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ELECTROSTATICS - GAUSS’ LAW
Fields and Waves Lesson 3.2 ELECTROSTATICS - GAUSS’ LAW Darryl Michael/GE CRD
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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
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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
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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:
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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)
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GAUSS’ LAW Use Gaussian surface to “pull” this out of integral Integral now becomes: Usually an easy integral for surfaces under consideration
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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
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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
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GAUSS’ LAW =0, since Evaluate LHS: These two integrals are equal
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GAUSS’ LAW Key Step: Take E out of the Integral Computation of enclosed charge
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GAUSS’ LAW (drop the prime) Do Problem 3a
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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’
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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
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GAUSS’ LAW Once again, For the region outside |z|>a
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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
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