The sheet of charge integral (technical)

In our discussion of the field produced by a flat sheet of uniform charge, we deferred showing how to actually find the correct result since it involved vector calculus. While that is true and just writing down the result and doing the result mathematically is a mess, we can use our understanding of what's happening in the physics to restructure "the mess" and bring the result down to something much simpler. (It's still somewhat messy, but the sort of thing you do in an introductory class on 1-variable integrals.) Let's look at it in some detail since it will give a good example of how understanding the physics helps you do the math.

The problem

We have a flat sheet (that we can consider infinite) covered by a uniform charge density $\sigma$ (Coulombs/m²). What is the electric field at a hight $d$ about the plane?

The physics

The electric field produced by a small bit of charge $q$ at a position with coordinates $\overrightarrow{r}_1$ at a position $\overrightarrow{r}_2$ is given by Coulomb's law (for the E field):

$$\overrightarrow{E}(\overrightarrow{r}_2) = \frac{k_C q}{|\overrightarrow{r}_2 - \overrightarrow{r}_1|^3} (\overrightarrow{r}_2 - \overrightarrow{r}_1)$$

(The denominator is cubed in order to convert the vector in the numerator into a unit vector of direction only. See the discussion on the page Coulomb's law -- vector character.)

But we don't have one small bit of charge here. We have a whole sheetload of it. So to get the field from all the charges, we have to break up the sheet into small charges, calculate the field produced by each at a particular observation point, and add all those results up.

Just doing the math

Now if you like just doing hairy math (I used to!), we can do this is a very straightforward if messy manner. Let's see how it works — even though we'll be able to make it simple and more understandable using the physics later. 

We can break the sheet up into small charges by using a coordinate system with the plane in the x-y plane. If we take the origin of the plane right below our observation point, the coordinates of our observation point $\overrightarrow{r}_2 = (0, 0, z).$ 

We'll break the plane up into little boxes of size $dx \times dy$ at position $\overrightarrow{r}_1 = (x, y, 0).$ 

The separation of the charge from the observation point will be $\overrightarrow{r}_2 - \overrightarrow{r}_1 = (-x, -y, z).$

Since the charge density on the plane is $\sigma,$ the charge on that little box is $dq = \sigma dx dy.$

The small contribution of the charge at $(x,y,0)$ to the field at $(0,0,z$ is therefore

$$d\overrightarrow{E}(\overrightarrow{r}) = \frac{k_C \sigma }{(x^2 + y^2 + z^2)^{3/2}} dx dy (-x\hat{i} - y\hat{j}+z\hat{k})$$

We argued from symmetry on the page discussing the sheet's field that the total E field had to point straight up — in the z direction. The the x direction and y direct integrals will cancel.

Taking the constants out front, we get

$$d\overrightarrow{E}(\overrightarrow{r}) = zk_C \sigma \frac{dx dy }{(x^2 + y^2 + z^2)^{3/2}}  \; \; \hat{k}$$

Adding up all the contributions means doing an integral in both x and y. Here's the result:

$$\overrightarrow{E}(\overrightarrow{r}) = (zk_C \sigma \; \; \hat{k}) \iint \frac{dx dy }{(x^2 + y^2 + z^2)^{3/2}} $$

integrated over the whole plane.

Using our math in physics toolbox to make some sense

I'm not really happy with this result. It's a messy integral and it depends on z, the height above the plane. I'd like to see how the answer depends on that but it's hidden in the current form.

Let's see if we can use dimensional analysis to separate the math from the physics and get the functional dependence of the integral on z.

The z (a constant in this example) in the integral is the only thing that sets the scale. All of our coordinates have units: they are distances. Let's change them into pure numbers by expressing them in terms of the distance scale z. What this means is writing

$$\chi = x/z \;\;\;\;\; \eta = y/z$$

These are pure numbers with no units.

We can simplify the integral (simpler as long as you don't get upset by a Greek "x" and "y") by then writing

$$x = z\chi \;\;\;\;\;y = z\eta$$

Putting these into the integral we get

$$\iint \frac{dx dy }{(x^2 + y^2 + z^2)^{3/2}} = \iint \frac{z d\chi z d\eta }{((z\chi)^2 + (z\eta)^2 + z^2)^{3/2}} $$

Pulling the z out of the denominator gives

$$\iint \frac{z d\chi z d\eta }{((z\chi)^2 + (z\eta)^2 + z^2)^{3/2}} = \iint \frac{z d\chi z d\eta }{z^2(\chi^2 + \eta^2 + 1)^{3/2}} = \frac{z^2}{z^3} \iint \frac{d\chi  d\eta }{((\chi^2 + \eta^2 + 1)^{3/2}}   $$

Phew! So the messy integral is proportional to 1/z which cancels the z in the numerator leaving the final result

$$\overrightarrow{E}(\overrightarrow{r}) = (k_C \sigma \; \; \hat{k}) \iint \frac{d\chi  d\eta }{((\chi^2 + \eta^2 + 1)^{3/2}}   $$

Independent of the height above the plane as claimed! The integral is just a number that you can look up in an integral table or have Mathematica™ do it for you.

