Question

A parallel plate capacitor is charged to a certain potential difference. A dielectric slab of thickness $4.8\text{mm}$ is inserted between the plates and it becomes necessary to increase the distance beteeen the plates by $4\text{mm}$ to maintain the same potential difference. The dielectric constant of the slab is :

• A. $2$
• B. $6$
• C. $4$
• D. $3$

Answer 1

The correct option is B $6$

Given, a dielectric slab of thickness $4.8\text{mm}$ is inserted between a parallel plate capicator and to maintain the potential difference the plates need to be separated by $4\text{mm}$.

As we know, electric field in the region between parallel plates of a capacitor is
$E=\frac{V}{d}\Rightarrow V=Ed=\frac{\sigma }{{\epsilon }_0}d.......\left(1\right)$

When dielectric is introduced between the plates, the electric field becomes,

$E=\frac{V}{(d^{\prime }-t)+\frac{t}{k}}\Rightarrow V=E(d^{\prime }-t+\frac{t}{k}).....\left(2\right)$

Where $d^{\prime }$ is the new distance between the plates and $t$ is thickness of the dielectric.

Given that potential must remains same,

From equation $\left(1\right)$ and equation $\left(2\right)$:

$\frac{\sigma }{{\epsilon }_0}d=\frac{\sigma }{{\epsilon }_0}(d^{\prime }-t+\frac{t}{k})d=d+4-4.8+\frac{4.8}{k}$
$0.8=\frac{4.8}{k}\Rightarrow k=6$