Displacement Current and air space

  • Thread starter aznkid310
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  • #1
aznkid310
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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
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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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?
 
  • #3
aznkid310
109
1
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
 
  • #4
alphysicist
Homework Helper
2,238
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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?
 

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