In the arrangement shown all the plates have equal area $A$ and spacing $d$ between them. The equivalent…

In the arrangement shown all the plates have equal area $A$ and spacing $d$ between them. The equivalent capacitance between point $P$ and $Q$ will be,

1. $\frac{\varepsilon_{0} A}{d}$
2. $\frac{\varepsilon_{0} A}{3 d}$
3. $\frac{\varepsilon_{0} A}{2 d}$
4. $\frac{2 \varepsilon_{0} A}{d}$

Solution

$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{2 C}+\frac{1}{C}+\frac{1}{2 C}=\frac{4}{2 C}$
$C_{\mathrm{eq}}=\frac{C}{2}=\frac{\varepsilon_{0} A}{2 d}$ ^

Asked in: JEE Mains - Capacitance - Test 1

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