A capacitor with the dielectric is charged completely and battery is disconnected . Now the dielectric is pulled out . Hence ??
a)Charge stored will decrease
b)energy stored will increase

Open in App

Solution

Dear student,
Charge stored will decrease.

The net field inside the capacitor and the potential difference across the capacitor is decreased when the electric field from the charge of capacitor plates are cancelled by the electric field from the polarized dielectric.More charge is needed by the capacitor to return to its original potential difference.

The net effect of the dielectric is to increase the amount of charge a capacitor can store for a given potential difference. The whole point of using a capacitor is to store charge, so coming up with a way to store more charge for the same amount of effort is a good .

Regards

Suggest Corrections

Similar questions

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.

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

Q. A potential difference is applied to the plate of a capacitor filled with an insulator, with energy stored in the capacitor as U. The capacitor is disconnected and the insulator is pulled out now. The work done in pulling out the insulator against electric field is 4U. Find the dielectric constant of the insulator. ___

Q. A parallel plate capacitor is charged by a battery.

The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.

How will (i) its capacitance, (ii) electric field between the plates and (iii) energy stored in the capacitor be affected ? Justify your answer giving necessary mathematical expressions for each case.

Q. Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

Join BYJU'S Learning Program

A capacitor with the dielectric is charged completely and battery is disconnected . Now the dielectric is pulled out . Hence ?? a)Charge stored will decrease b)energy stored will increase

A capacitor with the dielectric is charged completely and battery is disconnected . Now the dielectric is pulled out . Hence ??
a)Charge stored will decrease
b)energy stored will increase