A) \[\frac{{{\varepsilon }_{0}}({{k}_{1}}+{{k}_{2}})}{2d}\]
B) \[\frac{{{\varepsilon }_{0}}A}{2d}\left( \frac{({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}} \right)\]
C) \[\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{({{k}_{1}}{{k}_{2}})}{{{k}_{1}}+{{k}_{2}}} \right)\]
D) \[\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{({{k}_{1}}{{k}_{2}})}{{{k}_{1}}+{{k}_{2}}} \right)\]
Correct Answer: A
Solution :
Key Idea: The arrangement is equivalent to two capacitors connected in parallel. The first plate of the capacitor is connected to one point and the second plate to the other. Hence, the arrangement is equivalent to two capacitors connected in parallel. In each capacitor the area of each plate is \[\frac{A}{2}.\] \[\therefore \] Equivalent capacitance is \[C={{C}_{1}}+{{C}_{2}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A/2}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}A/2}{d}\] \[=\frac{({{K}_{1}}+{{K}_{2}})}{2}\frac{{{\varepsilon }_{0}}A}{d}\] Note: Capacitors connected in parallel have same potential difference between their plates, but the charge is distributed in proportion to their capacitances.
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