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Evaluating electric field

  • Thread starter Niles
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[SOLVED] Evaluating electric field

Homework Statement


I have an electric field in the z-direction given by: (where sigma is charge per area and z is a distance)

[tex]
{\bf{E}} = \frac{{\sigma z}}{{2\pi \varepsilon _0 }}\left( {\frac{1}{z} - \frac{1}{{\sqrt {R^2 + z^2 } }}} \right){\bf{z}}
[/tex]

I have to evaluate this for z >> R.

The Attempt at a Solution



Do I just insert R=0 or what? I overheard someone talk about Taylor-expanding it, but I don't see how/why?
 

Answers and Replies

  • #2
George Jones
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Take a factor [itex]z^2[/itex] out of the [itex]R^2 + z^2[/itex] term.
 
  • #3
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Thanks, but why?

What is the method when I am asked to evalute an expression for some variable >> some other variable?
 
  • #4
George Jones
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Thanks, but why?

What is the method when I am asked to evalute an expression for some variable >> some other variable?
After you do it, I'll give the motivation for the step.
 
  • #5
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I get 1/R instead of 1/sqrt(R^2+z^2).
 
  • #6
George Jones
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I get 1/R instead of 1/sqrt(R^2+z^2).
I was looking for

[tex]R^2 + z^2 = z^2 \left( \frac{R^2}{z^2} + 1 \right).[/tex]

What can you say about the first term inside the brackets?
 
  • #7
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It goes to 0 if z >> R.
 
  • #8
George Jones
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It goes to 0 if z >> R.
Right.

Now you something of the form

[tex]\left( 1 + a \right)^{-1/2},[/tex]

with [itex] |a| << 1. [/itex]

Can you write down the Taylor series expansion of the above expression?
 
  • #9
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I think I get:

1-x/2 - the first two orders. How does that sound?
 
  • #10
George Jones
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I think I get:

1-x/2 - the first two orders. How does that sound?
Yes.

Now use all this in the original expression.
 
  • #11
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Then I get:

1/z - (1-z/2).

This is for inside the brackets. Then I multiply out and finish?
 
  • #12
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What is the motivation for the step?
 
  • #13
Redbelly98
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Then I get:

1/z - (1-z/2).

This is for inside the brackets. Then I multiply out and finish?
Not quite. Recall that the "x" in your earlier equation is actually R/z.
Also, you have lost a factor of [tex]z^2[/tex] somewhere (the one that appears outside the brackets on the right-hand-side in George's post #6).
 
  • #14
Redbelly98
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What is the motivation for the step?
The motivation is to get something other than zero when you approximate this part of the expression:

[tex]
\left( {\frac{1}{z} - \frac{1}{{\sqrt {R^2 + z^2 } }}} \right)
[/tex]
 
  • #15
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Cool, I get it now. Thanks for being so kind.
 

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