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Question

# A capacitor of capacitance 2 μF is charged with a 10 V battery. Now the battery is disconnected and a dielectric of dielectric constant K=4 is inserted between the plates. What is the potential difference between the plates of the capacitor ?

A
10 V
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B
25 V
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C
0.25 V
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D
2.5 V
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Solution

## The correct option is C 0.25 VGiven: C=2 μF; V=10 V; K=4 Charge on the plates of the capacitor is given by Q=CV=2×10=20 μC When dielectric is inserted, its capacitance becomes, C′=KC ∴C′=4×20=80 μC Since the battery is disconnected, charge on the capacitor will remain 20 μC. ∴Q=C′V ⇒20=80 V ⇒V=14=0.25 V Therefore, option (C) is right. Why this question ? Tip : Dielectric increases the capacitance of the capacitor

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