \frac{Q_{1}}{2\varepsilon_{0}A}$
where $A$ is area.
For charge $Q_{2}$ electronic field is
$E_{2} = \frac{Q_{2}}{2\varepsilon_{0}A}$
Resultant electronic field
$E = E_{1} - E_{2} = \frac{(Q_{1} - Q_{2})}{2\varepsilon_{0}A}$
Potential difference between plates when they are brought close together to form a parallel plate capacitor is
$V = Ed = \frac{(Q_{1} - Q_{2})}{2\varepsilon_{0}A}d$
Since, $C = \frac{\varepsilon_{0}A}{d}$ for parallel plate capacitor.
$\therefore\ V = \frac{(Q_{1} - Q_{2})}{2C}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.