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Q.

A $2 μ F$ capacitor $C_{1}$ is first charged to a potential difference 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_{2}$ of 8µF. The charge on $C_{2}$ on equilibrium condition is _______ $μ C$

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a

$4 μ C$

b

$10 μ C$

c

$8 μ C$

d

$16 μ C$

answer is C.

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Detailed Solution

$Q = C_{1} V = 10 \times 2 = 20 μ C$
Common potential $= \frac{20}{10} = 2 V$
$Q_{2} = C_{2} V = 8 \times 2 = 16 μ C$

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