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)) :III
In the first capacitor the three slabs are stacked one above the other, so they act like three capacitors in series, each of plate separation d/3: C11=ε0K1Ad/3+ε0K2Ad/3+ε0K3Ad/3, giving C1=d3ε0A⋅K1K2+K2K3+K3K1K1K2K3. In the second capacitor the slabs sit side by side, each over area A/3, so they are in parallel: C2=3dε0A(K1+K2+K3). Since both are charged to the same V, the stored energies are in the ratio of the capacitances, E2E1=C2C1=(K1+K2+K3)(K2K3+K3K1+K1K2)9K1K2K3.
JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option
