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 plate 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) On bringing the changed metal plates closer, electric field $\overrightarrow{E}$ in the intervening space is
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$E = \frac{\sigma_{1}}{2\varepsilon_{0}} - \frac{\sigma_{2}}{2\varepsilon_{0}} = \frac{\sigma_{1}}{2{A\varepsilon}_{0}} - \frac{\sigma_{2}}{2{A\varepsilon}_{0}}$
Or $\ \ E = \frac{Q_{1} - Q_{2}}{2A\varepsilon_{0}} = \frac{V}{d}\ or\ V = \frac{\left( Q_{1} - Q_{2} \right)d}{2A\varepsilon_{0}}$
$\Rightarrow \ \ \ \ \ V = \frac{Q_{1} - Q_{2}}{2C}$ $\left( \therefore C = \frac{\varepsilon_{0}A}{d} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.