# Gauss' Law differential form

1. Jan 24, 2009

### joker_900

1. The problem statement, all variables and given/known data
I just have a little question about Gauss' Law (differential form).

If divE = p/e0 where p is the charge density and e0 is permittivity of free space.

But if we had a sphere with a total net charge of Q, then outside the sphere, the field is E=k/r^2 I think.

Then at any r outside the sphere, divE evaluates to a non-zero value; but outside the sphere, p is zero at any r. So how does Gauss match up?

Thanks

2. Relevant equations

divE=p/e0

3. The attempt at a solution
In the first section

2. Jan 24, 2009

### AEM

This is a good question because it illustrates a fundamental point about mathematics and physics: It is important to understand exactly what an equation says. Gauss's law written as:

$$\nabla \cdot E = \frac {\rho}{\epsilon_o}$$

holds at a point (recall that derivatives are defined in terms of a limiting process at a point).

Pick a point, P, outside of your charged sphere. What the point form of Gauss's law says is that net divergence of electric field vectors from this point is proportional to the charge at that point. In your example, there is no charge at any point P outside of the charged sphere, so what gives? Well draw an imaginary small test sphere around the point P that you picked. There will be electric field lines (E vectors) from the charged sphere entering and leaving the test sphere. If you think about it carefully and maybe draw a careful picture, you'll be able to convince yourself that there are as many leaving as entering with out doing the required mathematics.

Gauss's law in the integral form reads:

$$\int_ S E \cdot dS = \int_V \frac {\rho}{\epsilon_o} dV$$

Applied to the test sphere this says in mathematical language what you figured out with a careful picture: If the charge inside is zero, than as many field lines leave as enter if rho = 0.

Now, Gauss's theorem from vector analysis says the integral over the surface on the left of the equation above can be replaced by a volume integral of the divergence:

$$\int_V \nabla \cdot E dV = \int_V \frac {\rho}{\epsilon_o} dV$$

Imagine your test sphere shrinking smaller and smaller around the point P that you picked earlier. There will still be the same number of electric field line leaving the sphere as entering (no charge there) and the mathematics says that if the two integrals in this last equation are equal the stuff inside has to be equal, so it has to be that

$$\nabla \cdot E = \frac{\rho}{\epsilon_o}$$

What I've tried to show you here is that the two forms of Gauss's law -- the integral form and the differential form-- actually say the same thing. Hopefully this long winded explanation has helped you to see that there is no inconsistency in the mathematics it's just a matter of understanding what the math symbolism says.

3. Jan 24, 2009

### AEM

I just reread my previous post an realized that I forgot to explicitly say: At a point P outside the charged sphere $$\nabla \cdot E = 0$$ since there is no charge at the point. Joker 900 is wrong about that.

4. Jan 24, 2009

### joker_900

But if the field is proportional to r^(-2) and in the r direction, then divE is just the differential of r^(-2) wrt r. Which is proportional to r^(-3). So at some point r=a outside the sphere, divE=a^(-3), not zero.

I still don't understand

5. Jan 24, 2009

### AEM

Let me try to rephrase things a little differently. Gauss's law in differential form is talking about the electric field arising from what is happening at a particular point . That's why I told you to surround your point with a test sphere --to isolate point P in your mind from the rest of space including the charged sphere.

You seem to indicate above that when you are doing the calculation shown, you are thinking in terms of the electric field from the nearby charged sphere, but that is not correct because the source of that field is elsewhere.