Question

In a capacitor of capacitance $20\mu F$ the distance between the plates is $2mm$. If a dielectric slab of width $1mm$ and dielectric constant $2$ is inserted between the plates, then the new capacitance will be

• A. $22\mu F$
• B. $26.6\mu F$
• C. $52.2\mu \text{F}$
• D. $13\mu \text{F}$

Answer 1

Answer: (B)

The capacitance $C$ of a capacitor of area $A$ and distance between plates is $d$ then $C=\frac{{\epsilon }_{0}A}{d}$

When a dielectric slab of thickness $t$ is placed between the plates, we have
$C^{{}^{\prime }}=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$
Given, $C=20\mu F=20\times 10^{-6}F$
$d=2mm=2\times 10^{-3}n;t=1mm$
$=1\times 10^{-3}m,K=2$
$\therefore \frac{C^{{}^{\prime }}}{C}=\frac{d}{d-t\left(1-\frac{1}{K}\right)}$
$=\frac{2\times 10^{-3}}{2\times 10^{-3}-1\times 10^{-3}\left(1-\frac{1}{2}\right)}=1.33$
$\Rightarrow C^{{}^{\prime }}=1.33\times 20\times 10^{-6}=26.6\mu F$