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)): |
[JEE Main Held on 12-4-2019 Morning] |
A) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{9{{K}_{1}}{{K}_{2}}{{K}_{3}}}{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}\]
B) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{K}_{1}}{{K}_{2}}{{K}_{3}}}{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}\]
C) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}{{{K}_{1}}{{K}_{2}}{{K}_{3}}}\]
D) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{({{K}_{1}}+{{K}_{2}}+{{K}_{2}})({{K}_{2}}{{K}_{3}}+{{K}_{3}}{{K}_{1}}+{{K}_{1}}{{K}_{2}})}{9{{K}_{1}}{{K}_{2}}{{K}_{3}}}\]
Correct Answer: A
Solution :
[a] | |
........(1) | |
.(2) | |
Now, | |
|
You need to login to perform this action.
You will be redirected in
3 sec