Homework Help: Charge Distribution along line

1. Jan 24, 2014

NATURE.M

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

Two line charges, of length L/2 and carrying equal and opposite charge density ±λ, are placed on the x-axis so that their ends just touch at the origin, as shown in Figure 1. They are separated by an insulating material with negligible width.

a. Find the magnitude and the direction of the electric fields at points P and P′. P is a point on the y-axis at (0, b), and P′ is a point on the x-axis: (a, 0). Assume b>>L/2

b.Find the electric dipole moment of this line charge system.

3. The attempt at a solution

a. So basically I solved the problem but was a bit confused about the signs.

At point P: For the line charge on the left of the origin, taking differential elements of charge we have dq = -λdx. If my understanding is correct, then the electric field vector at P would be directed towards the line charge, and so the x and y components of electric force point in the negative direction. And so the x and y components of dE are positive. Now, I am considering one to many negative signs? So really I'm just a bit confused about how to treat $\pm$$\lambda$.

Intuitively, if you consider the other line charge on the right of the origin the electric field vector it generates at P should be directed away from the line charge.

But after taking the integral from -L/2 to 0 for the electric field generated by the first line charge, and then taking the integral from 0 to L/2 for the electric field generated by the second line charge, and using b>>L/2, the x component of electric field reduced to ≈0 and the y component is given by: $\frac{λL}{4\pi\epsilon_{0}b^{2}}$. This seems reasonable, but I'm unsure If I got all the signs right as expressed in my previous mark.

Last edited: Jan 24, 2014
2. Jan 25, 2014

haruspex

I would have thought the y component was zero at any b. The y component of the field at P from the positive charge would be equal and opposite to that from the negative charge. If the positive charge is on the positive x axis then the field at P would be parallel to the x axis, pointing in the negative x direction.

3. Jan 25, 2014

NATURE.M

Thats exactly what I got when drew the diagram. So I figured what i initially did was wrong, because i made a sign error somewhere.
For instance, from the negative line charge, I have

dE$_{x}$ = $\frac{kλdx}{x^{2}+b^{2}}sinθ$.

For this component since dq = -λdx, should it also be negative? Naturally, I would of thought the negative sign would cancel since the component is directed in the negative x direction.

4. Jan 25, 2014

haruspex

I'm not sure which way round the charges are. If the negative charge is on the negative x axis then the sign of each field should be negative in the x direction. The negative charge pulls that way, the positive charge pushes that way.

5. Jan 25, 2014

rude man

Maybe the problem stated that a >> L/2. It makes no difference whether b >> L/2 or not, as haruspex has pointed out.

If a >> L/2 then you can treat the two line segments as two point charges located at (-L/4,0) and (L/4,0), determine the dipole moment simply, and use your formula for the E field of a dipole.

6. Jan 25, 2014

NATURE.M

Thanks rude man, I figured you would have to concentrate each line charge at a single point in order to calculate the electric dipole for part b.

For part a though, I think my limits of integration might be wrong. For the negative line charge, I integrated from -L/2 to 0 and positive line charge from 0 to L/2. I know I should find the net electric field at P is pointing in the negative x direction. I'm getting the y components from each line charge to cancel, but the x components are also cancelling….

7. Jan 25, 2014

haruspex

It's a bit hard figuring out where you're going wrong without seeing your detailed working.

8. Jan 25, 2014

NATURE.M

Ill show you what I did to calculate the electric field of the negative line charge at point P.

So, dE = $\frac{kdq}{x^2 + b^2}$, where dq=-λdx.
Then the components of the electric field are:
dE$_{x}= dEsinθ = \frac{-kλdx}{x^2 + b^2}sinθ = \frac{-kxλdx}{(x^2+b^2)^{3/2}}$

dE$_{y} = dEcosθ = \frac{-kλdx}{x^2 + b^2}cosθ = \frac{-kbλdx}{(x^2+b^2)^{3/2}}$

Then integrating: $E_{x} = \int dE_{x} = \int^{0}_{-L/2} \frac{-kxλdx}{(x^2+b^2)^{3/2}} = kλ(\frac{1}{b} - \frac{1}{(b^2 + \frac{L^2}{4})^{1/2}})$

and similarly $E_{y} = \frac{-kλL}{2b(b^2 + \frac{L^2}{4})^{1/2}}$

So thats the first part. If you want me to post more I can.

