The plates of a parallel plate capacitor each have an area of 0.40 m 2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated.
-Suppose that a dielectric sheet is inserted to completely fill the space between the plates and the potential difference between the plates drops to 1000 V. Determine the dielectric constant.

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor with no dielectric is given by:

C = ε0(A/d)

Where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between them.

Plugging in the given values, we get:

C = (8.85 x 10^-12 F/m) * (0.40 m^2 / 0.02 m) = 1.77 x 10^-8 F

When the dielectric is inserted, the capacitance increases by a factor of κ, the dielectric constant:

C' = κε0(A/d)

We can solve for κ by using the ratio of the initial and final potential differences:

Vf/Vi = Ci/Cf = κ

Plugging in the given values, we get:

κ = (3000 V) / (1000 V) = 3

Therefore, the dielectric constant is 3, which corresponds to choice C.

The Plates of a Parallel Plate Capacitor Each Have an Area