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Approximation of electric field of uniform charged disk

  • Thread starter yoran
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  • #1
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Hi,

Homework Statement


The electric field of a uniform charged disk at a point on its axis at a distance x from the disk is given by
[tex]E = 2k_e\pi\sigma(1-\frac{x}{\sqrt{x^2+R^2}})[/tex]
where R the radius of the disk and [tex]\sigma[/tex] the surface charge density.
In my notes it says that when [tex]x\gg R[/tex], that is when the distance x to the disk is much bigger than the radius of the disk, then
[tex]E\approx k_e\frac{Q}{x^2}[/tex]
with the Q the total charge on the disk. How do they come to that result?

Homework Equations



The Attempt at a Solution


When [tex]x\gg R[/tex], then [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}=x[/tex]. But then I get that [tex]E=0[/tex], which is not obviously not correct.
I guess the approximation that [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}[/tex] is wrong. But how is the approximation then?

Thank you.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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When [tex]x\gg R[/tex], then [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}=x[/tex]. But then I get that [tex]E=0[/tex], which is not obviously not correct.
I guess the approximation that [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}[/tex] is wrong. But how is the approximation then?

Thank you.
Hi yoran! :smile:

Always divide so that you get a "1" at the beginning:

√(x²/(x² + R²)) = 1/√(1 + R²/x²) ≈ 1 - R²/2x². :smile:
 
  • #3
Also Yoran your mistake was that basically you did a 0th order expansion!

If you are not comfy with binomial expansions, you can always MacLaurin expand the function in R/x. You are still using the same thing though-- expanding in R/x instead of x because you want a variable that is small i.e. much less than one (so you can truncate the expansion and still have it legitimately approximate the original function), which R/x satisfies, while just straight up x is very large.
 
  • #4
118
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Allright then thanks a lot!
 

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