JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2019
- Hard
- MCQ
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 dielectrics, of equal thickness and dielectric constants \(K_1\) , \(K_2\) and \(K_3\) . The first capacitor 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 capacitor \((I)\) and \(E_2\) to capacitor \((II)\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{{{E_1}}}{{{E_2}}} = \frac{{9{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}\)
Step-by-step Solution
Detailed explanation
\(C_{1}=\frac{3 \varepsilon_{0} A K_{1}}{d}\)
\(C_{2}=\frac{3 \varepsilon_{0} A K_{2}}{d}\)
\(C_{3}=\frac{3 \varepsilon_{0} A K_{3}}{d}\)
\(\frac{1}{C_{e q}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)
\(\Rightarrow \quad C_{e q}=\frac{3 \varepsilon_{0} A K_{1} K_{2} K_{3}}{d\left(K_{1} K_{2}+K_{2} K_{3}+K_{3} K_{1}\right)}.........(i)\)
\(C_{1}=\frac{\varepsilon_{0} K_{1} A}{3 d}\)
\(C_{2}=\frac{\varepsilon_{0} K_{2} A}{3 d}\)
\(C_{3}=\frac{\varepsilon_{0} K_{3} A}{3 d}\)
\(C_{e q}^{\prime} =C_{1}+C_{2}+C_{3} \)
\( = \frac{{{\varepsilon _0}A}}{{3d}}({K_1} + {K_2} + {K_3}).........(ii)\)
Now,
\(\frac{{{E_1}}}{{{E_2}}} = \frac{{\frac{1}{2}{C_{eq}} \cdot {V^2}}}{{\frac{1}{2}C_{eq}^\prime {V^2}}} = \frac{{9{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}}\)
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