Question

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

• A. $3{\epsilon }_{0}Ad$
• B. ${\epsilon }_{0}Ad$
• C. $2{\epsilon }_{0}Ald$
• D. $\frac{3}{2}{\epsilon }_{0}Ad$

Answer 1

Answer: (C)

The given circuit is equivalent to parallel combination of two identical capacitors Each having capacitance $C=\frac{{\epsilon }_{0}A}{d}$
Thus: $C_{eq}=2C=\frac{2{\epsilon }_{0}A}{d}$