# E Field Due to Infinite Lines of Charge Where r --> 0

1. Feb 2, 2016

### Jake 7174

1. The problem statement, all variables and given/known data
Two infinitely long lines of uniform charge λ lay parallel on the xy plane (0, ±a) What is max E field in the xz plane.

No values are given. Symbolic answer is expected.

2. Relevant equations
equation for an infinite line of charge
E = λ / ( 2 π ε0 r)

3. The attempt at a solution
My first thought is max E field would be at a point that would be formed by an equilateral triangle. At this point we would have a radius of 2a and the following

E = 2λ / (4 π ε0) = λ/ (2 π ε0 a) (all in +z direction)

I then think that to maximize E field I need to minimize r. So what if I observe at a distance where r approaches 0. At this point I am on the surface of one of the lines. If r = 0 the equation falls apart due to div by 0. If I take the limit as x→0 it goes undefined. If I say r=1 that's great, but 0.5 is better, and 0.005 is even better yet. I understand that I will have some field due to the other wire to figure in with my magnitude but if I am very very close to one of the lines of charge I think this will be insignificant.

Am I conceptualizing this properly?

Is there an equation for E field where r = 0?

2. Feb 2, 2016

### TSny

If you stay in the $xz$ plane, then it is impossible for $r$ to approach zero for either line charge.

3. Feb 2, 2016

### Jake 7174

Point well taken. I guess my original equilateral idea was the way to go.

My question still remains. What is E field on the surface of an object. Everything I have ever seen has r in the denominator. This wont help me with my homework, I just really like this stuff and want to understand it better.

Last edited: Feb 2, 2016
4. Feb 2, 2016

### TSny

I wouldn't try to answer this question based on intuition. I don't think the answer corresponds to an equilateral triangle. Also, your result for the case of an equilateral triangle is incorrect because you did not add the individual fields as vectors.

Use calculus to derive the maximum E.

As you let r approach zero for an ideal line charge, then E would approach infinity. But, the same thing happens for the ideal point charge. In real life, we do not have ideal line charges. We might have a long, straight rod of small radius R with charge uniformly spread over the surface. In that case, the maximum value of E would occur on the surface of the wire where r = R.

5. Feb 2, 2016

### Jake 7174

What vectors did I not account for? Wouldn't everything except for the positive z components cancel due to interaction from the other wire? The closer I get to z = 0 for my reference point wouldn't I have more cancelation and less z magnitude? Conversely, the higher z I have then the larger r is and therefore smaller E. If it is not an equilateral then I really need help because I am lacking in fundamental understanding.

6. Feb 2, 2016

### TSny

Here's what you wrote in your OP:
I don't see anything here that suggests that you used vector addition. Can you explain how you got your first expression for E:

E = 2λ / (4 π ε0 a) ?

(I put in a factor of "a" in the denominator that I think you meant to be there.)

7. Feb 2, 2016

### Jake 7174

I see what you are saying. I need to break it in to components. I simply doubled the equation for infinite lines. I think my solution should then be

sin(60) * λ / (4 π ε0 a)

this is valid for one wire. for the other I should be able to multiply by 2 shouldn't I?

8. Feb 2, 2016

### TSny

Yes, that will give you the net electric field at a point corresponding to an equilateral triangle.
But, you haven't shown that the equilateral triangle corresponds to the maximum electric field. (I don't think it does.)

Pick an arbitrary point on the z axis at a distance z from the origin. Find an expression for the net E at that point. Your result will depend on z. Then you can try to find the value of z that gives the maximum E.

9. Feb 2, 2016

### Jake 7174

you are right. I played with some numbers and found that the max is somewhere around 60% of what the equilateral gives. How do I prove this. The first derivative test comes to mind but that seems kind of painful given both z and the angle change. Do you have any suggestions?

10. Feb 2, 2016

### TSny

Follow the suggestion in my previous post. Express E as a function of z.

11. Feb 2, 2016

### Jake 7174

ok, follow my logic for a second, I think have it.

I recognize that everything but z cancels and r = sqrt(z^2 + a^2). I still need the sine to isolate my z component. I can express sin as [ z / sqrt(z^2 + a^2) ].

If I plug these back in to the expression and after a bit of simplification I will get

E = λz / [π ε0 (z^2 + a^2) ]

I took the derivative with respect to z and got something messy. I set it = 0 and found that slope is 0 at ±a. This should be my max value correct?

12. Feb 2, 2016

### TSny

Good work. You have found the value(s) of z corresponding to max E. You still need to find the value of max E.

13. Feb 2, 2016

### Jake 7174

If I plug a in for z it cleans up nicely

λ / (2πε0a)

14. Feb 2, 2016

### TSny

I believe that's the correct answer.

15. Feb 2, 2016

### Jake 7174

Thank you for your help. It wasn't that bad once I got rid of a poor assumption.