A) \[\frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}}{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}}\]
B) \[\frac{{{K}_{1}}{{K}_{2}}+{{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}}\]
C) \[\frac{2\,\,{{K}_{1}}\,{{K}_{2}}\,{{K}_{3}}}{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}}\]
D) None of these
Correct Answer: A
Solution :
Capacitance of capacitor is given by \[C=\frac{K\,\,{{\varepsilon }_{0}}\,\,A}{d}\] According to question, these capacitors are in parallel arrangement. \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,{{C}_{1}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[\frac{{{K}_{eq}}\,{{\varepsilon }_{0}}A}{d}=\frac{{{K}_{1}}\,{{\varepsilon }_{0}}{{A}_{1}}}{d}+\frac{{{K}_{2}}\,{{\varepsilon }_{0}}{{A}_{2}}}{d}+\frac{{{K}_{3}}\,{{\varepsilon }_{0}}{{A}_{3}}}{d}\] \[{{K}_{eq}}=\frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}}{A}\] \[=\,\,\,\frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}}{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}}\]You need to login to perform this action.
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