MCQ
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
    $\frac{3}{4}\frac{{{\varepsilon _0}AE}}{d}$
  • $\frac{2}{3}\frac{{{\varepsilon _0}AE}}{d}$
  • C
    $\frac{{4{\varepsilon _0}AE}}{d}$
  • D
    $\frac{{{\varepsilon _0}AE}}{{2d}}$

Answer

Correct option: B.
$\frac{2}{3}\frac{{{\varepsilon _0}AE}}{d}$
b
Net capacitance

$=\frac{\left(\frac{2 \in 0 A}{d}\right)\left(\frac{\in 0 A}{d}\right)}{\frac{2 \in_{o} A}{d}}$

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

Charge flown $=C E$

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

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