Question

Two parallel plate capacitors of capacitances C1 and C2 such that C1=2C2 are connected across a battery of V volts as shown in the figure. Initially the key (k) is kept closed to fully charge the capacitors. The key is now thrown open and a dielectric slab of dielectric constant 'K' is inserted in the two capacitors to completely fill the gap between the plates. The ratio of the energies stored in the combination, before and after the introduction of the dielectric slab:

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Solution

Before slab inserted, net capacitance is C=C1+C2=2C2+C2=3C2 (as they are in parallel)

Energy stored U=12CV2=32C2V2

After insert slab C1 becomes C′1=KC1 and C2 becomes C′2=KC2

now net capacitance C′=C′1+C′2=KC1+KC2=K(C1+C2)=3KC2

When switch is opened ,but the capacitors fully charged by voltage V and the energy stored U′=12C′V2=32KC2V2

∴UU′=(3/2)C2V2(3/2)KC2V2=1K

Energy stored U=12CV2=32C2V2

After insert slab C1 becomes C′1=KC1 and C2 becomes C′2=KC2

now net capacitance C′=C′1+C′2=KC1+KC2=K(C1+C2)=3KC2

When switch is opened ,but the capacitors fully charged by voltage V and the energy stored U′=12C′V2=32KC2V2

∴UU′=(3/2)C2V2(3/2)KC2V2=1K

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