Potential of charged disk for x R

  • Thread starter qamptr
  • Start date
  • #1
10
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I understand the answer to this part of the problem intuitively but there's definitely something in the math that I'm missing--

Homework Statement


Along the axis through the center of a charged disk, for a point P at a distance x from the center, the potential will be

[tex]\frac{Q}{2\pi\epsilon_{0}R^{2}}\left[\sqrt{R^{2}+x^{2}}-x\right][/tex]

Homework Equations


For x>>R, I'm supposed to understand that this reduces to

[tex]V\approx\frac{Q}{2\pi\epsilon_{0}R^{2}}\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right]=\frac{Q}{4\pi\epsilon_{0}x}[/tex]

The Attempt at a Solution


As I said, I understand that this must work out, but I'm misunderstanding something about the math. It seems as though the term under the radical above ought to become just [tex]x^{2}[/tex] for x>>R, and thus inside the brackets x-x=0 and then V=0, which is untrue. Could someone please explain how [tex]\left[\sqrt{R^{2}+x^{2}}-x\right][/tex] becomes [tex]\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right][/tex] ?
 

Answers and Replies

  • #2
548
2
Could someone please explain how [tex]\left[\sqrt{R^{2}+x^{2}}-x\right][/tex] becomes [tex]\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right][/tex] ?

Bring the x^2 term outside the radical, then apply the general binomial expansion formula for (1+y)^(1/2) = 1 + (1/2)y - ....
 
  • #3
1,860
0
Binomial approximation will get you there.

[tex]\sqrt{R^2+x^2}=\sqrt{\frac{R^2 x^2}{x^2}+x^2}[/tex]

Now use the approx
[tex]x \sqrt{\frac{R^2}{x^2}+1}=x(1+\frac{1}{2} \frac{R^2}{x^2})[/tex]
 
  • #4
10
0
Binomial approximation will get you there.

[tex]\sqrt{R^2+x^2}=\sqrt{\frac{R^2 x^2}{x^2}+x^2}[/tex]

Now use the approx
[tex]x \sqrt{\frac{R^2}{x^2}+1}=x(1+\frac{1}{2} \frac{R^2}{x^2})[/tex]

*thumps forehead* thanks.
 

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