A) \[\frac{{{\varepsilon }_{0}}A}{d+t\left( 1-\frac{1}{k} \right)}\]
B) \[\frac{{{\varepsilon }_{0}}A}{d+t\left( 1+\frac{1}{k} \right)}\]
C) \[\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{k} \right)}\]
D) \[\frac{{{\varepsilon }_{0}}A}{d-t\left( 1+\frac{1}{k} \right)}\]
Correct Answer: C
Solution :
Potential difference between the plates \[V={{V}_{air}}+{{V}_{medium}}\] \[=\frac{\sigma }{{{\varepsilon }_{0}}}\times (d-t)+\frac{\sigma }{K{{\varepsilon }_{0}}}\times t\] Þ \[V=\frac{\sigma }{{{\varepsilon }_{0}}}(d-t+\frac{t}{K})\] \[=\frac{Q}{A{{\varepsilon }_{0}}}(d-t+\frac{t}{K})\] Hence capacitance \[C=\frac{Q}{V}=\frac{Q}{\frac{Q}{A{{\varepsilon }_{0}}}(d-t+\frac{t}{K})}\] \[=\frac{{{\varepsilon }_{0}}A}{(d-t+\frac{t}{k})}=\frac{{{\varepsilon }_{0}}A}{d-t\,\left( 1-\frac{1}{k} \right)}\]You need to login to perform this action.
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