Electrostatics Ques 118
- Two identical metal plates are given positive charges $Q_{1}$ and $Q_{2}\left(<Q_{1}\right)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$, the potential difference between them is
(1999, 2M)
(a) $\left(Q_{1}+Q_{2}\right) / 2 C$
(b) $\left(Q_{1}+Q_{2}\right) / C$
(c) $\left(Q_{1}-Q_{2}\right) / C$
(d) $\left(Q_{1}-Q_{2}\right) / 2 C$
Show Answer
Answer:
Correct Answer: 118.(d)
Solution:
Formula:
Distribution of Charges on Connecting Two Charged Capacitors:
- Electric field within the plates $\mathbf{E}=\mathbf{E}_ {Q_ {1}}+\mathbf{E}_ {Q_ {2}}$
$\therefore$ Potential difference between the plates
$$ V_{A}-V_{B}=E d=\frac{Q_{1}-Q_{2}}{2 A \varepsilon_{0}} \quad d=\frac{Q_{1}-Q_{2}}{2 \frac{A \varepsilon_{0}}{d}}=\frac{Q_{1}-Q_{2}}{2 C} $$
BETA
