## Electric Fields

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?
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 Recognitions: Gold Member Science Advisor Staff Emeritus Looks fine you just need to plug the numbers in.
 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)

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## Electric Fields

E will just be 2.57 as stated in the problem. Why are you worried about lambda?
 I wasn't sure if lambda = 1.47*10^-10
 Recognitions: Gold Member Science Advisor Staff Emeritus probably more likely to be x10-10 than the other way round.

 Quote by Kurdt probably more likely to be x10-10 than the other way round.
Thats what I had before 1.47 x 10^-10
 Recognitions: Gold Member Science Advisor Staff Emeritus What is it in the question?

 Quote by Kurdt What is it in the question?
The original question was if lambda = 1.47 x 10^-10.

You said yes, so I just wanted to make sure.
 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??
 Recognitions: Gold Member Science Advisor 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 $$k=\frac{1}{4\pi \epsilon_0}$$
 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.
 Recognitions: Gold Member Science Advisor 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus Where did you get that value of epsilon nought from? It should be: 8.85x10-12

 Quote by Kurdt Where did you get that value of epsilon nought from? It should be: 8.85x10-12
I need to be more careful from paper to online input. I have so much scratch work, some how I changed the epsilon value =[
 Recognitions: Gold Member Science Advisor Staff Emeritus No problem. Try a bit of latex for your equations it should help in future to diagnose problems quicker http://www.physicsforums.com/showthread.php?t=8997

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