# Electric Fields

by stylez03
Tags: electric, fields
 P: 139 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?
 Emeritus Sci Advisor PF Gold P: 4,975 Looks fine you just need to plug the numbers in.
 P: 139 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)
Emeritus
PF Gold
P: 4,975

## Electric Fields

E will just be 2.57 as stated in the problem. Why are you worried about lambda?
 P: 139 I wasn't sure if lambda = 1.47*10^-10
 Emeritus Sci Advisor PF Gold P: 4,975 probably more likely to be x10-10 than the other way round.
P: 139
 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
 Emeritus Sci Advisor PF Gold P: 4,975 What is it in the question?
P: 139
 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.
 P: 139 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??
 Emeritus Sci Advisor PF Gold P: 4,975 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 $$k=\frac{1}{4\pi \epsilon_0}$$
 P: 139 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.
 Emeritus Sci Advisor PF Gold P: 4,975 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.
 Emeritus Sci Advisor PF Gold P: 4,975 Where did you get that value of epsilon nought from? It should be: 8.85x10-12
P: 139
 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 =[
 Emeritus Sci Advisor PF Gold P: 4,975 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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