A) \[\frac{{{\varepsilon }_{0}}A}{2d}\]
B) \[\frac{{{\varepsilon }_{0}}A}{d}\]
C) \[\frac{3{{\varepsilon }_{0}}A}{d}\]
D) \[\frac{4{{\varepsilon }_{0}}A}{3d}\]
Correct Answer: D
Solution :
The net capacitance\[C=\frac{{{\varepsilon }_{0}}A}{\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}}}\] \[=\frac{{{\varepsilon }_{0}}A}{\frac{d}{2}\left( \frac{1}{1}+\frac{1}{2} \right)}=\frac{4{{\varepsilon }_{0}}A}{3d}\]
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