# Evaluating electric field

1. Apr 30, 2008

### Niles

[SOLVED] Evaluating electric field

1. The problem statement, all variables and given/known data
I have an electric field in the z-direction given by: (where sigma is charge per area and z is a distance)

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

I have to evaluate this for z >> R.

3. 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?

2. Apr 30, 2008

### George Jones

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

3. Apr 30, 2008

### Niles

Thanks, but why?

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

4. Apr 30, 2008

### George Jones

Staff Emeritus
After you do it, I'll give the motivation for the step.

5. Apr 30, 2008

### Niles

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

6. Apr 30, 2008

### George Jones

Staff Emeritus
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?

7. Apr 30, 2008

### Niles

It goes to 0 if z >> R.

8. Apr 30, 2008

### George Jones

Staff Emeritus
Right.

Now you something of the form

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

with $|a| << 1.$

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

9. Apr 30, 2008

### Niles

I think I get:

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

10. Apr 30, 2008

### George Jones

Staff Emeritus
Yes.

Now use all this in the original expression.

11. Apr 30, 2008

### Niles

Then I get:

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

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

12. May 1, 2008

### Niles

What is the motivation for the step?

13. May 2, 2008

### Redbelly98

Staff Emeritus
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).

14. May 2, 2008

### Redbelly98

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

$$\left( {\frac{1}{z} - \frac{1}{{\sqrt {R^2 + z^2 } }}} \right)$$

15. May 3, 2008

### Niles

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