Themes > Science > Physics > Electromagnetism > Electrostatics > Gauss's law > Implications of Gauss' Law Gauss' Law is a powerful method of calculating electric fields. If you have a solid conducting sphere (e.g., a metal ball) that has a net charge Q on it, you know all the excess charge lies on the outside of the sphere. 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. That's a pretty neat result. The result for the sphere applies whether it's solid or hollow. Let's look at the hollow sphere, and make it more interesting by adding a point charge at the center. What does the electric field look like around this charge inside the hollow sphere? How is the negative charge distributed on the hollow sphere? To find the answers, keep these things in mind: The electric field must be zero inside the solid part of the sphere Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone We know that the electric field from the point charge is given by kq / r2. Because the charge is positive, the field points away from the charge. If we took the point charge out of the sphere, the field from the negative charge on the sphere would be zero inside the sphere, and given by kQ / r2 outside the sphere. The net electric field with the point charge and the charged sphere, then, is the sum of the fields from the point charge alone and from the sphere alone (except inside the solid part of the sphere, where the field must be zero). This is shown in the picture: How is the charge distributed on the sphere? The electrons must distribute themselves so the field is zero in the solid part. This means there must be -5 microcoulombs of charge on the inner surface, to stop all the field lines from the +5 microcoulomb point charge. There must then be +2 microcoulombs of charge on the outer surface of the sphere, to give a net charge of -5+2 = -3 microcoulombs. Information provided by: http://physics.bu.edu