A) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{K+3}{4K} \right)\]
B) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{2K}{K+3} \right)\]
C) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{K}{K+3} \right)\]
D) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{4K}{K+3} \right)\]
Correct Answer: D
Solution :
(d) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{4K}{K+3} \right)\] Here, thickness of the slab, \[t=\frac{3}{4}\,d\] Capacitance, \[C=\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{K} \right)}\] \[=\frac{{{\varepsilon }_{0}}A}{d-\frac{3}{4}d\left( 1-\frac{1}{K} \right)}=\frac{{{\varepsilon }_{0}}A}{\frac{d}{4}+\frac{3}{4}\frac{d}{K}}=\frac{{{\varepsilon }_{0}}A}{\frac{d}{4}\left( 1+\frac{3}{K} \right)}\] \[=\frac{{{\varepsilon }_{0}}A}{d}\frac{4K}{\left( K+3 \right)}\]You need to login to perform this action.
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