Find the electric field a distance z above the center of a square loop (with side length a) carrying a uniform line charge.

I mostly need help with setting up the integral here. The tricky part is changing the unit vector r^ to something I can integrate. This is a physics question but I figured that asking on here isn't too ridiculous.

Question

Find the electric field a distance z above the center of a square loop (with side length a) carrying a uniform line charge.

I mostly need help with setting up the integral here. The tricky part is changing the unit vector r^ to something I can integrate. This is a physics question but I figured that asking on here isn't too ridiculous.

nuttyliaczar · Answer

|dw:1445476917772:dw|

nuttyliaczar · Answer

This gives a visual of what the problem is. R is the shortest length r can be (thus it is a constant). x is the base of the arbitrary triangle.

nuttyliaczar · Answer

$$E=\frac{ 1 }{ 4\pi \epsilon }\int\limits_{y}^{z}\frac{ \lambda }{ r ^{2} }r _{v}dr$$

nuttyliaczar · Answer

This is the equation for the electric field given some distance r for anyone who doesn't know. $$r _{v}$$ is the unit vector of r.

anonymous · Answer

this might help https://www.physicsforums.com/threads/electric-field-of-a-continuous-charge-distribution.475357/

nuttyliaczar · Answer

Can anyone explain this more thoroughly to me? The person asking the question in that link seems to have done most of the problem already, but I don't know how he got it or how much of it is correct.

anonymous · Answer

the solve from link is correct

anonymous · Answer

@nuttyliaczar Hi, I've found a good solution for it. Do you still like to discuss this problem?

anonymous · Answer

Did you find a way to make your integral?

anonymous · Answer

As this question was very beautiful and i liked it so much I tried to solve it. Making the integral wasn't that hard. The hard problem for me was calculating it. I asked for help and guidance on calculations here: http://openstudy.com/study#/updates/567b4085e4b0f8b8229fbf99

anonymous · Answer

after a very long time, i solved that hard integral with some big help.
But let me know what you found when solving it.

anonymous · Answer

@nuttyliaczar  seems u're not around.

the integral can be written as below:
$$\int\limits dE=\frac{8 k.q }{ l } \int\limits_{0}^{\frac{ a }{ 2 }}[(x-\frac{ a }{ 2 })^2+B]^{-\frac{ 3 }{ 2 }}dx$$
Where:
k=9*10^9
q=total static charge
l= total length of the frame
$$B=z^{2}+\frac{a ^{2} }{ 4 }$$

anonymous · Answer

its limits are x=0 and x=a/2

The Intergral above gives the total E at the point P.

We can talk about it later if you still like to know how this can be derived.