Four plates of the same area of cross-section A are joined as shown in the figure. The distance between each plate is d. The equivalent capacity across A and B will be

Question

Four plates of the same area of cross-section A are joined as shown in the figure. The distance between each plate is d. The equivalent capacity across A and B will be (A) (2 ε0 A/d) (B) (3 ε0 A/d) (C) (3 ε0 A/2 d) (D) (ε0 A/d)

Tardigrade Admin · Accepted Answer

The given arrangement is equivalent to the parallel combination of three identical capacitors. Hence equivalent capacitance =3 C=3 (ε0 A/d)

Q. Four plates of the same area of cross-section $A$ are joined as shown in the figure. The distance between each plate is d. The equivalent capacity across $A$ and $B$ will be

$\frac{2 ε _{0} A}{d}$

$\frac{3 ε _{0} A}{d}$

$\frac{3 ε _{0} A}{2 d}$

$\frac{ε _{0} A}{d}$

The given arrangement is equivalent to the parallel combination of three identical capacitors. Hence equivalent capacitance $= 3 C = 3 \frac{ε _{0} A}{d}$

Q. Four plates of the same area of cross-section A are joined as shown in the figure. The distance between each plate is d. The equivalent capacity across A and B will be

d2ε0​A​

d3ε0​A​

2d3ε0​A​

dε0​A​