$\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}$ are conduct- ing plates each of…

$\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}$ are conduct- ing plates each of area $\mathrm{A}$ and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch $\mathrm{S}$ is closed is

1. $\frac{3 \varepsilon_{0} \mathrm{~A}}{2 \mathrm{~d}} \mathrm{~V}^{2}$
2. $\frac{5 \varepsilon_{0} \mathrm{~A}}{12 \mathrm{~d}} \mathrm{~V}^{2}$
3. $\frac{\varepsilon_{0} \mathrm{~A}}{2 \mathrm{~d}} \mathrm{~V}^{2}$
4. $\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}} \mathrm{~V}^{2}$

Solution

$\mathrm{C}_{\mathrm{eff}}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ since effective capacitance between plates $A$ and $E$ is zero.
$\therefore \quad \mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\frac{\varepsilon_{0} \mathrm{~A}}{2 \mathrm{~d}} \mathrm{~V}^{2}$
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Asked in: JEE Mains - Capacitance - Test 1

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