Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dieletrics, of equal thickness and dielectric constants K_{1},K_{2}andK_{3}. The first capacitors is filled as shown in fig. I, and the second one is filled as shown in fig II. 

If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E_{1} refers to capacitors (I) and E_{2} to capacitor (II) ): 

  • Option 1)

    \frac{E_{1}}{E_{2}}=\frac{9K_{1}K_{2}K_{3}}{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}

  • Option 2)

    \frac{E_{1}}{E_{2}}=\frac{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}{9K_{1}K_{2}K_{3}}

  • Option 3)

    \frac{E_{1}}{E_{2}}=\frac{K_{1}K_{2}K_{3}}{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}

  • Option 4)

    \frac{E_{1}}{E_{2}}=\frac{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}{K_{1}K_{2}K_{3}}

 

Answers (1)

Arrangement I \rightarrow capacities in series

\frac{1}{C_1}=\frac{d/3}{K_{1}\varepsilon _{0}A}+\frac{d/3}{K_{2}\varepsilon _{0}A}+\frac{d/3}{K_{3}\varepsilon _{0}A}

\Rightarrow C_{1}=\frac{3K_{1}K_{2}K_{3}\varepsilon _{0}A}{d(K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1})}

Arrangement II \rightarrow capacitors in parallel

C_{2}=\frac{K_{1}\varepsilon _{0}(A/3)}{d}+\frac{K_{2}\varepsilon _{0}(A/3)}{d}+\frac{K_{3}\varepsilon _{0}(A/3)}{d}

       =\frac{(K_{1}+K_{2}+K_{3})\varepsilon _{0}A}{3d}

\frac{E_{1}}{E_{2}}=\frac{ \frac{1}{2}c_{1}V^{2} }{ \frac{1}{2}c_{2}V^{2}}=\frac{c_{1}}{c_{2}}=\frac{9K_{1}K_{2}K_{3}}{(K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1})(K_{1}+K_{2}+K_{3})}


Option 1)

\frac{E_{1}}{E_{2}}=\frac{9K_{1}K_{2}K_{3}}{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}

Option 2)

\frac{E_{1}}{E_{2}}=\frac{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}{9K_{1}K_{2}K_{3}}

Option 3)

\frac{E_{1}}{E_{2}}=\frac{K_{1}K_{2}K_{3}}{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}

Option 4)

\frac{E_{1}}{E_{2}}=\frac{(K_{1}+K_{2}+K_{3})(K_{2}K_{3}+K_{3}K_{1}+K_{1}K_{2})}{K_{1}K_{2}K_{3}}

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