Question

Between the plates of a parallel plate capacitor of capacitance $C$, two parallel plates of the same metal and area are placed. If the thickness of each plate is equal to $\frac{1}{5}$th of the distance between the parallel plates, the new capacitance is

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

Answer 1

The correct option is A $\frac{5}{3}C$


Let the plate area be $A$ and separation be $d$. So, the initial capacitance will be,

$C=\frac{{\epsilon }_{0}A}{d}$

Given,
the combined thickness of the metal plates

$t=\frac{d}{5}+\frac{d}{5}=\frac{2d}{5}$

Since, the metal plates are the good conductors,the dielectric constant of metal plates will be $K=\infty$.

$C^{\prime }=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$

$\Rightarrow C^{\prime }=\frac{{\epsilon }_{0}A}{d-\frac{2d}{5}+\frac{2d}{5\times \infty }}$

$\Rightarrow C^{\prime }=\frac{5{\epsilon }_{0}A}{3d}=\frac{5}{3}C$

Hence, option (a) is the correct answer.

Note: Capacitance increases even on inserting conducting plates. Key concept: Capacitance with conductor as dielectric medium.