A) \[\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}+{{K}_{2}}}{2} \right)\]
B) \[\frac{2{{\varepsilon }_{0}}A}{d}\left( {{K}_{1}}+{{K}_{2}} \right)\]
C) \[\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}} \right)\]
D) \[\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right)\]
Correct Answer: D
Solution :
In the arrangement shown, two capacitors each of area A and separation d/2 are in series. \[\therefore \] \[{{C}_{1}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A}{d/2}=\frac{2{{K}_{1}}{{\varepsilon }_{0}}A}{d}\] \[{{C}_{2}}=\frac{{{K}_{2}}{{\varepsilon }_{0}}A}{d/2}=\frac{2{{K}_{2}}{{\varepsilon }_{0}}A}{d}\] \[\therefore \]Effective capacitance C is given by \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\] \[=\frac{d}{2{{K}_{1}}{{\varepsilon }_{0}}A}+\frac{d}{2{{K}_{2}}{{\varepsilon }_{0}}A}\] \[=\frac{d}{2{{\varepsilon }_{0}}A}\left( \frac{1}{{{K}_{1}}}+\frac{1}{{{K}_{2}}} \right)\] \[\therefore \] \[C=\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right)\]
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