A capacitor of capacitance $2 μ F$ is charged with a $10 V$ battery. Now 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?

10 V

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5 V

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0.25 V

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1 V

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Solution

The correct option is C $0.25 V$
Charge of capacitor, $Q = C V$
$\Rightarrow Q = 2 \times 10 = 20 μ C$
when dielectric is inserted, its capacitance becomes, $C^{1} = K C$
$\Rightarrow C = 4 \times 20 = 80 μ F$
since battery is disconnected, charge on the capacitor will be $20 μ C$
Now, $Q = C^{1} V$
$\Rightarrow 20 = 80 V$
$\Rightarrow V = \frac{1}{4} = 0.25 V$

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A capacitor of capacitance 2μF is charged with a 10 V battery. Now 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?

The correct option is C 0.25 VCharge of capacitor, Q=CV ⇒Q=2×10=20 μC when dielectric is inserted, its capacitance becomes, C1=KC ⇒C=4×20=80 μF since battery is disconnected, charge on the capacitor will be 20 μC Now, Q=C1V ⇒20=80V ⇒V=14=0.25 V

A capacitor of capacitance $2 μ F$ is charged with a $10 V$ battery. Now 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?