Question

Between the plates of parallel plate capacitor of capacitance C, two parallel plates, of the same material and area same as per the plate of the original capacitor, are placed. If the thickness of these plates is equal to ${\frac{1}{5}}^{th}$ of the distance between the plates of the original capacitor, then the capacitance of the new capacitor is

• A. $\frac{1}{5}$ C
  
• B. $\frac{3}{5}$ C
  
• C. $\frac{3}{10}$ C
  
• D. $\frac{10}{3}$ C

Answer 1

The correct option is A

$\frac{1}{5}$ C

Refer the following fig. Let d be the distance between the plates of the original capacitor. If two plates are introduced between these plates, we have a series arrangement of three capacitors, each having a plate separation of $d^{\prime }=\frac{d}{5}$

Given $C=\frac{{\epsilon }_{0}A}{d}$
The capacitance of each of the three capacitors is
$C^{\prime }=\frac{{\epsilon }_{0}A}{d/5}=\frac{5{\epsilon }_{0}A}{d}=5C$
The equivalent capacitance of the series combination is
$\frac{1}{C^{\prime \prime }}=\frac{1}{C^{\prime }}+\frac{1}{C^{\prime }}+\frac{1}{C^{\prime }}=\frac{3}{C^{\prime }}or{C}^{\prime \prime }=\frac{C}{3}=\frac{5C}{3},$ which is choice (a).