Four plates of equal area A are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity is

Question

Four plates of equal area A are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity is (A) (2 ε0 A/d) (B) (3 ε0 A/d) (C) (3 ε0 A/2d) (D) (ε0 A/d)

Tardigrade Admin · Accepted Answer

The given circuit is equivalent to a parallel combination two identical capacitors. Hence, equivalent capacitance between A and B is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/11de444bc090c7da156c86a1471ad778-.png" /> C =(ε0 A/d)+(ε0 A/d) =(2 ε0 A/d)

Q. Four plates of equal area $A$ are separated by equal distances $d$ and are arranged as shown in the figure. The equivalent capacity is

$\frac{2 ε _{0} A}{d}$

$\frac{3 ε _{0} A}{d}$

$\frac{3 ε _{0} A}{2 d}$

$\frac{ε _{0} A}{d}$

The given circuit is equivalent to a parallel combination two identical capacitors.
Hence, equivalent capacitance between $A$ and $B$ is

$C = \frac{ε _{0} A}{d} + \frac{ε _{0} A}{d}$
$= \frac{2 ε _{0} A}{d}$

Q. Four plates of equal area A are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity is

d2ε0​A​

d3ε0​A​

2d3ε0​A​

dε0​A​