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  4. 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 K1, K2 and K3. 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 (E1 refers to capacitor (I) and E2 to capacitor (II) :

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Q. 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) :Physics Question Image

JEE MainJEE Main 2019Electrostatic Potential and Capacitance

Solution:

$
\begin{aligned}
\text { I. } C_{1} &=\frac{3 \epsilon_{0} AK _{1}}{ d } \text { (see fig. I) } \\
C _{2} &=\frac{3 \in_{0} AK _{2}}{ d } \\
C _{3} &=\frac{3 \in_{0} AK _{3}}{ d } \\
\frac{1}{ C _{ eq }} &=\frac{1}{ C _{1}}+\frac{1}{ C _{2}}+\frac{1}{ C _{3}} \\
\Rightarrow C _{ eq } &=\frac{3 \in_{0} AK _{1} K _{2} K _{3}}{ d \left( K _{1} K _{2}+ K _{2} K _{3}+ K _{3} K _{1}\right)}
\end{aligned}
$
II. $C_{1}=\frac{€_{0} K_{1} A}{3 d}$ (see fig. II)
$
\begin{array}{l}
C_{2}=\frac{\in_{0} K _{2} A }{3 d } \\
C _{3}=\frac{\in_{0} K _{3} A }{3 d } \\
C _{ eq }^{\prime}= C _{1}+ C _{2}+ C _{3} \\
=\frac{\in_{0} A }{3 d }\left( K _{1}+ K _{2}+ K _{3}\right)
\end{array}
$
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)}
$
so $\frac{ E _{1}}{ E _{2}}=\frac{\frac{1}{2} C _{ eq } \cdot V ^{2}}{\frac{1}{2} C _{ eq }^{\prime} V ^{2}}=\frac{9 K _{1} K _{2} K _{3}}{\left( K _{1}+ K _{2}+ K _{3}\right)\left( K _{1} K _{2}+ K _{2} K _{3}+ K _{3} K _{1}\right)}$
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