Five identical plates of equal area A are placed parallel to each other and at equal distance d from each other as shown in the figure. The effective capacity of the system between the terminals A and B is

Question

Five identical plates of equal area A are placed parallel to each other and at equal distance d from each other as shown in the figure. The effective capacity of the system between the terminals A and B is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-0v94iso1k7z9.png" /> (A) (3/5)(ϵ 0 A/d) (B) (5/4)(ϵ 0 A/d) (C) (5/3)(ϵ 0 A/d) (D) (4/5)(ϵ 0 A/d)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-0v94iso1qj6m.png" /> Plates 1-2, 2-3, 3-4 and 4-5 form a total of four capacitors. The capacity of each capacitor is, C0 = (ϵ 0 A/d) From figure, it is clear that, Ce q = (5/3)C0 = (5/3)(ϵ 0 A/d)

Q. Five identical plates of equal area $A$ are placed parallel to each other and at equal distance $d$ from each other as shown in the figure. The effective capacity of the system between the terminals $A$ and $B$ is

$\frac{3}{5} \frac{ϵ _{0} A}{d}$

$\frac{5}{4} \frac{ϵ _{0} A}{d}$

$\frac{5}{3} \frac{ϵ _{0} A}{d}$

$\frac{4}{5} \frac{ϵ _{0} A}{d}$

Plates 1-2, 2-3, 3-4 and 4-5 form a total of four capacitors.
The capacity of each capacitor is, $C_{0} = \frac{ϵ _{0} A}{d}$
From figure, it is clear that, $C_{e q} = \frac{5}{3} C_{0} = \frac{5}{3} \frac{ϵ _{0} A}{d}$

Q. Five identical plates of equal area A are placed parallel to each other and at equal distance d from each other as shown in the figure. The effective capacity of the system between the terminals A and B is

53​dϵ0​A​

45​dϵ0​A​

35​dϵ0​A​

54​dϵ0​A​