Now let's examine the effects of placing a dieletric film between the plates of a capacitor. The plates of an air-filled parallel-plate capacitor each have area 2.00103cm22.00103cm2 and are 1.00 cmcm apart. The capacitor is connected to a power supply and charged to a potential difference of V0=3.00kVV0=3.00kV. The capacitor is then disconnected from the power supply without any charge being lost from its plates. After the capacitor has been disconnected, a sheet of insulating plastic material is inserted between the plates, completely filling the space between them. When this is done, the potential difference between the plates decreases to 1.00 kVkV, while the charge remains constant. (a) What is the capacitance of the capacitor before and after the dielectric is inserted? (b) What is the dielectric constant of the plastic? (c) What is the electric field between the plates before and after the dielectric is inserted?

1. Suppose we repeat the example, with V0V0 = 3.00 kVkV, but this time keep the capacitor connected to the power supply while inserting the dielectric, so that the potential difference across the capacitor remains at 3.00 kVkV. Find the charge on the plate before the dielectric is inserted.

2. Find the charge on the plate after the dielectric is inserted.

3. Find the electric field between the plates before the dielectric is inserted.

4. Find the electric field between the plates after the dielectric is inserted.