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 the plates 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

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

The inner plates 2 and 3 are connected together. Hence they act as a single conductor. Since the outer plates 1 and 4 are connected together, there are effectively two capacitors (between plates 1 and 2 and plates 3 and 4) in parallel, each of capacitance $C={\epsilon }_{0}A/d$ as shown in figure. Thus the equivalent capacitance is
$C^{\prime }=2C=\frac{2{\epsilon }_{0}A}{d},whichischoice\left(c\right).$