Question

When a dielectric slab of thickness $4cm$ is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by $3cm$ to restore the capacity to it's original value. The dielectric constant of the slab is

• A. $\frac{1}{4}$
• B. $4$
  
• C. $3$
  
• D. $1$
  

Answer 1

The correct option is B $4$

Distance is increased.
We know $C=\frac{{ϵ}_{0}A}{d}$
When dielectric is added
$C_1=\frac{{ϵ}_{0}A}{d-t(1-\frac{1}{k})}$
$\to$ given distance has to be increased by 3 cm to restore original value.
$\Rightarrow \frac{{ϵ}_{0}A}{d+3-4(1-\frac{1}{k})}=\frac{{ϵ}_{0}A}{d}$
$\Rightarrow \frac{3}{4}=1-\frac{1}{k}$
$\Rightarrow K=4$