Electrostatics 5 Question 30
32. 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) $\left(Q _1-Q _2\right) / 2 C$
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Solution:
- Electric field within the plates $\mathbf{E}=\mathbf{E} _{Q _1}+\mathbf{E} _{Q _2}$
$\therefore$ Potential difference between the plates
$$ V _A-V _B=E d=\frac{Q _1-Q _2}{2 A \varepsilon _0} \quad d=\frac{Q _1-Q _2}{2 \frac{A \varepsilon _0}{d}}=\frac{Q _1-Q _2}{2 C} $$
