# Electric Fields of straight wire

1. Jan 26, 2007

### stylez03

1. The problem statement, all variables and given/known data
A very long, straight wire has charge per unit length 1.47×10^10

At what distance from the wire is the electric field magnitude equal to 2.57 N/C

2. Relevant equations

E = lambda / (2*pi*E_o*r)

E_o = 8.85*10^-9

3. The attempt at a solution

2*pi*E_o*E / lambda = r

Is this correct so far?

2. Jan 26, 2007

### Kurdt

Staff Emeritus
Looks fine you just need to plug the numbers in.

3. Jan 26, 2007

### stylez03

r = 2*pi*(8.85*10^-9)*(2.57) / (1.47*10^-10)

I'm not sure if lambda is represented correctly and is E just 2.57 or should it be 10^(something)

4. Jan 26, 2007

### Kurdt

Staff Emeritus
E will just be 2.57 as stated in the problem. Why are you worried about lambda?

5. Jan 26, 2007

### stylez03

I wasn't sure if lambda = 1.47*10^-10

6. Jan 26, 2007

### Kurdt

Staff Emeritus
probably more likely to be x10-10 than the other way round.

7. Jan 26, 2007

### stylez03

Thats what I had before 1.47 x 10^-10

8. Jan 26, 2007

### Kurdt

Staff Emeritus
What is it in the question?

9. Jan 26, 2007

### stylez03

The original question was if lambda = 1.47 x 10^-10.

You said yes, so I just wanted to make sure.

10. Jan 26, 2007

### stylez03

2*pi*(8.85*10^-9)*(2.57) / (1.47*10^-10) = r

This evaluated to:
r = 972

The online program says I'm off by an additive constant??

11. Jan 26, 2007

### Kurdt

Staff Emeritus
I've just noticed you have E on top of the fraction and lambda below. You need to swap these two so the equation is:

$$r=\frac{2k\lambda}{E}$$

Like I said in a previous thread, try manipulating equations with just their symbols until the very last moment. Its a lot easier to spot problems that way.

EDIT: Sorry $$k=\frac{1}{4\pi \epsilon_0}$$

Last edited: Jan 26, 2007
12. Jan 26, 2007

### stylez03

2* (1/4*pi* 8.85*10^-9) * (1.47*10^-10) / 2.57

= 1.03×10−3

Still says I'm off by a additive constant.

13. Jan 26, 2007

### Kurdt

Staff Emeritus
I think you've made a mistake in the calculation as i get a different answer. Try it again you're 3 orders of magnitude out.

14. Jan 26, 2007

### Kurdt

Staff Emeritus
Where did you get that value of epsilon nought from? It should be: 8.85x10-12

15. Jan 26, 2007

### stylez03

I need to be more careful from paper to online input. I have so much scratch work, some how I changed the epsilon value =[

16. Jan 26, 2007

### Kurdt

Staff Emeritus