How did it get to that? Stretching Coulomb's law...


by zimo
Tags: coulomb, stretching
zimo
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#1
Jun13-11, 02:22 PM
P: 45
1. The problem statement, all variables and given/known data

The charge is distributed with uniform
surface density σ on the disk of radius R. Find the potential
at the axis of the disk.

2. Relevant equations

Coulomb's law and the definition of a electric potential at point x

3. The attempt at a solution

I have a solution in front of me but can't understand some step inside it:

The potential can be defined now

phi(x)= (1/4pi*epsilon0)Integral[(sigma(x')/|x-x'|)dS'] and the solution for the integral from 0 to R is:

(sigma/2*epsilon0)(sqrt(x^2+R^2)-z)

Now, the electric field at this point is:

E(z)=(sigma/2*epsilon0)(1-(z/sqrt(z^2+R^2))

I can clearly follow until now, but then the book says that for z>>R we get

E=Q/(4pi*epsilon0*z)

where Q is the total charge of the disc.

How can it be proportional to 1/z?? when I take z>>R - I get E(z)=0...
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Doc Al
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Jun13-11, 02:52 PM
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Quote Quote by zimo View Post
I can clearly follow until now, but then the book says that for z>>R we get

E=Q/(4pi*epsilon0*z)
Are you sure that z isn't z?

How can it be proportional to 1/z?? when I take z>>R - I get E(z)=0...
The point is to see how the field drops off. Hint: Do a binomial expansion.
zimo
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Jun14-11, 12:44 AM
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Quote Quote by Doc Al View Post
Are you sure that z isn't z?


The point is to see how the field drops off. Hint: Do a binomial expansion.
1. Yeah, I'm sure. I double checked it.

2. what kind of a binomial expansion can I possibly make here? please direct me some more...

Doc Al
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Jun14-11, 04:09 AM
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How did it get to that? Stretching Coulomb's law...


Quote Quote by zimo View Post
1. Yeah, I'm sure. I double checked it.
I'd say that was wrong. What book are you using?

2. what kind of a binomial expansion can I possibly make here? please direct me some more...
For example:
[tex]\sqrt{z^2 + R^2} = z\sqrt{1 + (R/z)^2}[/tex]

Since R/z << 1, can you see how to use a binomial approximation now?
zimo
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Jun14-11, 05:58 AM
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Quote Quote by Doc Al View Post
I'd say that was wrong. What book are you using?
This one, third page.

Quote Quote by Doc Al View Post
For example:
[tex]\sqrt{z^2 + R^2} = z\sqrt{1 + (R/z)^2}[/tex]

Since R/z << 1, can you see how to use a binomial approximation now?
You mean the approximation is ~= 1+z^2/2*R^2?
Doc Al
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#6
Jun14-11, 06:36 AM
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Quote Quote by zimo View Post
This one, third page.
OK, but I think it's a typo. As you get far enough away, its field should be that of a point charge.
You mean the approximation is ~= 1+z^2/2*R^2?
Something like that: √(1 + R/z) ≈ 1 + R/z


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