question_answer

In a capacitor or capacitance \[20\,\mu F,\] the distance between the plates is 2 mm. If a dielectric of width 1 mm and dielectric constant 2 is inserted between the plates, then new capacitance will be :

A)  \[32\,\mu F\]         

B)  \[26.7\,\mu F\]

C)  \[13.3\,\mu F\]       

D)  \[3\,\mu F\]

Correct Answer: B

Solution :

The capacitance of a parallel plate capacitor having charge q, plate area A, and separation d is \[C=\frac{A\,{{\varepsilon }_{0}}}{d}\] ... (i) When dielectric slab is placed \[C=\frac{A{{\varepsilon }_{0}}}{d-t\,\left( 1-\frac{1}{K} \right)}\] ?. (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{C}{20\times {{10}^{-6}}}=\frac{2\times {{10}^{-3}}}{2\times {{10}^{-3}}-1\times {{10}^{-3}}\left( 1-\frac{1}{2} \right)}\] \[=\frac{2\times {{10}^{-3}}}{2\times {{10}^{-3}}-(0.5\times {{10}^{-3}})}\] \[\Rightarrow \] \[C=1.333\times 20\times {{10}^{-6}}\mu F\] \[C=26.7\,\mu F\]. Note: Effective capacitance increases on inserting a dielectric slab between the plates.