Question

The introduction of a metal plate between the plates of a parallel plate capacitor increases its capacitance by 4.5 times. If d is the separation of the two plates of the capacitor, the thickness of the metal plate introduced is

• A. $\frac{d}{3}$
  
• B. $\frac{5d}{9}$
  
• C. $\frac{7d}{9}$
  
• D. d

Answer 1

The correct option is C

$\frac{7d}{9}$

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

When a metal plate of thickness t is introduced in between the capacitor plates there is no electric field inside the metal plates and the electric field due to the metal paltes outside it is also zero. So the electric field is only in the gap between capacitor plates and metal plate (d - t); and is only due to the capacitor plates .
So, the capacitance becomes $C^{\prime }=\frac{{\epsilon }_{0}A}{(d-t)}.$

Given C'=4.5 C:
$\Rightarrow$ The value of t is $\frac{7}{9}d$