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. $52.2\mu F$
• D. $13\mu F$

Answer 1

The correct option is B $26.6\mu F$

Solutions So initially,

$C=\frac{{ϵ}_{0}A}{d}$

$\Rightarrow 2O\mu F=\frac{{ϵ}_{0}A}{2mm}$

After dielectric is inserted,

$C^{\prime }=\frac{{ϵ}_{0}A}{(d-t)+t/k}$

$t=$ thickness of slab

$K\to$ dielectric constant

So,

$c^{\prime }=\frac{{ϵ}_{0}A}{1mm+\frac{1mm}{2}}=\frac{2{ϵ}_{0}A}{3mm}$

$C^{\prime }=\frac{2\times 2}{3}\left(\frac{{ϵ}_{0}A}{2mm}\right)=\frac{4}{3}\times 20\mu F$

$C^{\prime }=26.6\mu F$

Option - B is correct.