Electric Field Over A Uniformly Charged Area- Please help

[maherelharake]

Electric Field Over A Uniformly Charged Area---- Please help!

Homework Statement

Imagine a square in the xy plane with each side of length a. It has a uniform charge per unit area (σ). The center of that square is at the origin. Determine the electric field at a distance z above the center.

Homework Equations

The Attempt at a Solution

I have attached my work. However, I feel like I have made a mistake somewhere since I don't know how to solve my integral. He told me to find Ez and Ex, and that Ex should equal 0 when I work it out. Thanks in advance!

 

[vela]

I can't really follow what you did. Could you explain in words what you were doing in your attempt?

 

[maherelharake]

First of all, thanks for answering.

I wanted to treat this problem as a square wire with a uniform charge, then integrate over the area.

My first equation is an equation that our book derived for Ez above a line of charge. I attempted to convert that to a form that fits an area (rather than a length) and to break it down into Ex and Ez (as my teacher told me to). Hope this helps, if not let me know. I will respond quickly.

 

[vela]

I'm not sure what you mean by "square wire." Do you mean you're breaking the area up into narrow strips of length a and width dx, where each strip is thought of as a wire?

 

[maherelharake]

Hmm, let's see if I can make what I did more clear.
I was told to use the idea of starting off with a square loop of wire (that might be more descriptive), and later integrating over the area between the boundaries of that square loop (from -a/2 to a/2).

 

[vela]

OK, I looked a bit more closely at what you did, and you're not using square loops of wire. You're breaking up the area into strips of length a and width dx and treating each strip as a wire.

You found earlier that the magnitude of the electric field at a point a distance d above the center of a wire of length L and charge density λ is given by

E = \frac{2kq}{d\sqrt{d^2+L^2}}

where q=λL is the total charge on the wire. The electric field points directly away from the wire.

In this problem, you want to find the electric field at the point (0,0,z) above the charged square which lies in the xy plane and is centered at the origin. Imagine a narrow strip of this square that lies along a line of constant x. The distance from this strip to the point is equal to d=\sqrt{x^2+z^2}, and the length of the strip is a. The charge of the strip is the charge density σ multiplied by the area of the strip dA=a dx. Using the earlier result, you get the magnitude of the electric field at the point due to the strip is

dE = \frac{2k\sigma a\,dx}{\sqrt{x^2+z^2}\sqrt{x^2+z^2+a^2}}

This is almost exactly what you have except for one small, yet critical mistake. Now to find just the x-component, you multiply by x/\sqrt{x^2+z^2}, integrate to add up the contributions from all the strips, and finally get

E = \int_{-a/2}^{a/2} \frac{2k\sigma ax\,dx}{(x^2+z^2)\sqrt{x^2+z^2+a^2}}

This integral you should be able to see is equal to 0 without actually doing the integral. (Hint: the interval of integration is symmetric.)

 

[maherelharake]

Thanks a lot.
Ok I followed everything you said, but I still have a couple questions.

First off, what you have at the bottom is for Ex, but what about for Ez (which is what we are ultimately trying to find)? Is it the same, except for instead of multiplying by x/sqrt(x^2+z^2)
I should multiply by z/sqrt(x^2+z^2)?

Secondly, you said I should be able to evaluate this integral just by looking at it, due to the symmetry. I think I am visualizing it properly, but is there some other way to write it down and explain it without having to actually work through the integral? I think he may want more than just me saying that it's 0 due to symmetry (even though I can see this and agree with it).

Thanks again. I think I am almost there.

 

[vela]

Thanks a lot.
Ok I followed everything you said, but I still have a couple questions.

First off, what you have at the bottom is for Ex, but what about for Ez (which is what we are ultimately trying to find)? Is it the same, except for instead of multiplying by x/sqrt(x^2+z^2)
I should multiply by z/sqrt(x^2+z^2)?

Yes.

