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

Question

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 capaciitor is filled as shown in figure $I,$ and the second one is filled as shown in figure $I I .$
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 $(I I)$ ):

TestHub · Accepted Answer

Energy stored in a capacitor E = 1 2 C V 2 ⇒ E 1 E 2 = C 1 C 2 C 1 = d 3 ε 0 k 1 A + d 3 ε 0 A k 2 + d 3 ε 0 A k 3 - 1 = d 3 A ε 0 - 1 1 k 1 + 1 k 2 + 1 k 3 - 1 = d 3 A ε 0 - 1 k 1 k 2 + k 2 k 3 + k 3 k 1 k 1 k 2 k 3 - 1 C 1 = 3 A ε 0 d k 1 k 2 k 3 k 1 k 2 + k 2 k 3 + k 3 k 1 C 2 = ε 0 A 3 k 1 d + ε 0 A 3 k 2 d + ε 0 A 3 k 3 d ∴ C 1 C 2 = E 1 E 2 = 9 k 1 k 2 k 3 k 1 + k 2 + k 3 k 1 k 2 + k 2 k 3 + k 3 k 1

Physics - Capacitor Question with Solution | TestHub

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