9. Jan 25, 2014

haruspex

The sign is already wrong there. You're measuring θ 'backwards' from the y axis, so x = - √(x2+b2)sin θ.

10. Jan 25, 2014

NATURE.M

I dont understand x= - √(x2+b2)sin θ. What does this mean?
Edit: Nevermind I see where it comes from.

11. Jan 25, 2014

NATURE.M

So then the negatives for dE$_{x}$ should cancel. I missed that point.
With this the y components of electric field at P still cancel and the x components don't which is as desired.
Thanks haruspex!

12. Jan 25, 2014

NATURE.M

I was just wondering.
If b>>L, then E ≈ 0 at point P. Is there a better way To express this approximation? I don't think using taylor expansion will help here. Then, is it just best to leave it as is?

Last edited: Jan 26, 2014
13. Jan 26, 2014

rude man

My point was that you should start by postulating point charges for the two line charges. Once you have figured out the locations and magnitudes of the charges you can do part (a) but you have to assume a >> L/2. If you don't do that you have to integrate as you & haruspex are doing.

But I suspect the idea is to treat the line segments as a dipole even for part (a). The fact that they said b >> L/2, which is meaningless since the field anywhere along the y axis is zero, led me to believe they really meant a >> L/2. So you decide.

14. Jan 26, 2014

NATURE.M

For the second part of a., I found that the electric field at point P' on the x-axis to be 0 after taking the net considerations from the negative and positive line charge. Now I don't think this is right, but the math seems to suggest it. ?

15. Jan 26, 2014

rude man

Well, after consulting my textbook and reactivating my tired brain it turns out that the E field is not zero even along the y axis. The y component is zero but the x component is not. This is obvious once you think about it. My apology for missing that until now.

I still think though that you should replace the two line charges by a dipole and then compute the field on that basis rather than do an integration of the line charges. I say that based on the way the problem is stated. It seems to emphasize the concept of a dipole. If you don't assume a >> L/2 then you do have to integrate for the (a,0) charge but for the (0,b) charge the problem statement clearly allows this.

Bottom line, I would draw the dipole based on the charged lines and compute the field for a,b >> L/2 with no integration.

So, describe the magnitude and position of the equivalent dipole? That would be my first step.

16. Jan 26, 2014

NATURE.M

I solved for the electric field at Point P on the y axis already using integration methods, and obtained the desired result. I find if b>>L then E ≈ 0.

And now I'm unsure of the electric field at Point P' on x-axis. Should the negative line charge and positive line charge cancel out, because according to my results they do. Thus, E = 0 at x = a, and if a>>L, E is still 0. I'm trying to make sense of this result.

17. Jan 26, 2014

rude man

The problem is what is meant by "far away". Clearly, if you go far enough away, the answer is Ex = Ey → 0 for both P and P'.

However, this is probably not what the problem wants you to come up with. To give you a hint, express the fields in terms of p and r3. p is the dipole moment and r is the distance between the dipole center and P or P'. The result is simple expressions for the field at P and P'.

18. Jan 27, 2014

NATURE.M

I was wondering part b simply asks for the electric dipole moment of the line charges. Normally,
p = qd, so in this case would we have P = L/2 * q, where q is the line charge concentrated at a single point.

19. Jan 28, 2014

rude man

That is exactly right. The charges are concentrated at the half-way point for each line segment so the charge separation is L/4 + L/4 = L/2. And q of course = λL/2.

Getting to part (a) yet?

20. Jan 28, 2014

NATURE.M

Yeah I'm pretty sure I got part (a), it turned have to integrate to find the electric field in both cases, and apply the approximation conditions after.