Solving the Electrostatic Conundrum: Brute Force vs Symmetry

[Christopher314]

Electrostatic conundrum: How "brute force" and symmetry arg. give different answers!

Hello everyone, this is my first post -- and an interesting one (or so I think).
**Note to Mods: this is NOT me asking for a solution to a HW type problem -- I know and have the solution already -- rather, this is an investigation on how two valid methods to answering a physics problem can have two complete different answers.**

Background: I do private tutoring, and a college student client of mine wanted help on this problem. I looked up the problem in my own copy of Griffiths and solved it and got the right answer. The thing is: I couldn't find anything wrong with how she did the problem. The problem is from Griffiths E&M book, page 106 problem 2.41: "Find E at a height z above the center of a square sheet (side a) carrying a uniform surface charge s. Answer: (s/2*e0)[(4/Pi)Arctangent[Sqrt[1+(a^2/(2z^2))]]]."

Here's the methods:
"Elegant": We can use a previous result of the field due to a finite line of constant line charge, except, now, we give it a constant surface charge s and an infinitesimal thickness. The point is to construct the square from thin square "frames." To do this we divide the square into four equal regions along the diagonals using the functions y = x and y = -x. So we have, essentially, line charges of varying lengths. And we use that by definition a square has all sides that are equal, so that if we vary x by dx, we inadvertently vary y by dy, i.e. since x = y, then dx = dy. Integrate and voila. This was my way and led to the correct answer.

**Edit: Using my client's notes here's her way exactly**
"Brute Force": My client, on the other hand, decided to go the "brute force" way and use the integral equation for the field. Since our field point is at z in the z-direction, r = (0, 0, z). On the x-y plane, any element of area dA is located at r' = (x, y, 0). So our |R| = |r - r'| = Sqrt[x^2 + y^2 + z^2]. Our element of area is dA = dxdy, a tiny square. By symmetry, the field points in the z-direction, so R = z cos Theta, where cos Theta = |r|/|R|. So then E = (zs/4 Pi e0)*Double Integral[dxdy/(x^2 + y^2 + z^2)^(3/2), where the limits of integration are -a/2<x<+a/2, and -a/2<y<+a/2 since we're integrating over a rectangle. This looks fine, right? Well if you compute this using Mathematica (or integral tables) you get a TOTALLY different answer!

The Plot Thickens: What's MORE interesting is that if, instead of integrating over the whole square and you just consider the contribution from 1/4 of the square cut along the diagonals (as in my derivation), where the limits of integration change to -a/2<x<+a/2, -x<y<+x and you multiply back in the factor of 4, then you DO get the correct answer.

The conundrum is: How can the solution change simply by changing the manner in which the surface integral is being calculated? Since this is a finite region, both integrals should be equal...but aren't.

Let's see what the forum's thoughts are. And if you'd like to compute the integrals yourself but don't have Mathematica, use this free Mathematica tool: http://integrals.wolfram.com/index.jsp"

 

[Euclid]

In the client's method, you seem to be solving for the potential, rather than the field. Does that settle the problem?

 

[Christopher314]

Oh man! I just called her on the phone and we laughed for about TWO minutes. I don't know how I didn't notice it! This has actually racked my brain -- I thought I was integrating wrong and even got out my old vector calculus book from the garage.
Sometimes you need a new pair of eyes to see the GLARING error.

Thanks a lot, Euclid.

**Edit** Wrong! This was a typo on my part in the initial thread. The conundrum lives!

 

[qbert]

with a little bit of work, you can get Griffith's answer
doing it the hard way.

starting from where you left it you have.
<br /> E_z = \frac{\sigma z}{4 \pi \epsilon_0} \int_{-a/2}^{a/2} <br /> \int_{-a/2}^{a/2} ( x^2 + y^2 + z^2 )^{-3/2} dx \; dy

which we can rewrite as
\frac{\sigma}{4 \pi \epsilon_0} \int_{-a/2}^{a/2} \int_{-a/2}^{a/2}<br /> ( (x/z)^2 + (y/z)^2 + 1)^{-3/2} \frac{dx}{z} \frac{dy}{z}

Letting u = x/z, v =y/z, and p = a/(2z) gives
\frac{\sigma}{4 \pi \epsilon_0} \int_{-p}^p \int_{-p}^p<br /> (u^2 + v^2 + 1)^{-3/2} du dv

Integrating this with respect to u gives
\frac{\sigma}{4 \pi \epsilon_0} \int_{-p}^p <br /> \frac{2p}{(v^2 + 1) \sqrt{p^2 + v^2 + 1}} dv

Then Integrating with respect to v gives
\frac{\sigma}{4 \pi \epsilon_0}<br /> 4 \tan^{-1} \left( \frac{p^2}{\sqrt{1+2p^2}} \right)

Now there is a little bit of trig required to get this
in the form that Griffiths answer is.

I need these results
1. arctan(x) + arctan(1/x) = Pi/2.
2. arctan(-x) = -arctan(x)
3. 2*arctan(a) = arctan(2a/(1-a^2))

The first two are pretty trivial. An easy
proof is subtract the left from the right,
differentiate wrt x. This gives 0 in both
cases. thus lhs = rhs + c. use x=1 to
fix the constant.

The third follows since
tan(a+b) =( tan(a) + tan(b) )/(1- tan(a)tan(b))
so, a + b = arctan( (tan(a) + tan(b)) / (1- tan(a)tan(b)) ).
If we Let a = arctan(u) and b = arctan(v) then we have
arctan(u) + arctan(v) = arctan( (u + v)/(1-uv) ).
Finally if u = v = a we get 3.Now we use them to rearrange the answer.
arctan(x) = arctan(-1/x) + pi/2:
<br /> \frac{\sigma}{2 \epsilon_0} (\frac{2}{\pi} \tan^{-1}( \frac{p^2}{\sqrt{1+2p^2}} ))<br /> = \frac{\sigma}{2 \epsilon_0}(\frac{2}{\pi} \tan^{-1}( \frac{-\sqrt{1+2p^2}}{p^2} ) - 1)<br />

Then we can put
= \frac{\sigma}{2 \epsilon_0}(\frac{2}{\pi} \tan^{-1}( \frac{-2\sqrt{1+2p^2}}{2p^2 +1 - 1} ) - 1)

which gives
= \frac{\sigma}{2 \epsilon_0}(\frac{2}{\pi} \tan^{-1}( \frac{2\sqrt{1+2p^2}}{1 - (2p^2 +1)} ) - 1)

then by 3. this is
= \frac{\sigma}{2 \epsilon_0}(\frac{4}{\pi} \tan^{-1}( \sqrt{1+2p^2} ) - 1)

with p = a/2/z, this is Griffith's answer

 

[Christopher314]

Thanks a lot, qbert. The so-called (or badly deemed) "conundrum" is solved. It's all due to those Arctangent trig relations. I guess instead of my vector calc book, I should have dug out my high school trig book from the garage!