# Electric field generated by a charged wire

1. Feb 8, 2013

### Renaldo

1. The problem statement, all variables and given/known data

A charge per unit length +λ is uniformly distributed along the positive y-axis from y = 0 to y = +a. A charge per unit length −λ is uniformly distributed along the negative y-axis from y = 0 to y = −a. Write an expression for the electric field at a point on the x-axis a distance x from the origin. (Use the following as necessary: k, λ, x, and a.)

2. Relevant equations

E = k|q|/r2

I derived this equation by treating the ends of the wire as point charges.
Ey = 2kqa/(y2+x2)3/2

3. The attempt at a solution

I quickly determined that the field vectors x and k would be 0.

For the y vector:

dEy = 2ak(dq)/(y2+x2)3/2

dq/dy = Q/2a

∫dEy = (4akQ/2a)∫dy/(y2+x2)3/2
= 2kQy/[x2(y2+x2)1/2]

2kλy/[x2(y2+x2)1/2]

The limits of integration are from 0 to a.

I am pretty sure this is not correct, and have been working at it for a while, so I would appreciate some help.

2. Feb 9, 2013

### haruspex

On the RHS, part of the expression is valid for any y in the range, but part of it is only valid at the endpoints. You later treat it as though it's valid over all y.

3. Feb 9, 2013

### Renaldo

I think I understand what you are talking about. Here is my revised process.

Ey = kqy/[(x2+y2)3/2]

dq = λdy

dEy = kλydy/[(x2+y2)3/2]

The limits of integration are from -a to a. I choose to integrate from 0 to a and multiply the result by two.

2∫dEy = 2kλ∫ydy/[(x2+y2)3/2]
= -2kλ/[(x2+y2)1/2]
=-2kλ/[(x2+a2)1/2]

4. Feb 9, 2013

### haruspex

Better. But it's a definite integral. y should have turned into a.

5. Feb 9, 2013

### Renaldo

=-2kλ/[(x2+a2)1/2]

How much better? Am I on the right track or am I already there?

6. Feb 9, 2013

### Renaldo

Actually, I just input that answer to webassign and it didn't accept it. I am kind of frustrated. lol

7. Feb 9, 2013

### haruspex

... and a definite integral has two bounds.

8. Feb 9, 2013

### Renaldo

ok. New answer for limits of integration 0 to a:

-2kλ/[(a2+x2)1/2]+2kλ/x

9. Feb 9, 2013

### Renaldo

Here is an attachment showing all of my work.

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• ###### CCI02092013_00000.jpg
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10. Feb 9, 2013

### haruspex

That's what I get.

11. Feb 9, 2013

### Renaldo

Yes, this worked. Thanks for your help.