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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 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)) :

E1/E2=K1K2K3/(K1+K2+K3)(K2K3+K3K1+K1K2)

E1/E2=(K1+K2+K3)(K2K3+K3K1+K1K2)/K1K2K3

E1/E2=9K1K2K3/(K1+K2+K3)(K2K3+K3K1+K1K2)

E1/E2=(K1+K2+K3)(K2K3+K3K1+K1K2)/9K1K2K3

Solution:

Correct option is A.
E1/E2=9K1K2K3(K1+K2+K3)/(K2K3+K3K1+K1K2)
I. C1 = 3ε₀AK1/d (see fig. I)
C2 = 3ε₀AK2/d
C3 = 3ε₀AK3/d
1/Ceq = 1/C1 + 1/C2 + 1/C3 ⇒ Ceq = 3ε₀AK1K2K3/d(K1K2 + K2K3 + K3K1).. (1)
II. C1 = ε₀K1A/3d (see fig. II)
C2 = ε₀K2A/3d
C3 = ε₀K3A/3d
Ceq' = C1 + C2 + C3 = ε₀A/3d(K1 + K2 + K3).. (2)
Now, E1/E2 = (1/2Ceq.V²)/(1/2Ceq'V²) = 9K1K2K3(K1+K2+K3)/(K1K2+K2K3+K3K1)