Question

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

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

Answer 1

The correct option is B $26.6\mu \text{F}$

Given:

Capacitance without dielectric, $C=\frac{{\epsilon }_{0}A}{d}=20\mu \text{F}$

Distance between plates, $d=2\text{mm}$

Thickness of dielectric, $t=1\text{mm}=\frac{d}{2}$

Hence, new capacitance is given by;

$C^{\prime }=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}=\frac{{\epsilon }_{0}A}{d-\frac{d}{2}+\frac{d}{2\times 2}}$

$\Rightarrow C^{\prime }=\frac{4{\epsilon }_{0}A}{3d}$

$\text{Or},C^{\prime }=\frac{4}{3}\times 20=26.6\mu \text{F}$

Why this question ? Key Concept: Capacitance of a capacitor with space filled partially by a dielectric slab is given by $C=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$