Parallel plate capacitor and magnetic field

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Homework Help Overview

The discussion revolves around a parallel plate capacitor being charged, with specific focus on calculating the displacement current density, the rate of change of the electric field, and the induced magnetic fields at various distances from the axis. The problem involves concepts from electromagnetism, particularly relating to capacitors and magnetic fields.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts calculations for displacement current density, electric field change, and induced magnetic fields, but expresses uncertainty about the correctness of their results.
  • Some participants question the exponent used in the calculations and suggest corrections for the equations applied.
  • Others inquire about the application of Ampere's law and the inclusion of both conduction and displacement currents in the calculations.
  • Further suggestions involve clarifying the use of area in the equations for the rate of change of the electric field.

Discussion Status

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for the problem. The original poster has indicated that some of their answers are incorrect, prompting further exploration of the underlying principles.

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Homework Statement


A parallel plate air-filled capacitor is being charged as in figure. The circular plates have a radius of 4.0 cm, and at a particular instant the conduction current in the wires is .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 2.0 cm from the axis?
D. What is the induced magnetic field between the plates at a distance of 1.0 cm from the axis?


Homework Equations


an image is attached



The Attempt at a Solution


This is my solution as is...

A. i = .280 A
r = 4cm (.04m)
n = 3.14
jD=i/nr^2 (thats n*r squared)
=.28/3.14*.04^2
= 55.73 A

B. dE/dt=jD/ [tex]\epsilon[/tex] (permitivitty constant)
= 55.73/8.85*10^-12
= 6.3*10^-12 N/Cs

C. B= ur/2nR^2 *i (u = mu) = (1.256*10^-6)(.04m)/2(3.14)(.02*.02)
= 2*10^-5 (.28)
= 5.6*10^-6

D. B=ur/2nR^2 (i) =(1.256*10^-6)(.04)/2(3.14)(.01*.01) (.28)
= 2.24*10^-13

answer a is correct but b, c, and d are not. I have tried this multiple times and can't figure out where I am going wrong. Any help would be appreciated.
 

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Your exponent for part b is wrong... it should be +12.
 
I don't understand how you're doing c and d. Did you try to apply Ampere's law? Can you show the details?

You should use the conduction current + displacement current.
 
Also for part b), the equation should be:

[tex]\frac{d\phi}{dt}=\frac{jD}{\epsilon}[/tex]

[tex]\frac{d(EA)}{dt}=\frac{jD}{\epsilon}[/tex]

[tex]A\frac{dE}{dt}=\frac{jD}{\epsilon}[/tex]

You know the area [tex]\pi{R}^2[/tex] where R = 0.04m, so you can calculate dE/dt.
 
correction: using B= (uo id/2piR^2) r
to find induced magnetic field inside a circular capacitor.

uo=mu
id=displacement current
pi=3.14
R^2= radius of plate squared
r= radius from center (point inside capacitor)

I have values
uo=1.256810^-6 (permeability constant)
id= .280
R= 4 cm
r(c)= 2cm
r(d)= 1 cm

plugging in the given values i get
part c= 7*10^-9
part d= 3.5*10^-9

does this look correct?
 

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