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

by zimo
Tags: coulomb, stretching
 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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P: 41,326
 Quote by zimo 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 z²?

 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.
P: 45
 Quote by Doc Al Are you sure that z isn't z²? 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...

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P: 41,326
How did it get to that? Stretching Coulomb's law...

 Quote by zimo 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:
$$\sqrt{z^2 + R^2} = z\sqrt{1 + (R/z)^2}$$

Since R/z << 1, can you see how to use a binomial approximation now?
P: 45
 Quote by Doc Al I'd say that was wrong. What book are you using?
This one, third page.

 Quote by Doc Al For example: $$\sqrt{z^2 + R^2} = z\sqrt{1 + (R/z)^2}$$ 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?
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P: 41,326
 Quote by zimo 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 + R²/z²) ≈ 1 + ½R²/z²

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