Prof. David R. Jackson ECE Dept. Spring 2016 Notes 10 ECE 3318 Applied Electricity and Magnetism 1

Gauss’s Law Assume q produces N f flux lines A charge q is inside a surface. Hence N S = N f ( all flux lines go through S ) 2 N S  # flux lines that go through S x y z q E S (closed surface) From the picture:

Gauss’s Law (cont.) NS = 0NS = 0 q S The charge q is now outside the surface (All flux lines enter and then leave the surface.) Hence 3

To summarize both cases, we have Gauss’s law: By superposition, this law must be true for arbitrary charges. Gauss’s Law (cont.) Carl Friedrich Gauss 4 (his signature)

Example Note: All of the charges contribute to the electric field in space. Note: E  0 on S 2 ! 5 - q q 2q2q S1S1 S2S2 S3S3

Using Gauss’s Law Gauss’s law can be used to obtain the electric field for certain problems. The problems must be highly symmetrical.  The problem must reduce to one unknown field component (in one of the three coordinate systems). 6 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 Guideline: Pick S to be  to E as much as possible Rule 2: S should go through the observation point (usually called r ). (This simplifies the dot product calculation.) 7 E S

Example Assume Find E Assume Point charge q S r r x y z 8 (only a function of r ) (only an r component) or Then

Example (cont.) We then have Hence 9 so

Note About Spherical Coordinates Note: In spherical coordinates, the LHS is always: 10 Assumption:

Example Find E everywhere Hollow shell of uniform surface charge Case a ) r < a x y z  s =  s0 a 11 r r a s0s0 Hence so LHS = RHS

Example (cont.) Case b) r > a The electric field outside the sphere of surface charge is the same as from a point charge at the origin. 12 r a r s0s0 Hence LHS = RHS

Example (cont.) r > a r < a r < a Note: A similar result holds for the force due to gravity from a shell of material mass. Summary x y z  s =  s0 a 13

Example (cont.) Note: The electric field is discontinuous as we cross the boundary of a surface charge density. x y z  s =  s0 a 14 a Q/(4  0 a 2 ) r ErEr

Example x y z  v =  v0 a Find E ( r ) everywhere Solid sphere of uniform volume charge Case a ) r < a 15 r a r V v0v0 Gaussian surface S

Example (cont.) Calculate RHS: 16 r a r V v0v0 Gaussian surface S LHS = RHS

Example (cont.) Hence, we have The vector electric field is then: 17 x y z  v =  v0 a r

Example (cont.) so Case b ) r > a Hence, we have 18 r a r  v =  v0 Gaussian surface S

Example (cont.) We can write this as: The electric field outside the sphere of charge is the same as from a point charge at the origin. where 19 Hence x y z  v =  v0 a r

Example (cont.) a  v0 a / (3  0 ) r ErEr (r < a)(r < a) (r > a)(r > a) Summary Note: The electric field is continuous as we cross the boundary of a volume charge density. x y z  v =  v0 a 20