Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a .. 29.47 particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) W

Answer 1

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis? (d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?

Given Information:

Radius of parallel-plate capacitor = 4.10 cm = 0.041 m

Conduction current = Ic = 0.276 A

Required Information:

a) Displacement current density = JD = ?

b) Rate of change of electric field = dE/dt = ?

c) Magnetic field between plates at r = 2 cm = ?

d) Magnetic field between plates at r = 1 cm = ?

Answer:

a) Displacement current density = JD = 52.27 A/m²

b) Rate of change of electric field = dE/dt = 5.904×10¹² V/m.s

c) Magnetic field between plates at r = 2 cm = 6.568×10⁻⁷ Tesla

d) Magnetic field between plates at r = 1 cm = 3.284×10⁻⁷ Tesla

Explanation:

a) What is the displacement current density JD in the air space between the plates?

Displacement current density is given by

JD = Id/A

Where Id is the conduction current and A is the area of capacitor given by

A = πr²

A = π(0.041)²

A = 0.00528 m²

As you can notice in the diagram, conduction current has equal displacement between the capacitor plates therefore, Id = Ic

JD = 0.276/0.00528

JD = 52.27 A/m²

b) What is the rate at which the electric field between the plates is changing?

The rate of change of electric field is given by

dE/dt = JD/ε₀

Where JD is the displacement current density and ε₀ is the permittivity of free space and its value is 8.854×10⁻¹² C²/N.m²

dE/dt = 52.27/8.854×10⁻¹²

dE/dt = 5.904×10¹² V/m.s

c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis?

The induced magnetic field between the plates can be found using Ampere's law

B = (μ₀/2)*JD*r

Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷ T.m/A, JD is the displacement current density and r is the distance from the axis.

B = (4π×10⁻⁷/2)*52.27*0.02

B = 6.568×10⁻⁷ Tesla

d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?

B = (μ₀/2)*JD*r

B = (4π×10⁻⁷/2)*52.27*0.01

B = 3.284×10⁻⁷ Tesla