Question

A parallel plate capacitor has an area of each plate of 20 cm2, and the distance between plates is 1.0 mm. The space between the plates is filled with a film of dielectric material whose dielectric constant is 30. The voltage between the plates is measured to be 9 V.

a) Calculate the charge stored in the capacitor and the electric field between the plates.
b) If the dielectric film is removed, what will the new voltage between the plates be?

Answer 1

The charge stored in the capacitor is 5.306 x 10^-8 C and the electric field between the plates is 9000 V/m. When the dielectric film is removed, the new voltage between the plates would be the same as the initial voltage, which is 9 V.

Step-by-step explanation:

To calculate the charge stored in the capacitor, we can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. The capacitance of a parallel plate capacitor with dielectric material is given by the formula C = εrA/d, where εr is the dielectric constant, A is the area of each plate, and d is the distance between the plates.

Using the given values, we can calculate the charge stored in the capacitor: Q = (30)(8.854 x 10^-12 C^2/Nm^2)(0.0002 m^2)/(0.001 m) = 5.306 x 10^-8 C.

The electric field between the plates can be calculated using the formula E = V/d, where E is the electric field, V is the voltage, and d is the distance between the plates. Using the given values, we can calculate the electric field: E = 9 V/(0.001 m) = 9000 V/m.

b) When the dielectric film is removed, the new voltage between the plates would be the same as the initial voltage, which is 9 V.