Using the physics to simplify the integral

While this wasn't too awful (if you're a physics major), we can use what we know about the physics to restructure the problem and do the summing up of the effects of the charges in a way that simplifies the integral and helps us build some intuition about what's happening.

What we know from the physics is this:

Because of Coulomb's law, we know that the contribution to the E field of all charges that are the same distance from the observation point will have the same magnitude.

Maybe something will then be constant and we can simplify things if we first integrate over all charges that are the same distance from the observation point.

The charges that are all the same distance from the observation point form a ring in the plane around the origin as shown in Fig. 1.

The contribution from each bit of charge $dq$ on the ring is a little blue arrow of the same magnitude. And we can see that charges on the opposite sides of the ring produce little arrows that add so that their horizontal components cancel and their vertical components add.

So we have to decide how big each little blue arrow ($dE$) is in magnitude and what trig factor we have to use to get the vertical components. Then we can add them up.

The magnitude of the small charge contributions

In order to figure out the magnitude of the contribution of each little charge $dq,$ let's use polar coordinates. Looked at from above the ring looks like this figure. The area corresponding to the charge in the ring is shaded in gray. The bit of charge we are considering, $dq,$ is colored red.

Let the ring have a radius $r,$ a small thickness, $dr.$ Let the bit of charge on the ring that we are considering be at an angle $\theta$ and correspond to a small angular increment, $d\theta.$ The length of the area of the small red box along the circumference will be $r d\theta.$

So the total area of the little box will be $r dr d\theta$ and the total charge on it will be

$$dq = \sigma r dr d\theta$$

The distance that each bit of charge in the ring is from the observation point can easily be found from Fig. 1 using the Pythagorean theorem:

$$R = \sqrt{r^2 + z^2}$$

so the magnitude of the E field from that small bit of (red) charge is

$$dE = \frac{k_C \sigma dq}{R^2} = \frac{k_C \sigma r dr d\theta}{r^2+z^2}$$

Getting the angle factor

That's the whole magnitude of the little vector $dE.$ We  only want the vertical component of it. It's easy to see what that is from the pair of triangles extracted from Fig. 1 and shown in Fig. 3.

The triangle made of the little vector  $dE$ and its horizontal and vertical components, $dE_h, dE_v,$ is similar to the triangle made of the distances $z, r, R.$

Since we know $dE$ and want $dE_v,$ the similarity ratios we want are

$$\frac{dE_v}{dE} = \frac{z}{R}$$

so 

$$dE_v = \frac{z}{R} dE = \frac{z}{\sqrt{r^2+z^2}} \frac{k_C \sigma r dr\;d\theta}{r^2+z^2}$$

or 

$$dE_v = k_C \sigma z \frac{\; r\; dr\; d\theta}{(r^2 + z^2)^{3/2}}$$

Summing it all up

We can now add up all the little charges around the ring by integrating over $\theta$ and add up all the rings by integrating over $r$.

The result is

$$E_v = k_C z \int \frac{\; r\; dr} {(r^2 + z^2)^{3/2}} \int d\theta$$

where the $\theta$ integral goes from 0 to $2\pi$ and the $r$ integral from  to $\infty.$

The nice thing is that the integrals separate and the angular integral can be done to get $2\pi.$ This leave us with only a single integral

$$E_v = 2\pi k_C \sigma z \int \frac{\; r\; dr} {(r^2 + z^2)^{3/2}}$$

If we play our dimensional analysis trick and replace our dimensional variable $r$ by a dimensionless one $\rho$ by defining

$$\rho = r/z\;\;\;\;\;r = z\rho$$

we pull out a factor of 1/z and get the result

$$E_v = 2\pi k_C \sigma \int \frac{\rho d\rho} {(\rho^2 + 1)^{3/2}}$$

This is still a somewhat messy integral, but it's only in 1 variable and you learn to do this with variable substitutions in a first year calculus class. No vectors or multiple dimensions required!

Both of these methods are messy, but the physicist in me feels that that second approach integrates physical thinking in with the math and both gives insight into whats happening and results in a simpler integral to do using calculus.

The integral over $\rho$ turns out to have the value 1, giving our result

$$E_v = 2\pi k_C \sigma $$

Joe Redish 2/23