A $60 μ F$ parallel plate capacitor whose plates are separated by 6 mm is charged to 250V, and then the charging source is removed. When a slab of dielectric constant 5 and thickness 3 mm is placed between the plates, find the potential difference across the capacitor?

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250 V

75 V

100 V

150 V

answer is C.

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Detailed Solution

$60 μ F = \frac{\in_{o} A}{6} . . . . . . (1)$

$for second case C' = \frac{\in_{o} A}{6 - 3 (1 - \frac{1}{5})} = \frac{\in_{o} A (5)}{18} = \frac{\in_{o} A \times 5}{6 \times 3} = \frac{60 μ \times 5}{3} C' = 100 μF$

as the battery is removed charge remains constant

New potential $V' = \frac{q}{C'} = \frac{60 μ \times 250}{100 μ} = 150 V$

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