Electric Fields of straight wire

AI Thread Summary
The discussion centers on calculating the distance from a charged wire where the electric field magnitude equals 2.57 N/C, using the formula E = λ / (2πE₀r). Participants confirm that the charge per unit length (λ) is correctly represented as 1.47×10^-10. There is confusion regarding the value of the permittivity of free space (E₀), which should be 8.85×10^-12 instead of 8.85×10^-9. After correcting the calculations, one participant finds a significant discrepancy in the results, suggesting a mistake in the calculations. The conversation emphasizes the importance of careful manipulation of equations and double-checking values to avoid errors.
stylez03
Messages
139
Reaction score
0

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


Homework Equations



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

E_o = 8.85*10^-9


The Attempt at a Solution



2*pi*E_o*E / lambda = r

Is this correct so far?
 
Physics news on Phys.org
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)
 
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
 
probably more likely to be x10-10 than the other way round.
 
Kurdt said:
probably more likely to be x10-10 than the other way round.

Thats what I had before 1.47 x 10^-10
 
What is it in the question?
 
Kurdt said:
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.
 
  • #10
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
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:
  • #12
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
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
Where did you get that value of epsilon nought from? It should be: 8.85x10-12
 
  • #15
Kurdt said:
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 =[
 
Back
Top