Electric Fields of straight wire

In summary, the homework statement is that 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. The Attempt at a Solution states that if lambda = 1.47 x 10^-10, then r = 972. However, when evaluated, r actually equals 1.03×10−3.
  • #1
stylez03
139
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?
 
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  • #2
Looks fine you just need to plug the numbers in.
 
  • #3
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)
 
  • #4
E will just be 2.57 as stated in the problem. Why are you worried about lambda?
 
  • #5
I wasn't sure if lambda = 1.47*10^-10
 
  • #6
probably more likely to be x10-10 than the other way round.
 
  • #7
Kurdt said:
probably more likely to be x10-10 than the other way round.

Thats what I had before 1.47 x 10^-10
 
  • #8
What is it in the question?
 
  • #9
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:

[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:
  • #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 =[
 
  • #16

1. What is an electric field?

An electric field is a region around a charged object in which other charged objects experience a force. It is a fundamental concept in physics that helps explain the behavior of electrically charged particles.

2. How is an electric field created by a straight wire?

An electric field is created by a straight wire when it has a net charge, either positive or negative. This charge creates an electric field that extends outwards from the wire and can interact with other charged objects.

3. How does the strength of an electric field change with distance from the wire?

The strength of an electric field created by a straight wire decreases as the distance from the wire increases. This is because the electric field follows an inverse square law, meaning that it becomes weaker as the distance increases.

4. Can electric fields from multiple straight wires interact with each other?

Yes, electric fields from multiple straight wires can interact with each other. When two or more wires with net charges are placed close to each other, their electric fields can overlap and create a combined electric field that is the sum of the individual fields.

5. How can the direction of an electric field around a straight wire be determined?

The direction of an electric field around a straight wire is determined by the direction of the force that a positive test charge would experience if placed at a certain point. The direction of the electric field is always perpendicular to the wire and points towards or away from the wire, depending on its charge.

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