The capacity of the capacitor is
A) \[\frac{A{{\varepsilon }_{0}}({{K}_{1}}\times {{K}_{2}})}{d({{K}_{1}}+{{K}_{2}})}\]
B) \[\frac{A{{\varepsilon }_{0}}({{K}_{1}}-{{K}_{2}})}{d}\]
C) \[\frac{A{{\varepsilon }_{0}}{{K}_{1}}{{K}_{2}}}{({{K}_{1}}+{{K}_{2}})}\]
D) \[\frac{A{{\varepsilon }_{0}}({{K}_{1}}+{{K}_{2}})}{d}\]
Correct Answer: D
Solution :
In each capacitor the area of the plate will be\[\frac{A}{2}\]. In the given combination the arrangement is equivalent to two capacitors connected in parallel. Also in each capacitor the area of the plate will be. \[\therefore \,\,\,\,Equivalent\,\,capacitance\,\,C'={{C}_{1}}+{{C}_{2}}\] \[=\,\,\frac{{{K}_{1}}{{\varepsilon }_{0}}A/2}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}A/2}{d}=\frac{({{K}_{1}}+{{K}_{2}}){{\varepsilon }_{0}}A}{d}\]
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