Question

In a capacitor of capacitance $20 \mu F$ the distance between the plates is $2\, mm$. If a dielectric slab of width $1 \,mm$ 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\text{ }\mu \text{F} $
• D. $ ~13\text{ }\mu \text{F} $

Answer 1

Answer: (B)

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

When a dielectric slab of thickness $t$ is placed between the plates, we have
$C^{'}=\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}} $
Given, $C=20\, \mu F =20 \times 10^{-6}\, F$
$d=2\, mm =2 \times 10^{-3}\, n ; t=1 \,mm$
$=1 \times 10^{-3}\, m , K=2$
$\therefore \frac{C^{'}}{C}=\frac{d}{d-t\left(1-\frac{1}{K}\right)}$
$=\frac{2 \times 10^{-3}}{2 \times 10^{-3}-1 \times 10^{-3}\left(1-\frac{1}{2}\right)}=1.33$
$\Rightarrow C^{'}=1.33 \times 20 \times 10^{-6}=26.6 \,\mu F$