Evaluating Electric Field at a Distance of z >> R

[Niles]

[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)

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

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?

 

[George Jones]

Take a factor z^2 out of the R^2 + z^2 term.

 

[Niles]

Thanks, but why?

What is the method when I am asked to evalute an expression for some variable >> some other variable?

 

[George Jones]

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.

 

[Niles]

I get 1/R instead of 1/sqrt(R^2+z^2).

 

[George Jones]

I get 1/R instead of 1/sqrt(R^2+z^2).

I was looking for

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

What can you say about the first term inside the brackets?

 

[Niles]

It goes to 0 if z >> R.

 

[George Jones]

It goes to 0 if z >> R.

Right.

Now you something of the form

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

with |a| &lt;&lt; 1.

Can you write down the Taylor series expansion of the above expression?

 

[Niles]

I think I get:

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

 

[George Jones]

I think I get:

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

Yes.

Now use all this in the original expression.

 

[Niles]

Then I get:

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

This is for inside the brackets. Then I multiply out and finish?

 

[Niles]

What is the motivation for the step?

 

[Redbelly98]

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 z^2 somewhere (the one that appears outside the brackets on the right-hand-side in George's post #6).

 

[Redbelly98]

What is the motivation for the step?

The motivation is to get something other than zero when you approximate this part of the expression:

<br /> \left( {\frac{1}{z} - \frac{1}{{\sqrt {R^2 + z^2 } }}} \right)<br />

 

[Niles]

Cool, I get it now. Thanks for being so kind.