Electric Field Above a Square Loop with Linear Charge Density?

[Vaentus]

Homework Statement

Problem 2.4 from Griffiths Intro to Electro[/B]
Find the electric field a height z above the centre of a square loop with sides a and linear charge density λ.

height is given to be z and sides given to be a, ∴ distance from origin to side is given by a/2

Homework Equations

E = \frac{1}{4 \pi \epsilon_0} \int {\frac{1}{r^2} \hat {r}dq}

dq = \lambda dl

The Attempt at a Solution

Considering the side of the square perpendicular to the positive y axis

First using the position of point P (0, 0, z) and position on the side (x, a/2, z) where z and a/2 are constants the direction from origin to P is given by r=z\hat{z} and direction from origin to side is given by r' = (a/2)\hat{y} + x\hat{x}.

∴direction of electric field is given by r = r - r' =z\hat{z}-(a/2)\hat{y}-x\hat{x}

\hat{r}=\frac{r}{|r|}=\frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt{z^2+\frac{a^2}{4}+x^2}}

considering the "right" side of the square loop y and z values are constant while the x goes from -a/2 to a/2
∴ dq=\lambda dx

Plugging these equations into the integral we get
E = \frac{1}{4 \pi \epsilon_0} \int {\frac{1}{z^2+\frac{a^2}{4}+x^2} \frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt{z^2+\frac{a^2}{4}+x^2}}\lambda dx}

which can be simplified to
E=\frac{\lambda}{4\pi\epsilon_0} \int{\frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt[3]{z^2+\frac{a^2}{4}+x^2}}}

I know I can split the integral up into three different integrals w.r.t x y and z directions then group the y and z integrals because z and (a/2) are constants.while leaving the integral in the x direction due to its dependence on x which changes from -a/2 to a/2 but I am unsure whether that is the correct way to proceeed/what to do next.

Thanks for your help!

 

[TSny]

Welcome to PF!

Your work looks good to me. You now have three integrals that can be done separately. However, using symmetry arguments, you should see that you will only need to evaluate one of the three integrals to derive the answer for the entire square.

 

[BvU]

Hello Vaentus, and welcome to PF

First a few comments:

"P (0, 0, z) and position on the side (x, a/2, z)"

you mean (x, a/2, 0) I suppose ? (which is correct, and you do it correctly in the integral).​

"I know I can split the integral up into three different integrals w.r.t x y and z directions then group the y and z integrals because z and (a/2) are constants.while leaving the integral in the x direction due to its dependence on x which changes from -a/2 to a/2"

Weren't you integrating over x only ? and x appears in all three terms.
What do you mean with the y and z integrals ? I can imagine and integral over dy for the side (a/2, y,0) but I can't imagine a z integral !​

Then two tactical questions:

1) Which way do you think the electric field will point when you've done all the integrals ?
Or: is there a way to explore some of the symmetry here ?
(Physicists love symmetries: they can save loads of dull work...)

2) A slightly easier problem is the electric field at a distance $s$ above the midpoint of a straight line segment of charge.
Again: which way will the result point ?
(I use s because of course in your case $s^2 = z^2 + {a^2\over 4}$).
There's four of these that point in 4 different directions. Adding them up (as vectors) is relatively easy.

[edit] I got beaten by TSny; my fault: too verbose. Your integral setup is indeed just fine and quite usable.

 

[BvU]

Hey, in my 1999 version there's a very useful "[Hint: use the result of exercise 2.1]" ! Did you forget that part of the problem statement ?

 

[agnimusayoti]

Hey, in my 1999 version there's a very useful "[Hint: use the result of exercise 2.1]" ! Did you forget that part of the problem statement ?

Hei, can you explain how to use the result of example 2.1?

 

[BvU]

Hey , what did you find in example 2.1 (2.2 in a later edition) that might be useful here ?

 

[agnimusayoti]

Hey , what did you find in example 2.1 (2.2 in a later edition) that might be useful here ?

In example 2.1 the author Griffith use infinitesimal element on the left side and the right side of the wire. So that the electric field parallel to wire cancel each other.

For this problem, I try use this and get this answer:
$$E = \frac{4k\lambda a}{\sqrt{(z^2+\frac{a^2}{4})(z^2+\frac{a^2}{2})}}$$
But I get different answer if I try to use the simetry of two wire that parallel each other.

 

[BvU]

From 2.1 :
$$E={1\over 4\pi\varepsilon_0}\;{2\lambda L\over z\sqrt{z^2+L^2}} $$ for a point P at distance z above a linear charge:

You

1. translate to the case of the problem (use $z'$ in the expression of the example):$$z'{\,^2} = z^ 2 + \left (a\over 2\right ) ^2$$where I think you miss something already, but I'm not sure
2. maintain the ${1\over 4\pi\varepsilon_0}\;$
3. Consider the directions of the $z$ component and the $z'$ component
4. multiply by 4

If you do all that I think you'll come out right

 

[agnimusayoti]

Yes I do get the same thing where k is a constant equivalent with $\frac{1}{4 \pi \epsilon_0}$.

What if I choose infinitesimal element from two opposite wire? I get the different answer.

Then, what is the problem?

 

[agnimusayoti]

 

[BvU]

In case you didn't notice, the picture stands on it's side. My head does not
For this exercise, integration is not necessary -- that was already done in the example.

Yes I do get the same thing where k is

Sorry, I overlooked that .

Then, what is the problem?

I think you overlooked point 3

 

[agnimusayoti]

Uh, sorry for the picture.
Yes, if I use two opposite wire, then $\cos \theta = \frac{z}{\sqrt{a^2/4+y^2+z^2}}$. Meanwhile, if I use two different side of each wire, I got $\cos \theta = \frac{\sqrt{z^2+a^2/4}}{\sqrt{a^2/4+y^2+z^2}}$. Hmm I don't know what is the problem so the answer is different...