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Question

A parallel plate air capacitor with no dielectric between the plates is connected to the constant voltage source. What will be the new capacitance and charge on the capacitor if a dielectric with dielectric constant K=5 is inserted between the plates? Intially capacitance and charge on the capacitor are 50 μF and 200 μC respectively before the introduction of the dielectric.

A
50 μF and 200 μC
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B
250 μF and 1000 μC
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C
250 μF and 200 μC
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D
50 μF and 1000 μC
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Solution

The correct option is B 250 μF and 1000 μC
Given:
C0=50 μF; Q0=200 μC; K=5

In the presence of the dielectric completely filling the space between the plates, the capacitance will increase by the factor of the dielectric constant of the material.

Thus, C=KC0=5×50=250 μF

If the battery remains connected across the capacitor, the voltage will remain the same i.e. equal to the voltage of the source.

From, Q=CV, the charge will also become K times the previous charge on the capacitor.

So, Q=KQ0=5×200 μC=1000 μC

Hence, option (b) is correct.
Key concept - Capacitance & charge on parallel plate capacitor increases due to fully filled dielectric, if battery remains connected.

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