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Separation between the plates of a parallel plate capacitor is 5 mm. This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3 mm and dielectric constant K = 10 is placed between the plates as shown. Potential difference between the plates after the dielectric slab has been introduced is –

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18.5 V

13.5 V

11.5 V

6.5 V

answer is C.

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Detailed Solution

$q = C_{1} V_{1} = C_{2} V_{2}$ where $C_{1} = \frac{ε_{0} A}{5} a n d C_{2} = \frac{ε_{0} A}{d - t (1 - \frac{1}{K})} = \frac{ε_{0} A}{5 - 3 (1 - \frac{1}{10})} = \frac{10 ε_{0} A}{23}$

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