Question

A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?

1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)

1. Answer 1: the charge on the plates will increase

Explanation: placing a dielectric between two charged plate increases its capacitance

C = Q/V,

If the plates remain connected then the voltage remains the same.

Therefore for an increase in capacitance charge will increase.

Answer 2: electric field potential remains the same.

Explanation: if the plates are disconnected, charge on plates remains contant while voltage varies with change in distance, electric field intensity remains constant and this is proportional to the electric potential energy.

Answer 3: capacitance increases

Explanation: introducing a dielectric between two plates causes opposite charges to be induced on the faces of the dielectric. This also reduces the p.d across the capacitor.

Answer 4: voltage remains constant.

Explanation: A connected plate has a constant voltage across its field.

Answer 5: charge remains contant.

Explanation: capacitance will increase with introduction of dielectric, p.d across the plates will drop, the charge will remain constant.

Answer 6: capacitance increases

Explanation: placing a dielectric between plates always increase the capacitance.

Answer 7: electric potential energy falls.

Answer 8: voltage between plates decreases