Two infinite rods with charges

[Eric_meyers]

Homework Statement

Two infinite parallel rods are separated a distance 2d. One carries a uniformly distributed positive charge (lambda), the other an opposite charge (-lambda) the rods are parallel to the axis and intersect the x-y plane at x=0 y = +/- d. Find the electric field for a point on the positive x axis. How does E behavor for x>>d?

Homework Equations

E = 1/(2pi*epsilion0) * (lambda) / r

The Attempt at a Solution

Ok, so my solution is that the r = (x^2 + d^2)^1/2 using Pythagoreans theorem from any point down the x-axis to the two rods.

I'm also guessing that the two electric fields don't cancel each other and that actually they add to one another since the difference between the two rods is 2 * lambda so if I add lambda to both rods my negative lambda goes to zero and my positive 1 lambda goes to two..

So my answer for part one was

2[1/(2pi * epsilion0) * (lambda)/(x^2 + d^2)^1/2]

and then for x >> d I just took out the d term

2[1/(2pi * epsilion0) * (lambda)/x]

 

[Delphi51]

2[1/(2pi * epsilion0) * (lambda)/(x^2 + d^2)^1/2]

Doesn't look quite right. You should have a nice diagram showing the circular cross sections for the two wires, one a distance 2d below the other. From the middle of these, off to the right a distance x, you have your point P where you will find the E field. At P, there should be an E arrow going away from the positive wire and another E arrow going toward the negative wire. You need to add these two E vectors together. There is some symmetry that will simplify things but you will definitely have a sine or cosine in your final answer (which can be expressed in terms of d and x).

 

[w3390]

I believe after you draw the picture you will realize that the net E will point straight in the -y direction just using symmetry, if I understand the problem correctly. So I guess the sine or cosine Delphi51 was talking about will just be 1.

 

[Delphi51]

No, it won't be 1. The E's are at an angle from the y-axis that varies with x and d.

 

[w3390]

Yes, they do vary with x and d but after an infinite number of E vectors are drawn from an infinite number of dq's on the rod, the net E vector should point in the -y hat direction.

 

[Delphi51]

I agree that the total points in the -y direction. But there are only two E vectors, one from each line charge. That formula already includes all the infinite number of dq's on the line.

So, you only need to find the y component for each of the two E's.
However, these E's do not point in the y direction so you must do some trig to find the y component in each case.

 

[w3390]

Ok, I understand what you are saying now and it is correct.

 

[mrkmrk]

Could someone help me out with this? I can't, for the life of me, figure out where to even begin. I've got a great professor, and I understand everything he says in class, but when it comes to actually solving homework problems, I don't know what to do. Any help would be greatly, greatly appreciated.

EDIT: Like the OP, I'm thinking that the fields don't cancel, but add. Also, I don't understand why the field is in the y-direction... I have the same equation, but I can't really think of how to describe 'r' because there are two rods producing fields...

 

[LowlyPion]

Could someone help me out with this? I can't, for the life of me, figure out where to even begin. I've got a great professor, and I understand everything he says in class, but when it comes to actually solving homework problems, I don't know what to do. Any help would be greatly, greatly appreciated.

EDIT: Like the OP, I'm thinking that the fields don't cancel, but add. Also, I don't understand why the field is in the y-direction... I have the same equation, but I can't really think of how to describe 'r' because there are two rods producing fields...

Welcome to PF.

You are right the y-components of the sum of the e-fields do add. But the x components cancel. Since the y component basically vanishes at x>>d (due to the fact that you are dealing not only with the diminishing y component proportional to sinθ, which is d/x, but also the 1/R relationship).

 

[mrkmrk]

I don't understand how the x-components cancel... I could see that being the case if the rods had equal charges, but opposite? Wouldn't that make the field go doubly in the direction of negative charge?

 

[LowlyPion]

I don't understand how the x-components cancel... I could see that being the case if the rods had equal charges, but opposite? Wouldn't that make the field go doubly in the direction of negative charge?

If they were the same charge then the y-components would cancel.

But they are opposite. When you are on the x axis, one points toward the other away. Hence they cancel as they are equidistant.

(Note: the arrangement is ± d on the y-axis.)

 

[mrkmrk]

Oh, I think I get that, then. So, if the fields add in the y-direction, how to I denote that?

 

[LowlyPion]

Oh, I think I get that, then. So, if the fields add in the y-direction, how to I denote that?

From any point x on x, the y-component will be sinθ*|E|.

Sinθ = d/x

 

[mrkmrk]

I don't mean to sound like an idiot, but that's for /both/ charges? How does one come to that conclusion?

 

[mrkmrk]

I'm still trying... this is for one pole.

E = K*Lambda*Integrate[r-hat/r^2,{r,-Infinity,Infinity}]
E = K*Lambda*Integrate[<0,y,0>/(y*r^2),{r,-Infinity,Infinity}]
E = K*Lambda*Integrate[<y>/(((0^2+y^2+0^2)^(1/2))*r^2),{r,-Infinity,Infinity}]
E = K*Lambda*Integrate[<y>/(y*r^2),{r,-Infinity,Infinity}]

r = (x^2+d^2)^(1/2)

E = K*Lambda*Integrate[<y>/(x^2+d^2),{d,-Infinity,Infinity}]

This comes out to (K*Lambda*Pi/x^2)<y>. If I subtract the other E-field (subtract, as positives mut subtract to cancel, and I know these two add, so...), I get...

(K*Lambda*Pi/x^2)<y> - (K*-Lambda*Pi/x^2)<y> =2K*Lambda*Pi/x^2

Is that right? If not, where have I gone wrong, and by what terrible degree?

EDIT: Can't be right. Y in the denominator up and flew away, it would seem.

 

[LowlyPion]

Look at the problem as 2 charges, 1 + and the other -.

Now examine a point on the ⊥ bisector and write an expression for the field contribution of both charges at once.

E = ∑ E = ∑ i + ∑ j

From one charge you have (+i,+j) for the other you have (-i,-[-]j) = (-i,+j)

When you sum the pair wise contribution at that point you should recognize then that so long as you are on the ⊥ bisector, the components +,- along the direction of the bisector cancel. That means you can represent the field at that point as being

2*Sinθ*|E| j = 2*d/r*|k*q/r| j = 2*d*k*q/r² j

where Sinθ can be represented far away as d/r (small angle). (i.e. d << r )

Otherwise it is d/(d² + r²)^½