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Q.

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V by a battery. The battery is then disconnected and the space between the plate of capacitor OF CAPACITY C is completely filled with a material of dielectric constant K. The potential difference across the capacitor now becomes

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a

$\frac{3 V}{K + 2}$

b

$\frac{K + 2}{3 V}$

c

$\frac{V}{K + 2}$

d

$\frac{V}{K + 3}$

answer is A.

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Detailed Solution

The capacitors are connected in parallel. So, the resultant capacitance $= C +2 C =3 C$
$∴$ Total charge Q = 3 CV
When the capacitor of capacity C is filled with dielectric, its capacitance becomes KC. The new resultant capacitance becomes $C^{'} = KC + 2 C = (K + 2) C$ .As charge remains the same, Q = 3CV
$∴ \frac{Q}{C^{'}} = \frac{3 CV}{(K + 2) C} = \frac{3 V}{K + 2}$

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Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V by a battery. The battery is then disconnected and the space between the plate of capacitor OF CAPACITY C is completely filled with a material of dielectric constant K. The potential difference across the capacitor now becomes