(C) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$,where $K$ is the dielectric constant,$A$ is the area,and

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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$,where $K$ is the dielectric constant,$A$ is the area,and $d$ is the distance between the plates.Given that the new distance $d' = \frac{d}{2}$ and the new dielectric constant $K' = 3K$.The new capacitance $C'$ becomes $C' = \frac{(3K) \epsilon_0 A}{(d/2)} = 6 \left( \frac{K \epsilon_0 A}{d} \right) = 6C$.Thus,the Assertion is correct.The capacity of a capacitor depends on the dielectric constant $(K)$ of the material placed between the plates. Therefore,the Reason is incorrect.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

Easy

Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

AIIMS 1997 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

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If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now,the space between the plates is filled with a liquid of dielectric constant $K = 5$. The charge that flows from the battery to the capacitor is......$\mu C$.

If a dielectric is inserted into a charged capacitor (battery removed),then the quantity that remains constant is

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Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.