# Help needed with (the point form of) Gauss's law & divergence of E and

1. ### Lisa...

189
I need to show that

$$\vec{\nabla} \cdot \vec{E}= \frac{\rho}{\epsilon_0}$$
where $$\rho$$ is the volume charge density.

I know that if I can show that the net flux of the electric field (in three directions xyz) out of the a small gaussian surface in the shape of a cube with faces parallel to the xy, xz and yz planes is:

$$\phi_n_e_t = ( \displaystyle{\frac{\delta E_x}{\delta x}} + \displaystyle{\frac{\delta E_y}{\delta y}} + \displaystyle{\frac{\delta E_z}{\delta z}} ) \Delta V$$ = $$(\vec{\nabla} \cdot \vec{E}) \Delta V$$

with $$\Delta V$$ is the volume enclosed by the gaussian surface.

then according to

$$\phi_n_e_t = \displaystyle{\frac{\Delta q}{\epsilon_0}} = \displaystyle{\frac{\rho \Delta V}{\epsilon_0}}= (\vec{\nabla} \cdot \vec{E}) \Delta V$$ the two $$\Delta V$$ cancel, leaving:

$$\vec{\nabla} \cdot \vec{E}= \frac{\rho}{\epsilon_0}$$

The only question is: how do I show that $$(\vec{\nabla} \cdot \vec{E}) \Delta V$$ is a correct formula?

Perhaps it would be easier to show that if the electric field was only along the x axis the equation would be:

$$\phi_n_e_t = \displaystyle{\frac{\delta E_x}{\delta x}} \Delta V$$

but how will I do that?

Last edited: Feb 18, 2006
2. ### dicerandom

308
I think you're correct in your approach, i.e. that you need to calculate the flux of field lines through a surface surrounding the point, however the divergence of the electric field will not depend on the size of the volume (providing there's only the one point charge).

I think you want to set up a surface integral, argue that the flux of the electric field lines is proportional to the charge enclosed, and use the divergence theorem to convert that surface integral to the divergence.

Edit: Also, are you familliar with spherical coordinates and the $\nabla$ operator in that coordinate system? I think this problem will be much easier in spherical coordinates.

Last edited: Feb 18, 2006
3. ### nrqed

3,048
This is easy if you think of a very small cube with sides dx, dy and dz, let's say with one corner at the origin (so the edges are at (0,0,0), (dx,0,0) and so on) . Consider the flux through the face at x=0 in the yz plane (the left face if you will). The flux through that side depends only on $$E_x$$, obviously. So it is simply $$- E_x dy dz$$ where the E field is evaluated at the coordinate (0,0,0). The reason for the minus sign is that if E_x is positive, the flux is entering our little box so it is a negative flux. You might ask why we don't use the E_x evaluated at (0,dy,0) or (0,dy/2,dz/2) or any other point in the left surface. The reason is that the whole expression is already of order dy dz (because of the surface area) so we can neglect any dy or dz dependence in E_x.

Now, consider the flux on the right face (or area dy dz but located at x=dx). That flux is $$E_x dy dz$$ but now with E_x evaluated at the coordinates (dx, 0, 0) .

So finally, the flux through the left face plus the flux through the right face is
$$- E_x(0,0,0) dy dz + E_x (dx,0,0) dy dz = { \partial E_x \over \partial x } dx dy dz ={ \partial E_x \over \partial x } dV$$

Adding the results from the other 4 faces gives you your result. Since any volume can be divided into a large number of tiny cubes, this applies to any volume.

Hope this helps.

Patrick

Last edited: Feb 18, 2006
4. ### Lisa...

189
Thanks a hell of a lot for your great explanation Patrick!!!

Last edited: Feb 18, 2006
5. ### nrqed

3,048
You are very welcome! It's a pleasure to help when people appreciate.

Patrick

6. ### Lisa...

189
Hey! I just went over your explanation again, but this time I didn't seem to get the following step:

What are you actually doing in order to get to $${ \partial E_x \over \partial x } dx dy dz ={ \partial E_x \over \partial x } dV$$ ?

7. ### arildno

12,015
In Cartesian coordinates, dxdydz is the infinitesemal volume dV of a box.

8. ### Lisa...

189
I know ;) But how did he get to the partial derivative? And why do you keep a $$E_x$$ factor. I mean $$-E_x + E_x= 0$$ right ?

9. ### Lisa...

189
Never mind :) I figured it out already :)

10. ### arildno

12,015
No, note that the arguments are distinct!
We take a first term Taylor series approximation of $E_{x}(dx,0,0)$ :
We have:
$$E_{x}(dx,0,0)=E_{x}(0,0,0)+\frac{\partial{E}_{x}}{\partial{x}}dx$$

Okay, I see you have figured it out.

11. ### Lisa...

189
Just two more things :) ;)

1) Why do you need to multiply $$E_x$$ by it's coordinates (I know that's because $$E_x$$ is looked upon at the coordinates it's multiplied with but why don't you add the coordinates for example to $$E_x$$?
2) What are you exactly doing in the Taylor series approximation of
$E_{x}(dx,0,0)$ ?

12. ### nrqed

3,048
By writing $$E_x(0,0,0)$$ for example, we do NOT mean that we are multiplying by the coordinates! We mean that we are *evaluating* the x component of the E field at those coordinates. (it's like saying y(3) to mean the function y(x) evaluated at x=3).

As for the other question, you can do a Taylor expansion like the other poster suggested but if that confuses you, you don't have to. Just consider

$${ E_x(dx,0,0) - E_x(0,0,0) \over dx}$$

In the limit dx being infinitesimal, this is, by definition, the derivative of the function $$E_x$$ with respect to x (right?). This i sjust the definition of a partial derivative.

Patrick