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Question

# A parallel plate capacitor with a dielectric slab of dielectric constant 3, filling the space between the plates, is charged to a potential V. The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced by another dielectric slab of dielectric constant 2. If the energies stored in the capacitor before and after the dielectric slab is changed are E1 and E2, then E1/E2 is:

A
49
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B
23
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C
32
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D
95
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Solution

## The correct option is B 23Let the charge stored by capacitor with dielectric constant 3 be Q.Thus energy stored be Q22C1Since the charge remains the same after changing the dielectric, the new energy stored=Q22C2⟹E1E2=C2C1=K2K1=23 (C=KAϵ0d)

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