Notes 10 ECE 3318 Applied Electricity and Magnetism Gauss’s Law I Spring 2019 Prof. David R. Jackson Dept. of ECE Notes 10 Gauss’s Law I Notes prepared by the EM Group University of Houston

NS  # flux lines that go through S Gauss’s Law A charge q is inside a closed surface. S (closed surface) NS  # flux lines that go through S From the picture: NS = Nf (all flux lines go through S) Assume q produces Nf flux lines Hence

Gauss’s Law (cont.) NS = 0 The charge q is now outside the surface (All flux lines enter and then leave the surface.) Hence

Gauss’s Law (cont.) To summarize both cases, we have Gauss’s law: Carl Friedrich Gauss By superposition, this law must be true for arbitrary charges. (his signature)

Note: All of the charges contribute to the electric field in space. Example Note: E  0 on S2 ! Note: All of the charges contribute to the electric field in space.

Using Gauss’s Law Gauss’s law can be used to obtain the electric field for certain problems in a simple way. The problems must be highly symmetrical.  The problem must reduce to one unknown field component (in one of the three coordinate systems). Note: When Gauss’s law works, it is usually easier to use than Coulomb’s law (i.e., the superposition formula).

Choice of Gaussian Surface Rule 1: S must be a closed surface. Rule 2: S should go through the observation point (usually called r). Guideline: Pick S to be  to E as much as possible E S (This simplifies the dot product calculation.)

Example Point charge Find E

Example (cont.) S Assume Assume Then or (only an r component) (only a function of r) Then or

Example (cont.) We then have so Hence

Example (cont.) Summary

Note About Spherical Coordinates Note: In spherical coordinates, the LHS is always the same: Assumption: Spherical Gaussian surface Assumption

Hollow shell of uniform surface charge density Example Hollow shell of uniform surface charge density Find E everywhere

Example (cont.) Case a) r < a LHS = RHS so Hence

Example (cont.) Case b) r > a LHS = RHS Hence The electric field outside the sphere of surface charge is the same as from a point charge at the origin.

Example (cont.) Summary Note: A similar result holds for the force due to gravity from a shell of material mass.

Example (cont.) Note: The electric field is discontinuous as we cross the boundary of a surface charge density.

Solid sphere of uniform volume charge density Example Solid sphere of uniform volume charge density Find E (r) everywhere

Example (cont.) Case a) r < a Gaussian surface S

Example (cont.) Calculate RHS: Gaussian surface S LHS = RHS

Example (cont.) Hence, we have The vector electric field is then:

Example (cont.) Case b) r > a Gaussian surface S so Hence, we have

Example (cont.) We can write this as: Hence where The electric field outside the sphere of charge is the same as from a point charge at the origin.

Example (cont.) Summary Note: The electric field is continuous as we cross the boundary of a volume charge density.