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Parallel plate capacitor and magnetic field

  • Thread starter phys01
  • Start date
  • #1
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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 cant figure out where im going wrong. Any help would be appreciated.
 

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Answers and Replies

  • #2
learningphysics
Homework Helper
4,099
5
Your exponent for part b is wrong... it should be +12.
 
  • #3
learningphysics
Homework Helper
4,099
5
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.
 
  • #4
learningphysics
Homework Helper
4,099
5
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.
 
  • #5
2
0
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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