A) \[\frac{2{{\varepsilon }_{0}}A}{d}\,\left[ \frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}} \right]\]
B) \[\frac{{{\varepsilon }_{0}}A}{d}\,\left[ \frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}} \right]\]
C) \[\frac{2{{\varepsilon }_{0}}A}{d}\,\left[ {{k}_{1}}+{{k}_{2}}+{{k}_{3}} \right]\]
D) None of these
Correct Answer: D
Solution :
\[{{C}_{1}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}\frac{A}{2}}{\left( \frac{d}{2} \right)}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A}{d}\] \[{{C}_{2}}=\frac{{{K}_{2}}{{\varepsilon }_{0}}\left( \frac{A}{2} \right)}{\left( \frac{d}{2} \right)}=\frac{{{K}_{2}}{{\varepsilon }_{0}}A}{d}\]and \[{{C}_{3}}=\frac{{{K}_{3}}{{\varepsilon }_{0}}A}{2d}=\frac{{{K}_{3}}{{\varepsilon }_{0}}A}{2d}\] Now, \[{{C}_{eq}}={{C}_{3}}+\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]\[=\left( \frac{{{K}_{3}}}{2}+\frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right).\frac{{{\varepsilon }_{0}}A}{d}\]
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