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. $\frac{2{\epsilon }_{o}A}{d}$
• B. $\frac{3{\epsilon }_{o}A}{d}$
• C. $\frac{{\epsilon }_{o}A}{2d}$
• D. $\frac{{\epsilon }_{o}A}{d}$

Answer 1

The correct option is A $\frac{2{\epsilon }_{o}A}{d}$

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

C = $\frac{{\epsilon }_{o}A}{d}$+$\frac{{\epsilon }_{o}A}{d}$= 2$\frac{{\epsilon }_{o}A}{d}$