Question

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. $\frac{Q_1-Q_2}{2C}$

Answer 1

Answer: (D)

For charge $Q_1$ , electric field is
$E_1=\frac{Q_1}{2{ϵ}_{0}A}$
where $A$ is area.
For charge $Q_2$ electric field is
$E_2=\frac{Q_2}{2{ϵ}_{0}A}$
Resultant electric field
$E=E_1-E_2=\frac{(Q_1-Q_2)}{2{\left(ϵ\right)}_{0}A}$
Potential difference between plates when they are brought close together to form a parallel plate capacitor is
$V=Ed=\frac{(Q_1-Q_2)}{2{\left(ϵ\right)}_{0}A}d$
Since, $C=\frac{{ϵ}_{0}A}{d}$ for parallel plate capacitor.
$V=\frac{(Q_1-Q_2)}{2C}$