Three plates of common surface area A are connected as shown in figure. The effective capacitance will be

Question

Three plates of common surface area A are connected as shown in figure. The effective capacitance will be (A) 3 ε0 A / d (B) ε0 A / d (C) 2 ε0 A / d (D) (3/2) ε0 A / d

Tardigrade Admin · Accepted Answer

The given circuit is equivalent to parallel combination of two identical capacitors. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a6d012f955070d9174661cc8f655f087-.png" /> Each having capacitance, C=(ε0 A/d) Thus, Ce q=2 C=(2 ε0 A/d)

Q. Three plates of common surface area $A$ are connected as shown in figure. The effective capacitance will be

$3 ε_{0} A / d$

$ε_{0} A / d$

$2 ε_{0} A / d$

$\frac{3}{2} ε_{0} A / d$

The given circuit is equivalent to parallel combination of two identical capacitors.

Each having capacitance, $C = \frac{ε _{0} A}{d}$
Thus, $C_{e q} = 2 C = \frac{2 ε _{0} A}{d}$

Q. Three plates of common surface area A are connected as shown in figure. The effective capacitance will be

3ε0​A/d

ε0​A/d

2ε0​A/d

23​ε0​A/d