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 \[\mu F\]

D)  13 \[\mu F\]

Correct Answer: B

Solution :

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,   \[C=20\,\mu F=20\times {{10}^{-6}}F\], \[d=2\,mm=2\times {{10}^{-3}}m,\,t=1\,mm\]                  \[=1\times {{10}^{-3}}m,\,\,K=2\] \[\therefore \] \[=\frac{2\times {{10}^{-3}}}{2\times {{10}^{-3}}-1\times {{10}^{-3}}\left( 1-\frac{1}{2} \right)}=1.33\] \[\Rightarrow \] \[C=1.33\times 20\times {{10}^{-6}}=26.6\,\mu F\]