Question

Two plates (area = $S$) charged to $ +q_{1}$ and $ +q_{2}(q_{2} < q_{1})$ brought closer to form a capacitor of capacitance $C$. The potential difference across the plates is:

• A. $ \frac{q_{1}-q_{2}}{2C} $
• B. $ \frac{q_{1}-q_{2}}{C} $
• C. $ \frac{q_{1}-q_{2}}{4C} $
• D. $ \frac{2(q_{1}-q_{2})}{C}$

Answer 1

Answer: (A)

Capacitance of the parallel plate capacitor is
$C=\frac{\varepsilon_{0} A}{d}$
where $A$ is area of plates and $d$ is distance between them.
Potential difference across the plates $V=E d$
Here, $E=\frac{\left(q_{1}-q_{2}\right)}{2 \varepsilon_{0} A}$
$\therefore V=\frac{\left(q_{1}-q_{2}\right) d}{2 \varepsilon_{0} A}$
$=\frac{q_{1}-q_{2}}{2 C} .$