question_answer

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants \[{{K}_{1}},\text{ }{{K}_{2}},\text{ }{{K}_{3}},\text{ }{{K}_{4}}\] arranged as shown in the figure. The effective dielectric constant K will be: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{2}})({{K}_{2}}+{{K}_{4}})}{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}}}\]

B) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{2}})({{K}_{2}}+{{K}_{4}})}{2({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]

C) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{2}})({{K}_{3}}+{{K}_{4}})}{({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]

D) \[K\,\,=\,\,\frac{({{K}_{1}}+{{K}_{4}})({{K}_{2}}+{{K}_{3}})}{2({{K}_{1}}+{{K}_{2}}+{{K}_{3}}+{{K}_{4}})}\]

E) None of these

Correct Answer: E

Solution :

[Bonus] \[{{C}_{eq.}}\,=\,\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\,+\,\frac{{{C}_{3}}{{C}_{4}}}{{{C}_{3}}+{{C}_{4}}}\] \[{{C}_{1}}\,=\,\frac{{{k}_{1}}\,{{\in }_{0}}\,A/2}{d/2}\,=\,\frac{{{k}_{1}}\,{{\in }_{0}}\,A}{d}\] Similarly\[{{C}_{2}}\,=\,\frac{{{k}_{2}}\,{{\in }_{0}}\,A}{d},\,\,{{C}_{3}}\,\,=\,\,\frac{{{k}_{3}}\,{{\in }_{0}}\,A\,\,}{d}\],\[{{C}_{4}}\,=\,\frac{{{k}_{4}}\,{{\in }_{0}}\,A}{d}\]  \[{{C}_{eq.}}\,=\,\,\frac{{{k}_{eq.}}\,A}{d}\] \[{{C}_{eq.}}\,=\,\frac{\frac{{{k}_{1}}\,{{\in }_{0}}\,A}{d}\,\,.\,\,\frac{{{k}_{2}}\,{{\in }_{0}}\,A}{d}}{\frac{{{k}_{1}}\,{{\in }_{0}}\,A}{d}+\frac{{{k}_{2}}\,{{\in }_{0}}\,A}{d}}\,\,+\,\,\frac{\frac{{{k}_{3}}\,{{\in }_{0}}\,A}{d}.\,\frac{{{k}_{4}}\,{{\in }_{0}}\,A}{d}}{\frac{{{k}_{3}}\,{{\in }_{0}}\,A}{d}\,+\frac{{{k}_{4}}\,{{\in }_{0}}\,A}{d}}\]] \[{{C}_{eq.}}\,=\,\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\,\,\frac{{{\in }_{0}}A}{d}\,\,+\,\,\frac{{{k}_{3}}{{k}_{4}}}{{{k}_{3}}+{{k}_{4}}}\,\,\frac{{{\in }_{0}}A}{d}\] Now, \[{{k}_{eq.}}\,=\,\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\,\,+\,\,\frac{{{k}_{3}}{{k}_{4}}}{{{k}_{3}}+{{k}_{4}}}\,\,\]