The capacity of a parallel plate condenser with air as dielectric is 2µF. The space between the plates is filled with dielectric slab with K = 5. It is charged to a potential of 200V and disconnected from cell. Work done in removing the slab from the condenser completely

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0.6J

1.2J

1.6J

0.8J

answer is A.

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Detailed Solution

$U_{i} = \frac{1}{2} {KCV}^{2} = \frac{1}{2} \times 5 \times 2 \times 10^{- 6} \times 4 \times 10^{4} = 0 .2 J$
When we remove the dielectric slab is
C ↓ by k and V ↑ K times
$U_{f} = \frac{1}{2} C (KV)^{2} = 1 J ∴ W = U_{f} - U_{i} = 0.8 J$

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