Gauss’ Law.pdf

7/27/2019 Gauss Law.pdf

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Maxwells Equation

Reference: Ch 23 Fundamentals of Physics, Halliday and Resnick


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SurfaceS1. The electric field is outward for all

points on this surface. Thus, the flux of the

electric field through this surface is positive, as

well as the net charge within the surface.

SurfaceS2The electric field is inward for allpoints on this surface. Thus, the flux of the

electric field is negative and so is the enclosed

charge.

SurfaceS3. This surface encloses no charge.

Gauss' law requires that the net flux of the

electric field through this surface be zero. That isreasonable because all the field lines pass

entirely through the surface, entering it at the

top and leaving at the bottom.

SurfaceS4. This surface encloses no net charge,

because the enclosed positive and negative

charges have equal magnitudes. Gauss' lawrequires that the net flux of the electric field

through this surface be zero. That is reasonable

because there are as many field lines leaving

surface S4 as entering it.


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Since the plate is thin and very large, we can assume that essentially

all the excess charge is on the two large faces of the plate.

0

1

E

0

1

E

0

12

E


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From the symmetry of the situation, we

know that at any point the electric field is

also perpendicular to the surface and

directed outward from the interior. Thus,since the angle between Eand dA is zero,

we can Gauss' law as

Here qenc= q. Although E varies radially with distance from q, it

has the same value everywhere on the spherical surface.


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If we apply Gauss' law to surface S2, for which r R, we would

find that

Here q = qencby the Gaussian surface.

This field is the same as one set up by a point charge q at the

center of the shell of charge. Thus, the force produced by a shell

of charge q on a charged particle placed outside the shell is the

same as the force produced by a point charge q located at the

center of the shell.

This proves the first shell theorem.


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This Gaussian surface encloses no charge. Thus, if a charged

particle were enclosed by the shell, the shell would exert no

net electrostatic force on the particle.

This proves the second shell theorem.


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To find the electric field at points on this Gaussian surface, we

consider two sets of charged shellsone set inside the Gaussian

surface and one set outside.

the charge lying outside the Gaussian surface does not set up a

net electric field on the Gaussian surface.

the charge enclosed by the surface sets up an electric field as if

that enclosed charge were concentrated at the center.

Letting q'represent that enclosed charge, we can then write Eas


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If the full charge qenclosed within radius R

is uniform, then qenclosed within radius rin the figure is proportional to q:


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The figure shows two large, parallel, nonconducting sheets

with identical (positive) uniform surface charge densities, and a

sphere with a uniform (positive) volume charge density. Rank

the four numbered points according to the magnitude of the

net electric field there, greatest first.

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Gauss’ Law.pdf