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Electric Fields of straight wire

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1. Homework Statement
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. Homework 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?
 

Answers and Replies

Kurdt
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Looks fine you just need to plug the numbers in.
 
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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)
 
Kurdt
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E will just be 2.57 as stated in the problem. Why are you worried about lambda?
 
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I wasn't sure if lambda = 1.47*10^-10
 
Kurdt
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probably more likely to be x10-10 than the other way round.
 
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Kurdt
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What is it in the question?
 
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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.
 
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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??
 
Kurdt
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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:

[tex] r=\frac{2k\lambda}{E} [/tex]

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 [tex] k=\frac{1}{4\pi \epsilon_0} [/tex]
 
Last edited:
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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.
 
Kurdt
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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.
 
Kurdt
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Where did you get that value of epsilon nought from? It should be: 8.85x10-12
 
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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 =[
 
Kurdt
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