Secondly, you said I should be able to evaluate this integral just by looking at it, due to the symmetry. I think I am visualizing it properly, but is there some other way to write it down and explain it without having to actually work through the integral? I think he may want more than just me saying that it's 0 due to symmetry (even though I can see this and agree with it).

Thanks again. I think I am almost there.

You can say the integral is 0 by symmetry, but I'd expect you should explain exactly what symmetry you're using. You say you can see why it's 0. Can you explain what you mean?

 

[maherelharake]

Well When I try to visualize the problem, I can see that, since it's a square, all of the x and y components will cancel, thus only leaving Z components. It's hard for me to explain how I see it, but I just kind of visualize how for every x and y component in one direction, you have another x and y completely countering it.

Also, for the Ez question, is that the final form of my answer? I don't think I can integrate that by hand.

 

[vela]

Well When I try to visualize the problem, I can see that, since it's a square, all of the x and y components will cancel, thus only leaving Z components. It's hard for me to explain how I see it, but I just kind of visualize how for every x and y component in one direction, you have another x and y completely countering it.

Yeah, that's right. As far as the integral goes, is the integrand even or odd?

Also, for the Ez question, is that the final form of my answer? I don't think I can integrate that by hand.

I don't know. You should ask your instructor. I don't see a straightforward way of integrating it either.

 

[maherelharake]

Would it be considered an odd integrand, since if you replace 'x' with '-x' the result will be the same?

 

[maherelharake]

Wait I think I said that wrong. I meant if you substitute '-x' for 'x' you get the opposite result.

 

[maherelharake]

I am going to sleep soon. My assignment is due at 9:30 AM EST. If you respond to me any time between now and then, I will see it and respond to you. Thank you so much for your time. I have attached my final version of this problem, if you can glance at it and see what you think. Thanks again. I really appreciate it.

 

[vela]

Would it be considered an odd integrand, since if you replace 'x' with '-x' the result will be the same?

Wait I think I said that wrong. I meant if you substitute '-x' for 'x' you get the opposite result.

Right. It's an odd integrand, so when you integrate over a symmetric interval, you get 0.

 

[vela]

I don't know. You should ask your instructor. I don't see a straightforward way of integrating it either.

Just thought I'd mention Mathematica spit out a pretty clean result, so I imagine there must be a relatively straightforward way to do it by hand.

 

[Mindscrape]

Yikes, there are ugly integrals all over the place in this problem! I tried both your way of using strips, my way of simply making the integral from scratch, and finally the way of actually making square loop shells (which finally gave me an integral I could do). Since this problem is really tough, as far as getting a good integrand is concerned, let me give you a hand.

You can find the electric field above a square loop of area a to be, without showing the work (do it on your own if you haven't done this as an example or problem already), as such
E = k \frac{4 \lambda a z}{(z^2+a^2/4)\sqrt{z^2+a^2/2}}

So then to extend that square loop by another bit da (for a total of a+da as a side length) we can simply sum up all the square loops from a=0 to a = a' (the real a). Maybe I should have chosen something other than 'a' initially, but too late now, I'm not retyping, hopefully you get it. So now to make this a 2d summation then lambda goes to sigma*da/2 because that's how much the square grows. In all we get

E = k \int_0^{a=a'} \frac{2 \sigma z}{(z^2+a^2/4)\sqrt{z^2+a^2/2}} a da

Now I'll give you another major hint so that you can maybe get your HW done. You can get this in the form of arctan.

 

[maherelharake]

I don't have Matematica, and I don't think I have time to completely redo what I did. I guess I will just turn in what I attached, since it is due in an hour.

 

[Mindscrape]

For your own edification, I would suggest trying out the way I posted. Maybe at first don't look at what I did, and try it on your own. Then if you get stuck you can take a peek. Then when you get to actually evaluating the integral my hint should be a good help. This, however, is merely a suggestion to help you succeed in EM, a study in which you can get behind easily if you don't stay on top of it.

 

[maherelharake]

I will try to do it this evening and let you know if I get stuck. Thanks again.