ELECTROSTATICS - GAUSS’ LAW Fields and Waves Lesson 3.2 ELECTROSTATICS - GAUSS’ LAW Darryl Michael/GE CRD

MAXWELL’S FIRST EQUATION Enclosed Charge Differential Form Integral Form - ‘dv’ integral over volume enclosed by ‘ds’ integral For vacuum and air - think of D and E as being the same D vs E depends on materials constant

Use Gauss’ Law to find D and E in symmetric problems GAUSS’ LAW - strategy Do Problem 1 Use Gauss’ Law to find D and E in symmetric problems Get D or E out of integral Always look at symmetry of the problem - and take advantage of this

GAUSS’ LAW - use of symmetry - charges are infinite in extent on say x,y plane Example: A sheet of charge , is sum due to all charges Arbitrary Point P , points in all other components cancel only a function of z (not x or y) x y z Surface of infinite extent of charge Can write down:

GAUSS’ LAW Do Problem 2a Problem 2b ,is constant. For example a planar sheet of charge, where z is constant Problem 2c To use GAUSS’ LAW, we need to find a surface that encloses the volume GAUSSIAN SURFACE - takes advantage of symmetry - when r is only a f(r) - when r is only a f(z)

GAUSS’ LAW Use Gaussian surface to “pull” this out of integral Integral now becomes: Usually an easy integral for surfaces under consideration

Example of using GAUSS’ law to find Z = a -a < z < a Z > a ; z< -a Z = -a “a slab of charge” By symmetry: From symmetry If r0 > 0, then Z=0

in region |z| < a and create a surface at arbitrary z GAUSS’ LAW First get in region |z| < a and create a surface at arbitrary z Use Gaussian surface with top at z = z’ and the bottom at -z’ Note: Gaussian Surface is NOT a material boundary

GAUSS’ LAW =0, since Evaluate LHS: These two integrals are equal

GAUSS’ LAW Key Step: Take E out of the Integral Computation of enclosed charge

GAUSS’ LAW (drop the prime) Do Problem 3a

Back to rectangular, slab geometry example….. GAUSS’ LAW Back to rectangular, slab geometry example….. Need to find , for |z| > a Z = -a Z = a r0 Z = -z’ Z = z’

GAUSS’ LAW As before, Computation of enclosed charge Note that the z-integration is from -a to a ; there is NO CHARGE outside |z|>a

GAUSS’ LAW Once again, For the region outside |z|>a

Note: E-field is continuous GAUSS’ LAW -a z a Note: E-field is continuous Plot of E-field as a function of z for planar example