Electric field of finite line charge and 2 point charges

1. Jul 30, 2013

eminem14

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

A finite uniform linear charge ρ_L = 4 nC/m lies on the xy plane; start point and end point are (7,0,0) and (0,7,0) .While
point charges of 8 nC each are located at (0, 1, 1) and (0, -1, 1). Find E at (0, 0 ,0)

2. Relevant equations
dE=ρ_L *dz'/4∏ε * (r-r'/[magnitude(r-r')]^3)
where r'=z' *a_z and r=ρ * a_ρ and z=0, 0<=∅<=2∏
dE_ρ=ρ_L*dz'/4∏ε *ρ/R^3 dE_z= - ρ_L*dz'/4∏ε * z/R^3

3. The attempt at a solution

I obtained the E-field contributions of the two point charges using the E-field formula for a point charge. However, the line charge is tricky because if it were infinite there would only be the ρ component contributing. I know the length of the rod matters and also the vector from the point of observation (0,0,0) to the midpoint of the rod. However, can't quite connect the dots. Need some help asap.

Thank you in advance

2. Jul 31, 2013

haruspex

It might be easier to rotate the set-up 45 degrees, so the line of charge runs from (a, -a) to (a, a). Which way will the field act at O? What is the magnitude of the field an element of the line length dy at (a, y) exerts at O? What is the component of this in the direction of the net field?

3. Jul 31, 2013

eminem14

i got to a solution for the line charge:
Q=ρ_L * dL= 4*10^-9 *(dx + dy)
=> k*ρ_L /_Rmagnitude^3∫Rx dx + k*ρ_LRmagnitude^3∫Ry dy

where R is the vector pointing from the origin to the midpoint of the line charge
=> R= -3.5i -3.5j

but what are the limits of the integral? i did it over the length of the line charge √(7^2 + 7^2)= 7√2
it ddnt get the correct result. however when i did the integral over the L*(√2)/2= 7 it got the correct result. (i verify the result by running a MATLAB solution I have)

I dont know why 7 is the correct upper limit for the integral. Can u tell me why?

4. Jul 31, 2013

rude man

Not sure why you solved for Q.

Looking at this a couple of ways, I would suggest:

note symmetry so that the E field is in the direction of i + j , i.e. 45 deg. from x to y axis

then determine an element of E along this direction as dE = {kρdL/r2}cos(θ - π/4)

then use y = -x + 7
and dL = √2 dx
x = r cosθ
y = r sinθ

finally obtaining dE in terms of x and constants. Integrate from x = 0 to x = 7. (Ignore sign of result, it's obvious what the direction of the E field is.)

My answer, for what it's worth, is E = 7kρ(0.0408) = 10.28 V/m in direction i + j. Subject to all sorts of mistakes, of course

PS this is for the line charge only.

Last edited: Jul 31, 2013
5. Jul 31, 2013

haruspex

How do you get dx+dy?
It really is much easier my way. Set 2a = length of line = 7√2. Rotate the set-up so that the line runs from (a, -a) to (a, a). Consider an element dy at (a, y), y = a tan(θ), -π/4 < θ < π/4. This has charge ρ dy = ρ a sec2(θ) dθ. The field at O from it has magnitude k ρ a sec2(θ) dθ / (a sec(θ))2 = ρ dθ / a, where k = 1/4πε. The component of that in the x direction is (k ρ dθ / a) cos(θ). Integrating, $\int_{\theta=-\frac \pi 4}^{\frac \pi 4}\frac{\rho \cos(\theta)}a d\theta = \frac \rho a [\sin(\theta)]_{\theta=-\frac \pi 4}^{\frac \pi 4} = \frac \rho a \sqrt(2) = \rho/7$.
Now, it bothers me that both you and rude man get something like Lkρ where I'm getting kρ/L, but it makes more sense to me that it would be inversely proportional to the scaling of L. See also eqn 2 at http://www.eece.ksu.edu/~bala/notes/emt/finite-line-field.pdf [Broken]

Edit: the final step should read "= 2ρ/7"

Last edited by a moderator: May 6, 2017
6. Jul 31, 2013

eminem14

i know the correct e-field is -7.231 i - 7.231 j which is the result i got using the method i mentioned in the previous comment (integrating over L*(√2 / 2)). Im still confused regarding that solution if someone can clarify that for me

I got dx+ dy because in the formula sheet provided by the professor, dL in cartesian coordinates is dx*i + dy*j +dz*k. Q is ρ_L * dL and E= ∫kQ/R_mag^3 * R * dL= k*ρ_L/R_mag^3(∫dx +∫dy).

R is the vector pointing from the observation point to the mid point of the line = -3.5i -3.5j

Can any of you explain why that solution works?

Last edited by a moderator: May 6, 2017
7. Jul 31, 2013

rude man

My E = kρy0*∫from 0 to 7 of dx/(2x2 - 2y0x + y2)(3/2)
with y0 = 7 which wolfram alpha had no trouble solving.

Comparing with your answer, when the omitted k is included, it is interesting to observe that my answer is exactly twice yours (your E = 5.14 V/m).

I don't know how you came to conclude that my E scales with L. Looking at the indefinite form of my integral it doesn't seem to. Our respective answers seem to reinforce that viewpoint.

I also just took note of the fact that my magnitude agrees with the OP's stated 'correct answer'.

Last edited by a moderator: May 6, 2017
8. Jul 31, 2013

haruspex

That's correct if dL is a vector. But the charge on an element length |dL| is ρ|dL| = ρ√(dx2+dy2). Given that in this case dy=-dx, that reduces to ρ√2.dx. OTOH, dx+dy reduces to zero!
You effectively turned |dL| into 2dx, getting √2 times the correct answer.

9. Jul 31, 2013

haruspex

I guess I read too much into "E = 7kρ....".
There was an error in my working at the final step. It should say 2ρ/7.

10. Jul 31, 2013

eminem14

So the correct answer is which? And how would it work out using my method? (E(0,0,0) due to the linear charge as a vector?)
could u show me the correct steps (just a modification of what i did)?

11. Jul 31, 2013

haruspex

If you simply change your dx+dy to dx√2 (which you understand, yes?) I think you'll get the right answer. Whether the rest of your method is right I can't tell since you didn't post the details of what you did.

12. Jul 31, 2013

eminem14

E= k∫p_L*dL*R/(R_mag^3). so that would mean (√2*k*ρ_L)/(R_mag^3) * ∫R dx ?

1. Is the upper limit of the integral L which is √(7^2 + 7^2) = 7√2 ?

2. Since R = -3.5i -3.5j that would mean: ∫R dx= -3.5∫dx -3.5 ∫(-dx) (Because dy=-dx) →∫R dx=0

that doesnt make any sense. Where's the mistake?

13. Jul 31, 2013

haruspex

Isn't R the vector from the element of the line charge to the origin, -xi-yj? It varies as x varies, and so does its magnitude. How do you move the Rmag-3 term out of the integral?

14. Jul 31, 2013

eminem14

R is the vector pointing to the midpoint of the line charge from the origin. if the midpoint of the line charge is at (3.5, 3.5, 0) then R would be < 0 - 3.5, 0 - 3.5, 0 - 0 > ∴ R= -3.5 i -3.5 j

Regarding Rmag-3 , the magnitude of R is a constant so it can be out of the integral

thats how the code does it

15. Aug 1, 2013

haruspex

Then I am at a loss for how you arrive at the integral you posted. It makes the whole integrand a constant.

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