Displacement Current and air space

[aznkid310]

Homework Statement

A parallel-plate, air-filled capacitor is being charged. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.280 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 2.00 cm from the axis? (d) At 1.00 cm from the axis?

Homework Equations

Do i assume displacement current iD = 0.280?

If so, what is iC then?

The Attempt at a Solution

a) jD = iD/A = 0.280/[pi*(0.04)^2] = 55.7 A/m

b) dE/dt = jD/[sigma_0] = 55.7/(8.85*10^-12] = 6.3*10^12

c) B = ([u_o]/2pi)*(r/R^2)*iC where r = radius, R = distance

 

[alphysicist]

Hi aznkid310,

I believe part c is incorrect. The formula looks correct in its form, but the identification of r and R seem to be swapped. (R needs to be the radius of the plates (4cm).) Can you post some details about how you derived your answer for part c?

 

[aznkid310]

Using ampere's law: integral[ B * dl ] = B(2pi*r) = jD*A = u_0(r^2/r^2)iC

B = (u_0/2pi)(r/R^2)iC

 

[alphysicist]

Using ampere's law: integral[ B * dl ] = B(2pi*r) = jD*A = u_0(r^2/r^2)iC

B = (u_0/2pi)(r/R^2)iC

That's right (except you're missing an R in your first equation); so r is the distance of 2 cm, and R is the radius of 4 cm. Your original post had those values swapped:

c) B = ([u_o]/2pi)*(r/R^2)*iC where r = radius, R = distance

but perhaps it was just a mistake in typing?