question_answer

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)         \[~13\text{ }\mu \text{F}\]

Correct Answer: B

Solution :

The capacitance C of a capacitor of area A and distance between plates is \[C=\frac{{{\varepsilon }_{0}}A}{d}\] 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, \[C=20\,\mu F=20\times {{10}^{-6}}F,\] \[d=2\,mm\,=2\times {{10}^{-3}}m,\,t=1\,mm\]                 \[=1\times {{10}^{-3}}m,\,K=2\] \[\therefore \]  \[\frac{C}{C}=\frac{d}{d-t\left( 1-\frac{1}{K} \right)}\] \[=\frac{2\times {{10}^{-3}}}{2\times {{10}^{-3}}-1\times {{10}^{-3}}\left( 1-\frac{1}{2} \right)}=1.33\] \[\Rightarrow \]\[C=1.33\times 20\times {{10}^{-6}}=26.6\,\mu F\]