# Gauss' Law

Interactive 5.2.1

Let's start with a little geometric experiment, using the interactive image on the right. Start by dragging a positive charge onto the center of the white area; the electric field lines will then appear. Now click on a part of the white area and drag to form a box. We will define the electric flux through this box as the number of field lines that leave the box, minus the number of field lines that enter it. As you draw a box on the simulation, the points where field lines cross the box's border are marked with red arrows, and the total flux is shown below the figure.

Try dragging boxes in several regions of the figure. You should discover two basic cases:

• If the charge is inside the box you draw, then the total flux is 8: there are 8 field lines leaving the positive charge and they all must go to infinity, and the only way to get to infinity is to pass through the box.
• If the charge is outside the box you draw, then any field line that enters the box must leave it as well, and so the total flux is 0.
(The flux-calculating program isn't perfect, so you can get other results if a field line passes through the corner of a box, or if the charge sits on an edge of the box. These are spurious results; don't worry about them.)

Now add a negative charge, creating a dipole. You should note that drawing a box around the positive charge still yields a positive flux, and drawing a box around the negative charge gives a negative flux (because all the field lines are coming inward). However, if you draw a box around both charges, the net flux is zero.

These results are the consequence and illustration of Gauss' Law:
Gauss' Law: The total charge inside a closed surface is proportional to the electric flux through the surface.

Play around with this demonstration a little more: try adding more charges or increasing the magnitude of the charges there. Convince yourself that Gauss' Law is true. This demonstration is purely two-dimensional, but Gauss' Law applies in three dimensions as well, so long as we use a closed surface (like a sphere or cube).

You may have noticed the word proportional in the statement of Gauss' Law. When drawing a field line diagram, we can choose to draw as many field lines as we like, and the total flux depends on our choice. We must be sure to give each charge the proper number of field lines entering or leaving, however: if one charge is twice as large as another, it should have twice as many field lines. This can't always be done precisely (for example, if one charge is 1C and another is π C), and there is a more precise mathematical formulation of Gauss' Law which we will discuss in a future section.

Gauss' Law is a neat little result, but it might not seem very practical: it tells us about this quantity electric flux, but electric flux is not a particularly useful quantity. However, we will show that Gauss' Law can allow us to derive the electric field of certain symmetric charge distributions in a simple way. Before we can do that, however, we need to discuss symmetry.

Consider the field line diagram to the right, created by charges A-D.
Which charge has the largest magnitude?
Is the net charge of this system positive, negative, or neutral?
Is the total charge of A and B positive, negative, or neutral? In a previous section, we looked at the electric field of this charge distribution, and found that there was a point to the right of the charges (marked here by a star) where the electric field was zero. To the left of the star, the field line points towards the negative charge because it is closer, but on the right, the field line points away from the system because the net charge of the system is positive. As we move along the x axis, the electric field has to change directions somewhere, and at that point the electric field is zero.
Does the electric field immediately above the star point upwards or downwards? Use Gauss' Law to make your argument. Suppose we drew a Gaussian sphere around the star. There is no charge at the star so the net flux through the sphere must be zero. Since we have two field lines pointing out of the sphere, there must be some field lines in other places which are pointing into the sphere, and it stands to reason that this will include a spot immediately above or below the star: thus the field immediately above the star probably points downwards. This point where E=0 is sometimes called a saddle point: it is like a charge in that field lines start and end there, but unlike a charge it has some field lines which go in and field lines which go out.