Parallel plate capacitor and magnetic field

In summary, we have a parallel plate air-filled capacitor with circular plates of radius 4.0 cm being charged. At a specific instant, the conduction current in the wires is .280 A. We then calculated the displacement current density (jD) in the air space between the plates to be 55.73 A. Next, we found the rate at which the electric field between the plates is changing to be 6.3*10^-12 N/Cs. Moving on to the induced magnetic fields, we used the equation B= (uo id/2piR^2) r to calculate the magnetic field at a distance of 2.0 cm from the axis to be 7*10^-9 and at
  • #1
phys01
2
0

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.
 

Attachments

  • phys #2.jpg
    phys #2.jpg
    32.4 KB · Views: 3,922
Last edited:
Physics news on Phys.org
  • #2
Your exponent for part b is wrong... it should be +12.
 
  • #3
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
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
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?
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of electrical component that stores electrical energy by creating an electric field between two parallel conductive plates separated by a dielectric material. It is commonly used in electronic circuits for its ability to store and release energy quickly.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by accumulating electrical charge on its plates, creating an electric field between them. When a voltage is applied to the capacitor, electrons from one plate are attracted to the other plate, creating a potential difference between the two plates. The strength of the electric field depends on the distance between the plates and the amount of charge stored on them.

3. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by three main factors: the distance between the plates, the area of the plates, and the type of dielectric material between the plates. The closer the plates are, the larger the area of the plates, and the higher the dielectric constant of the material, the higher the capacitance will be.

4. Can a parallel plate capacitor store a charge indefinitely?

No, a parallel plate capacitor cannot store a charge indefinitely. Over time, the charge on the plates will dissipate due to leakage currents and the breakdown of the dielectric material. However, a parallel plate capacitor can store a charge for a relatively long time compared to other types of capacitors.

5. How does a magnetic field affect a parallel plate capacitor?

A magnetic field has no direct effect on a parallel plate capacitor. However, if the capacitor is placed in a changing magnetic field, it will experience a force due to the interaction between the electric and magnetic fields. This phenomenon is known as electromagnetic induction and is the basis for many electrical devices such as generators and transformers.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
304
  • Introductory Physics Homework Help
Replies
11
Views
378
  • Introductory Physics Homework Help
2
Replies
58
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
633
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
859
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
903
Back
Top