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.You need to login to perform this action.
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