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 b

Step-by-step solution

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