Question

The terminals of a 1.30 mF parallel-plate vacuum capacitor is connected to a battery that keeps a voltage of 29.6 V across the plates. Then, without disconnecting the battery, a piece of material with dielectric constant of 3.51 is inserted between the plates. After inserting the dielectric, how much energy (in Joules) is stored in the capacitor?

Answer 1

Answer:

The energy stored in the capacitor after replacing the dielectric material = 2.00 J

Explanation:

The energy stored in a capacitor is given by

E = ½CV²

where C = capacitance of the capacitor; initially with vacuum as dielectric material is equal to 1.30 mF

V = voltage across the capacitor = 29.6 V

The capacitance of a capacitor is also given as

C = (Aε/d)

A = Cross sectional Area of the capacitor

d = distance between the plates of the capacitor

ε = permissivity of material between the plates of the capacitor

ε = kε₀

k = relative permissivity

ε₀ permissivity of vacuum.

If the Cross sectional Area of the capacitor and the distance between the plates of the capacitor are constant, it is evident that the capacitance of the capacitor is directly proportional to the permissivity of the dielectric material.

C₀ = (Aε₀/d) = 1.30 mF

C = (Aε/d) = (kAε₀/d) = kC₀

For the is question, the relative permissivity = 3.51

C = 3.51 × 1.30 mF = 4.563 mF = 0.004563 F

Energy stored in a capacitor is still

E = ½CV²

C = 4.563 mF = 0.004563 F

V = 29.6 V

E = (1/2)(0.004563)(29.6²)

E = 1.99895904 J = 2.00 J

Hope this Helps!!!

Answer 2

Answer:

Energy stored in capacitor with new dielectric is 1.998 joules approximately 2 joules

Explanation:

Detailed explanation and calculation is shown in the image