Question

Q.32 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 (1) \( \frac{3}{4} \frac{\varepsilon_{0} AE }{ d } \) (2) \( \frac{2}{3} \frac{\varepsilon_{0} AE }{ d } \) (3) \( \frac{4 \varepsilon_{0} A E}{d} \) (4) \( \frac{\varepsilon_{0} AE }{2 d } \)

Answer 1

Correct option is (2) \(\frac 23 \frac{\epsilon_0A}d E\)

Net capacitance = \(\cfrac{\left(\frac{2\,\epsilon_0A}2{d}\right)\left(\frac{\epsilon_0A}d\right)}{\frac{2\,\epsilon_0A}d}\) 

\(= \frac 23 \frac{\epsilon_0A}d\)

Charge flow q = CV

\(q = \frac 23 \frac{\epsilon_0A}d E\)   ∵ V = E