# Confirmation of Gauss's Law

1. Jan 21, 2006

### Fledy

Okay, first off, some quick background. I am a student in AP Physics C, and I like to think I generally have a good grasp of physics (I read about it outside the class). We just learned about Gauss's Law this week, and I have a good grasp of it. I understand how it can be used to simplify some electric field equations and whatnot, and (at least from my point) this seems to be the primary purpose of the law: a generalization, and also a simplification.

So then.

I also think I have a fairly good grasp of multivariable calc (i'm taking a correspondence course). So of course, even though we don't deal with a full-fledged surface integral in AP Physics C, I understand the formulation of Gauss's law in the terms of a surface integral.

With that in mind, I have been trying to confirm Gauss's law for a point charge and a spherical gaussian surface centered at the charge using multivariable calculus techniques. I know, it defeats the purpose of Gauss's law, but I like to see these things. Problem is, I wind up finding an integral that evaluates to 0. Ahem.

Here's what I first did: (I won't use the Tex notation for now, but I'll use boldface for vectors)

Position vector x = xi + yj + zk

E(x) = k q x / x³

where x is just the magnitude of x.

Next, I parameterized a sphere with radius r as

A(u,v) = r (sin(u) cos(v), sin(u) sin(v), cos(u)).

(u runs from 0 to pi, v runs from -pi to pi)

Now, to evaluate the integral of E . dA, I had to have an expression for dA. I used the differential area element. This is given by

dA = (∂A/∂u × ∂A/∂v) du dv

Finally, I integrate:

E(A(u,v)) . (∂A/∂u × ∂A/∂v) du dv

with u running from 0 to pi, and v running from -pi to pi.

Problem: This whole integral evaluates to 0, and not the expected q / epsilon naught.

...

Okay...

Time for a different approach! If I take the divergence of E(x,y,z), I should get rho / epsilon nought, the charge density over epsilon nought.

I basically used the same definition of electric field above, expanding r and r in terms of x, y, and z. Simply put, replace r with (x, y, z) and r with sqrt(x² + y² + z²). Del = (∂/∂x, ∂/∂y, ∂/∂z).

Some basic rules of calculus let me calculate del . E
= ∂E/∂x + ∂E/∂y + ∂E/∂z

...

And it came out to 0. Again. ... Oops.

So basically, I want to know what I'm doing wrong. Why doesn't doing this yield the net charge over epsilon nought, or the charge density over epsilon nought as they supposedly should? Is this an incorrect application of multivariable calculus or gaussian surfaces? I really must know where I went wrong.

Thanks
Much love,
Steve

2. Jan 22, 2006

### Claude Bile

The problem lies with the point charge. Because it has no distribution in space (i.e. it is a discontinuity). Whenever you integrate the volume, you naturally omit the discontinuity (when you shouldn't), hence the seemingly contradictory answers.

The solution is to use a delta function to represent the point charge, or alternatively use a charge that is distributed throughout space.

Note that this is a classic EM problem and one that is discussed at length by any EM text worth its salt.

Claude.

3. Jan 22, 2006

### krab

What's wrong with that? The charge density at the location of the evaluation of the div as in fact zero, so you got the correct answer.

4. Jan 22, 2006

### krab

What is this A? It's certainly not an area you intend it to be.
What's wrong with that? The charge density at the location of the evaluation of the div as in fact zero, so you got the correct answer.

5. Jan 22, 2006

### Fledy

Okay, so I understand that part now. There is no charge density. I figured there was something fishy there.

So I suppose I see how to apply a different mathematical approach here, but it's just a bit disappointing. Ah well.

So what is a delta function? And how else can I just integrate in this space to get the charge?

6. Jan 22, 2006

### Staff: Mentor

Try this route instead: first split up $d \vec A$ into magnitude (area) and direction:

$$d \vec A = \hat n dA$$

where $\hat n$ is the unit vector normal to the surface. For a sphere this is radially outward, so you can construct it from the position vector of a point on the sphere:

$$\hat n = \frac {x \hat i + y \hat j + z \hat k}{r}$$

(I use r instead of your x for the magnitude of the position vector.)

For the area element, your u and v are what most books call $\theta$ (the latitude-like angle measured from the "north pole") and $\phi$ (the "longitude" or azimuthal angle). The area of a section of the sphere marked off by $d \theta$ and $d \phi$ is

[tex]dA = (r d \theta)(r \sin {\theta} d \phi) = r^2 \sin {\theta} d \theta d \phi[/itex]

Set up $\vec E \cdot \hat n dA$. It will include the dot product
$(x \hat i + y \hat j + z \hat k) \cdot (x \hat i + y \hat j + z \hat k)$ which equals...

Now integrate over the angles and things should work out.

Last edited: Jan 22, 2006
7. Jan 23, 2006

### Claude Bile

For a point charge, the charge density is zero except at the origin (where I assume the point charge is located), where the charge density is infinity.

The delta function is a mathematical function designed to represent this discontinuity. More info on the delta function can be found here;

http://mathworld.wolfram.com/DeltaFunction.html

Claude.

8. Jan 24, 2006

### jackiefrost

Uhhh... he's not (or shouldn't be) integrating at the charge location but rather at the spherical surface at some finite radius r > 0. So, I really don't see why the worry about the charge density of the point charge.

The problem as I see it is that E must be stated correctly (from Columb's law) for a typical surface element dA of that spherical surface. If we're going to integrate E dA then what is the expression for E in terms of charge and distance from source for a typical surface element dA? Once that's settled and the integration performed it won't yield rho/epsilon_0 (charge density) but rather Q/epsilon_0 (total flux - i.e. total charge of sphere).

Ok, ok... I only had four hours sleep and probably don't know what I'm talking about anyway :-)

Last edited: Jan 24, 2006
9. Jan 24, 2006

### Claude Bile

No, you're right. I also lack sleep. Well, that's my excuse anyway.

Claude.