Parallel plate capacitor and magnetic field

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SUMMARY

The discussion focuses on calculating the displacement current density, the rate of change of the electric field, and the induced magnetic fields in a parallel plate capacitor with a radius of 4.0 cm and a conduction current of 0.280 A. The correct formula for displacement current density (jD) is derived as jD = i/(πr²), yielding a value of 55.73 A/m². The rate of change of the electric field (dE/dt) is calculated using the permittivity constant, resulting in 6.3 x 10^-12 N/Cs. The induced magnetic fields at distances of 2.0 cm and 1.0 cm from the axis are found to be 7 x 10^-9 T and 3.5 x 10^-9 T, respectively, using Ampere's law.

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  • Understanding of Maxwell's equations
  • Familiarity with the concepts of displacement current and conduction current
  • Knowledge of the permittivity and permeability constants
  • Ability to apply Ampere's law in electromagnetic contexts
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  • Study the derivation and application of Maxwell's equations in capacitors
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  • Explore the use of Ampere's law in calculating magnetic fields in various geometries
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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/ \epsilon (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:

\frac{d\phi}{dt}=\frac{jD}{\epsilon}

\frac{d(EA)}{dt}=\frac{jD}{\epsilon}

A\frac{dE}{dt}=\frac{jD}{\epsilon}

You know the area \pi{R}^2 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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