Question

Two identical metal plates are given positive charges $Q_{1}$ and $Q_{2}\left(\right. < Q_{1}\left.\right)$ respectively. If they are now brought close together to form a parallel plates capacitor with capacitance $C$ , the potential difference between them is

• A. $\frac{Q_{1} + Q_{2}}{2 C}$
• B. $\frac{Q_{1} + Q_{2}}{C}$
• C. $\frac{Q_{1} - Q_{2}}{C}$
• D. $\frac{Q_{1} - Q_{2}}{2 C}$

Answer 1

Answer: (D)

For charge $Q_{1}$ , electric field is
$E_{1}=\frac{Q_{1}}{2 \epsilon _{0} A}$
where $A$ is area.
For charge $Q_{2}$ electric field is
$E_{2}=\frac{Q_{2}}{2 \epsilon _{0} A}$
Resultant electric field
$E=E_{1}-E_{2}=\frac{\left(\right. Q_{1} - Q_{2} \left.\right)}{2 \left(\epsilon \right)_{0} A}$
Potential difference between plates when they are brought close together to form a parallel plate capacitor is
$V=Ed=\frac{\left(\right. Q_{1} - Q_{2} \left.\right)}{2 \left(\epsilon \right)_{0} A}d$
Since, $C=\frac{\epsilon _{0} A}{d}$ for parallel plate capacitor.
$V=\frac{\left(\right. Q_{1} - Q_{2} \left.\right)}{2 C}$