question_answer

A parallel plate capacitor of capacitance C, area A and plate separation d is filled with three different dielectric materials having dielectric constants \[{{K}_{1}},\,\,{{K}_{2}}\,\,and\,\,{{K}_{3}}\] as shown in figure. If a single dielectric material is used to have same capacitance C in this capacitor, then its equivalent dielectric constant \[{{K}_{eq}}\] is given by

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. \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,{{C}_{1}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[\frac{{{K}_{eq}}\,{{\varepsilon }_{0}}A}{d}=\frac{{{K}_{1}}\,{{\varepsilon }_{0}}{{A}_{1}}}{d}+\frac{{{K}_{2}}\,{{\varepsilon }_{0}}{{A}_{2}}}{d}+\frac{{{K}_{3}}\,{{\varepsilon }_{0}}{{A}_{3}}}{d}\] \[{{K}_{eq}}=\frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}}{A}\] \[=\,\,\,\frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}}{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}}\]