Question

a parallel-plate capacitor is connected to a power supply that maintains a fixed potential difference between the plates. if a sheet of dielectric is then slid between the plates, what happens to (i) the electric field between the plates, (ii) the magnitude of charge on each plate, and (iii) the energy stored in the capacitor?

Answer 1

Introducing a dielectric reduces the electric field and, for an isolated capacitor, it also decreases the energy stored without changing the charge on the plates. If connected to a power supply, the charge increases to maintain the voltage, which may increase the stored energy.

Step-by-step explanation:

When a dielectric material is introduced between the plates of a parallel-plate capacitor, three things happen:

1. The electric field between the plates decreases because the dielectric material has a property known as dielectric constant, which reduces the effective electric field.
2. The magnitude of charge on each plate remains unchanged if the capacitor is isolated. If the capacitor is still connected to the power supply, the magnitude of charge will increase to restore the initial voltage across the plates.
3. The energy stored in the capacitor decreases when the capacitor is isolated due to the lowered electric field. However, if the capacitor is connected to a voltage source, the energy stored may increase because additional charge is moved onto the plates to sustain the voltage across the increased capacitance of the dielectric-filled capacitor.

Thus, the final effect on the stored energy depends on whether the capacitor is connected to a constant-voltage source during the insertion of the dielectric.