Electric Fields

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?

Kurdt

Looks fine you just need to plug the numbers in.

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)

Kurdt

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

stylez03

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

Kurdt

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

stylez03

Thats what I had before 1.47 x 10^-10

Kurdt

What is it in the question?

stylez03

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

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

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??

Kurdt

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]

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.

Kurdt

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

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

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 =[

Kurdt

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