A parallel plate capacitor has a paper dielectric having dielectric strength 8.00 kV/mm and dielectric constant 3.00. The plate area is 3000 cm² and the plate separation is 0.500 mm. If 10.0 V are applied to this capacitor, what is the charge on the capacitor?

Question

Quizplus · Accepted Answer

The electric field strength inside the dielectric material can be found using the formula:

E = V/d
where V is the voltage applied and d is the distance between the plates. Here, E = 10.0 V/0.500 mm = 20.0 kV/mm.

The maximum electric field that the dielectric can withstand before breaking down is given by:

Emax = dielectric strength
Therefore, Emax = 8.00 kV/mm.

Since the electric field inside the dielectric is less than its breakdown strength, the dielectric will not break down.

The capacitance of the parallel plate capacitor with a dielectric material can be found using the formula:

C = εA/d
where ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

Here, ε = ε0εr = 8.85 x 10^-12 F/m x 3.00, A = 3000 cm^2 = 0.3000 m^2, and d = 0.500 mm = 0.0005 m.

Therefore, C = (8.85 x 10^-12 F/m x 3.00) x (0.3000 m^2/0.0005 m) = 1.59 x 10^-8 F.

The charge on the capacitor can be found using the formula:

Q = CV
where V is the voltage applied.

Therefore, Q = (1.59 x 10^-8 F) x (10.0 V) = 1.59 x 10^-7 C = 159 nC.

A Parallel Plate Capacitor Has a Paper Dielectric Having Dielectric