Five conducting parallel plates having area A and separation between them d , are placed

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1. Electric Charges and Fields

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Five conducting parallel plates having area $A$ and separation between them $d$, are placed as shown in the figure . Plate number $2$ and $4$ are connected wire and between point $A$ and $B$, a cell of emf $E$ is connected . The charge flown through the cell is :-

A

$\frac{3}{4}\frac{{{\varepsilon _0}AE}}{d}$

B

$\frac{2}{3}\frac{{{\varepsilon _0}AE}}{d}$

C

$\frac{{4{\varepsilon _0}AE}}{d}$

D

$\frac{{{\varepsilon _0}AE}}{2d}$

Solution

Net capacitance

$=\frac{\left(\frac{2 \varepsilon_{0} A}{d}\right)\left(\frac{\varepsilon_{0} A}{d}\right)}{\frac{3 \varepsilon_{0} A}{d}}$

$=\frac{2}{3} \frac{\varepsilon_{0} A}{d}$

Charge flown $=\left(\frac{2}{3} \frac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}}\right) \mathrm{E}$

$=\frac{2}{3} \frac{\varepsilon_{0} A}{d} E$

Standard 12

Physics

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