Applying Gauss' Law

Gauss' Law is a powerful method for calculating electric fields. It states that the electric field passing through a surface is proportional to the charge enclosed by that surface. For instance, if you have a solid conducting sphere (e.g., a metal ball) with a net charge Q, all the excess charge lies on the outside. Gauss' law tells us that the electric field inside the sphere is zero, and the electric field outside the sphere is the same as the field from a point charge with a net charge of Q. This result is true for a solid or hollow sphere. So we can say:

• The electric field is zero inside a conducting sphere.
• The electric field outside the sphere is given by: E = kQ/r², just like a point charge.
• The excess charge is located on the outside of the sphere.

In the simulation you can use the buttons to show or hide the charge distribution. Notice that for the hollow sphere above the excess charge does lie on the outside.

Using the above facts plus what we know about superposition, we can find out what the electric field due to 2 concentric spheres looks like. The smaller sphere is positive with a net charge of +4 C and the larger sphere is negative with a net charge of -3 C. Notice that the electric field for both spheres is just as we predicted from Gauss' Law: inside the sphere there is no field and outside the sphere the field is one of a point charge with the same sign and magnitude placed at the center of the sphere. If you use the buttons below the simulation, you can also see that the excess charge lies on the outside of the spheres.

Now we need only use the principle of superposition to find the electric field at all points. This means that the net electric field is the vector sum of the field from the smaller sphere alone and the larger sphere alone. We define positive as pointing radially outward and negative as pointing radially inward.

• Inside the smaller sphere the field is zero:
  • E_net = E_small+E_large = 0 + 0 = 0.
• In between the two spheres, the field is that of a +4 C point charge located at the center of the two spheres:
  • E_net = E_small+E_large = +k(4 C) / r² + 0 = + k(4 C) / r².
• Outside the larger sphere, the field is that of a +1 C point charge located at the center of the two spheres:
  • E_net = E_small+E_large = +k(4 C) / r² - k(3 C) / r² = k(+4 - 3) C / r² = +k(1 C) / r² .

Here is the net electric field from the 2 concentric spheres.

Something interesting to note is that when the inner sphere is introduced, the charge distribution on the outer sphere changes. The excess charge no longer lies only on the outside. The charge must redistribute itself so that E = 0 inside the conductor. This is most easily seen using field lines. Since field lines begin on positive charges and end on negative charges, every field line generated by the inner sphere must terminate on the inner surface of the outer sphere in order for there to be no electric field inside the conductor. Since the number of field lines is proportional to the charge, this means that the both surfaces would have the same amount of charge. So in this case, since the inner sphere has +4 C, -4 C gathers on the inside surface of the outer sphere. Since the outer sphere has a net charge of -3 C, then +1 C gathers at the outer surface of the outer sphere.

Two charged spheres - Click on the milestone icon to answer a conceptual question that will appear in the milestone window at the upper right. Click the Explanation button to see a detailed solution to the milestone question.