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Question

Statement 1: A parallel plate capacitor is charged by a battery of voltage V. The battery is then disconnected. If the space between the plates is filled with a dielectric, the energy stored in the capacitor will decrease.

Statement 2: The capacitance of a capacitor increases due to the introduction of a dielectric between the plates.

Statement 2: The capacitance of a capacitor increases due to the introduction of a dielectric between the plates.

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Solution

The correct option is **A** Both the statements are correct.

Initially, let the capacitance of the capacitor be C and charge on it be Q on connecting to the battery.

So, initial energy stored in the capacitor,

U=Q22C

Now, upon insertion of the dielectric, the charge on the capacitor will not change (since it is isolated as the battery has been disconnected).

However, the capacitance will change and become equal to

C′=KC

Where K is the dielectric constant of the dielectric.

Since K>1, the value of capacitance increases on insertion of dielectric and statement 2 is correct.

Now, final energy,

U=Q22(KC)=UK, which is less than U.

So, statement 1 is also correct.

Initially, let the capacitance of the capacitor be C and charge on it be Q on connecting to the battery.

So, initial energy stored in the capacitor,

U=Q22C

Now, upon insertion of the dielectric, the charge on the capacitor will not change (since it is isolated as the battery has been disconnected).

However, the capacitance will change and become equal to

C′=KC

Where K is the dielectric constant of the dielectric.

Since K>1, the value of capacitance increases on insertion of dielectric and statement 2 is correct.

Now, final energy,

U=Q22(KC)=UK, which is less than U.

So, statement 1 is also correct.

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