A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

Question

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in (A) Reduction of charge on the plates and increase of potential difference across the plates (B) Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates (C) Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates (D) None of the above

Tardigrade Admin · Accepted Answer

Battery in disconnected so Q will be constant. As C propto K, so with introduction of dielectric slab capacitance will increase. Using Q=C V, V will decrease and using U=(Q2/2 C), energy will decrease.

Q. A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

Reduction of charge on the plates and increase of potential difference across the plates

Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

None of the above

Battery in disconnected so $Q$ will be constant. As $C \propto K$ , so with introduction of dielectric slab capacitance will increase. Using $Q = C V, V$ will decrease and using $U = \frac{Q ^{2}}{2 C}$ , energy will decrease.

Q. A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

Reduction of charge on the plates and increase of potential difference across the plates

Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

None of the above

Battery in disconnected so Q will be constant. As C∝K, so with introduction of dielectric slab capacitance will increase. Using Q=CV,V will decrease and using U=2CQ2​, energy will decrease.

Battery in disconnected so $Q$ will be constant. As $C \propto K$ , so with introduction of dielectric slab capacitance will increase. Using $Q = C V, V$ will decrease and using $U = \frac{Q ^{2}}{2 C}$ , energy will decrease.