Electric field at a distance from a charged disk

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Homework Help Overview

The discussion revolves around calculating the electric field at a distance from a charged disk with a specified radius and surface charge density. Participants are exploring the appropriate equations and methods to arrive at the correct electric field value.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations for calculating the electric field, including those for a charged disk and an infinite sheet of charge. There are questions about the correct substitution of variables, particularly regarding the surface charge density and the area of the disk.

Discussion Status

There is ongoing dialogue about the correct application of formulas and the values being used. Some participants have provided links to external resources, and there is an acknowledgment of confusion regarding the calculations. The discussion is active, with participants seeking clarification on their approaches and assumptions.

Contextual Notes

Participants note the absence of a figure for the problem and mention that the thread was moved to a homework help forum after initial discussion. There is also a reminder about the proper forum for such requests.

kb1408
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A disk of radius 2.4 cm carries a uniform surface charge density of 3.1 μ C/m2. Using reasonable approximations, find the electric field on the axis at the following distances.

I have used the equation E=(Q/ε0)(1/(4*pi*r2))
I also tried the equation E=(Q/2(ε0))(1-(z/(√(z2)+(r2)))

Thanks in advance for the help. Both equations have not led me to the correct answer.

*note, there is not a figure provided for this question*
 
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kb1408 said:
I have used the equation E=(Q/ε0)(1/(4*pi*r2))
This looks like the field from an infinite sheet of charge. You should write it as σ/2ε, where σ is the surface charge density. Not what you want.
I also tried the equation E=(Q/2(ε0))(1-(z/(√(z2)+(r2)))
That's the one you want, but you need to replace Q with σ.
 
Doc Al, thanks for your quick reply. Unfortunately I am still doing something wrong. I am using 3.1E-6 C/m2 for σ. Is that wrong?
 
How about you post all your working and the target answer?
 
E= (3.1E-6/(2*8.85E-12))(1-((.0001/(√(.00012)+(.0242))
so E= 1.79E5 N/C
where:
σ=3.1E-6 C/m2
ε0=8.85E-12 C2/N*m2
z=.01E-2 m
r= 2.4E-2 m
 
Did you simply replace Q by σ? What about the disc area?
 
Yes, that's what I did. Is it σ=Q/A then?
 
Yes, as in the link andrien posted.
 
  • #10
There was an advertisement above that link earlier so I ignored it thinking it was tied to that ad.

Alright, thanks for your help!

cheers!
 
  • #11
And thank you andrien for the link!
 
  • #12
Note: This thread had already developed quite a bit before I noticed that it really should have been in one of the homework help forums. Therefore I've simply moved it instead of deleting it and asking the original poster to start over, which is the normal practice.

In the future, please post requests for help on specific exercises like this in one of the homework help forums, even if they're not actually assignments for a class. The "normal" forums are more for conceptual questions and general discussion of their topics.
 

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