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(B) The arrangement consists of three parallel plate capacitors connected in parallel.Let the area of each plate be $A = l^2$.The capacitance of a par

Vedclass · Accepted Answer

$\frac{5 \varepsilon_0 l^2}{3 d}$

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(B) The arrangement consists of three parallel plate capacitors connected in parallel.Let the area of each plate be $A = l^2$.The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.From the figure,we identify three capacitors:$C_1$ formed by plates $1$ and $2$ with separation $3d$: $C_1 = \frac{\varepsilon_0 l^2}{3d}$.$C_2$ formed by plates $3$ and $4$ with separation $d$: $C_2 = \frac{\varepsilon_0 l^2}{d}$.$C_3$ formed by plates $5$ and $6$ with separation $3d$: $C_3 = \frac{\varepsilon_0 l^2}{3d}$.Since these capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is:$C_{eq} = C_1 + C_2 + C_3$$C_{eq} = \frac{\varepsilon_0 l^2}{3d} + \frac{\varepsilon_0 l^2}{d} + \frac{\varepsilon_0 l^2}{3d}$$C_{eq} = \frac{\varepsilon_0 l^2}{d} (\frac{1}{3} + 1 + \frac{1}{3}) = \frac{\varepsilon_0 l^2}{d} (\frac{1+3+1}{3}) = \frac{5 \varepsilon_0 l^2}{3d}$.Thus,the correct option is $(B)$.

Six identical square metallic plates are arranged as shown in the figure. The length of each plate is $l$. The equivalent capacitance of this arrangement would be . . . . . . .

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