A) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{9{{K}_{1}}{{K}_{2}}{{K}_{3}}}{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}\]
B) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{K}_{1}}{{K}_{2}}{{K}_{3}}}{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}\]
C) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}{{{K}_{1}}{{K}_{2}}{{K}_{3}}}\]
D) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}{9{{K}_{1}}{{K}_{2}}{{K}_{3}}}\]
Correct Answer: A
Solution :
\[\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\] \[\Rightarrow {{C}_{eq}}=\frac{3{{\varepsilon }_{0}}A{{K}_{1}}{{K}_{2}}{{K}_{3}}}{d({{K}_{1}}{{K}_{2}}+{{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}})}\] ........(1) \[C_{eq}^{'}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[=\frac{{{\varepsilon }_{0}}A}{3d}({{K}_{1}}+{{K}_{2}}+{{K}_{3}})\] ?.(2) Now,\[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{1}{2}{{C}_{eq}}.{{V}^{2}}}{\frac{1}{2}C_{eq}^{'}}=\frac{9{{K}_{1}}{{K}_{2}}{{K}_{3}}}{({{K}_{1}}+{{K}_{2}}+{{K}_{3}})({{K}_{1}}{{K}_{2}}+{{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}})}\] |
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