A parallel plate capacitor of capacitance $10 μ F$ is connected across a battery of e.m.f. $5 m V$ . Now, the space between the plates of the capacitor is filled with a dielectric material of dielectric constant $K = 5$ . Then, the charge that flows through the battery, till equilibrium is reached is

250 μ C

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250 n C

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200 n C

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200 μ C

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Solution

The correct option is C $200 n C$
Case-I

$Q_{1} = 10 μ F \times 5 m V$
$Q_{1} = 10 \times 10^{- 6} \times 5 \times 10^{- 3} C$
$Q_{1} = 50 \times 10^{- 9} C$

Case-II

$Q_{2} = 5 \times 10 μ F \times 5 m V$
$Q_{2} = 5 \times 10 \times 10^{- 6} \times 5 \times 10^{- 3} C$
$Q_{2} = 250 \times 10^{- 9} C$
Now, $\begin{matrix} Q_{2} - Q_{1} & = 250 \times 10^{- 9} - 50 \times 10^{- 9} C Q_{2} - Q_{1} & = 200 \times 10^{- 9} C = 200 n C \end{matrix}$

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A parallel plate capacitor of capacitance 10μF is connected across a battery of e.m.f. 5 mV. Now, the space between the plates of the capacitor is filled with a dielectric material of dielectric constant K=5. Then, the charge that flows through the battery, till equilibrium is reached is

The correct option is C 200 nCCase-I Q1=10μF×5mV Q1=10×10−6×5×10−3C Q1=50×10−9C Case-II Q2=5×10μF×5mV Q2=5×10×10−6×5×10−3C Q2=250×10−9C Now, Q2−Q1=250×10−9−50×10−9CQ2−Q1=200×10−9C=200 nC