Question

Four metal plates numbered 1, 2, 3 and 4 are arranged as shown in Fig. The area of each plate is A and the separation between adjacent paltes is d. The capacitance of the arrangement is

• A. $\frac{{\epsilon }_{0}A}{d}$
  
• B. $\frac{2{\epsilon }_{(}0)A}{d}$
  
• C. $\frac{3{\epsilon }_{0}A}{d}$
  
• D. $\frac{4{\epsilon }_{0}A}{d}$

Answer 1

The correct option is C

$\frac{3{\epsilon }_{0}A}{d}$

Plate 1 is connected to plate 3 and plate 2 is connected to plate 4. Thus, there are three capacitors in parallel, each of capacitance
$C=\frac{{\epsilon }_{0}A}{d}$
as shown in following fig. Hence, the equivalent capacitance is
$C^{\prime }=3C=\frac{3{\epsilon }_{0}A}{d},whichischoice\left(c\right)$