Electric field generated by a charged wire

[Renaldo]

Homework Statement

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.)

Homework Equations

E = k|q|/r²

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

The Attempt at a Solution

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

For the y vector:

dE_y = 2ak(dq)/(y²+x²)^3/2

dq/dy = Q/2a

∫dE_y = (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.

 

[haruspex]

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

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.

 

[Renaldo]

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.

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

E_y = kqy/[(x²+y²)^3/2]

dq = λdy

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

 

[haruspex]

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

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

 

[Renaldo]

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

=-2kλ/[(x²+a²)^1/2]

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

 

[Renaldo]

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

 

[haruspex]

=-2kλ/[(x²+a²)^1/2]

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

... and a definite integral has two bounds.

 

[Renaldo]

... and a definite integral has two bounds.

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

-2kλ/[(a²+x²)^1/2]+2kλ/x

 

[Renaldo]

Here is an attachment showing all of my work.

 

[haruspex]

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

-2kλ/[(a²+x²)^1/2]+2kλ/x

That's what I get.

 

[Renaldo]

That's what I get.

Yes, this worked. Thanks for your help.