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A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 3d/4, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given C0 = capacitance of capacitor with air as medium between plates.)

1. 4KC0/(3 + K)
2. 3KC0/(3 + K)
3. (3 + K)/(4KC0)
4. K/(4 + K)

Correct answer: 4KC0/(3 + K)

Solution

The capacitor is an air gap d/4 in series with a dielectric of thickness 3d/4: C=eps0*A/(d/4 + (3d/4)/K). With C0=eps0*A/d this gives C = C0/(1/4 + 3/(4K)) = 4K*C0/(3+K), option (A).

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