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