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Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Chapter 28. Gausss Law

Using Gausss law, we can

deduce electric fields,

particularly those with a

high degree of symmetry,

simply from the shape of the

charge distribution. Thenearly spherical shape of the

girls head determines the

electric field that causes her

hair to stream outward.Chapter Goal: To

understand and apply

Gausss law.


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The electric field inside aconductor in electrostatic

equilibrium is

A.uniform.B.zero.C.radial.D.symmetric.


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The electric field inside aconductor in electrostatic

equilibrium is

A.uniform.B.zero.C.radial.D.symmetric.


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The Electric Flux

The electric flux measures the flow of electric field. The

flux through a small surface of areaA whose normal to the

surface is tilted at angle from the field is:

Or using the dot-product:

We have assumed A is so small that E is uniform within it.


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When the electric field is parallel to the surface so perpendicular

to the area vector, the flux thru the surface vanishes. (angle is 90

degrees) . When the electric field is pperpendicular to the

surface so parallel to the area vector, the flux thru the surface is

maximal and EA. (angle is 0 degrees).

The Electric Flux


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EXAMPLE 28.1 The electric flux inside a

parallel-plate capacitor

QUESTION:


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EXAMPLE 28.1 The electric flux inside a

parallel-plate capacitor


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The Electric Flux for a finite surface

When the electric field is non

uniform or the surface curved,

approximate the total flux by

adding the contributions of

little pieces. This defines the

surface integral.

E

i=E

iA

icos=

E

i

A

i


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Gausss Law

For any closedsurface enclosing total charge Qin,the net

electric flux through the surface is

This result for the electric flux is known as Gausss Law.

Gausss law follows from Coulombs law but is more general. It

applies when charges moves and Coulombs law is invalid.


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Using Gausss Law

1.Gausss law applies only to a closedsurface, called aGaussian surface.

2.A Gaussian surface is not a physical surface. It is animaginary, mathematical surface in the space

surrounding one or more charges.

3.We cant generally find the electric field from Gaussslaw alone. We can apply Gausss law when from

symmetry and superposition, we already can guess the

shape of the field.


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Gausss Law for a point charge

12

Consider a closed spherical surface,centered on a positive point charge

The E-field is perpendicular to surface,directed outward

E-field magnitude on sphere:

All surface elements have the sameelectric field:

E=1

4o

q

R2

R

E =E d

A = EdA =E dA =EA = 1

4o

q

R2

4R2( ) = q /o


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13

13

q

Two spheres of radiusR and 2R are centered on thepositive charges of the same value q.

The electric flux through the spheres compare as

A)Flux(R)=Flux(2R)

B)Flux(R)=2*Flux(2R)C)Flux(R)=(1/2)*Flux(2R)D)Flux(R)=4*Flux(2R)E) Flux(R)=(1/4)*Flux(2R)

q

R

2R


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14

q

Two spheres of the same size enclose different positivecharges, q and 2q. The flux through these spheres

compare as

A)Flux(A)=Flux(B)

B)Flux(A)=2Flux(B)C)Flux(A)=(1/2)Flux(B)D)Flux(A)=4Flux(B)E) Flux(A)=(1/4)Flux(B)

2q

A B


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General Gausss Law

E =E d

A = Q

enclosed

o

Works for any closedsurface, not only sphere

q

Works for any chargedistribution (by

superposition)


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Field of a sphere of charge


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EXAMPLE 28.3 Outside a sphere of charge


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EXAMPLE 28.3 Outside a sphere of charge


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Line charge

19

Field direction: radially out from line charge Gaussian surface:

Cylinder of radius r Area where

Value of on this area:

Flux thru Gaussian surface:

Charge enclosed:

E d

A 0 :

2rL

E dAE

L

E2rL

Gauss law: E2rL = L /o E =

1

2o

r


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Plane of charge

20

Field direction: perpendicular to plane Gaussian surface:

Cylinder of radius r Area where

Value of on this area:

Flux thru Gaussian surface:

Charge enclosed:

E d

A 0 :

2r2

E dAE

r2

E2r2

Gauss law: E2r2 =r2/o E =

2o

Surface charge C/m2


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Conductors in Electrostatic Equilibrium

The electric field is zero at all

points within a conductor in

electrostatic equilibrium. Else

charge carriers would move.

----

-

-

+ +++++


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Apply Gausss Law to a pill box

spanning the surface of a

conductor. The charge enclosed is

A where is the surface charge

density of the conductor. The total

outward flux is EA from the outer

disk parallel to the local surface

surface and must be the charge

enclosed.

Conductor surface charge density


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EXAMPLE 28.7 The electric field at the

surface of a charged metal sphere

QUESTION:


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A uniformly charged rod has afinite

lengthL. The rod is symmetric under

rotations about the axis and under

reflection in any plane containing the

axis. It is notsymmetric under

translations or under reflections in a

plane perpendicular to the axis otherthan the plane that bisects the rod.

Which field shape or shapes match the

symmetry of the rod?

A.c and eB.a and dC.e onlyD.b onlyE.none of the above


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A uniformly charged rod has afinite

lengthL. The rod is symmetric under

rotations about the axis and under

reflection in any plane containing the

axis. It is notsymmetric under

translations or under reflections in a

plane perpendicular to the axis otherthan the plane that bisects the rod.

Which field shape or shapes match the

symmetry of the rod?

A.c and eB.a and dC.e onlyD.b onlyE.none of the above


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This box contains

A.a net positive charge.B.a net negative charge.C.a negative charge.D.a positive charge.E. no net charge.


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This box contains

A.a net positive charge.B.a net negative charge.C.a negative charge.D.a positive charge.E. no net charge.


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The total electric flux through this box isA.6 Nm2/C.B.4 Nm2/C.C.2 Nm2/C.D.1 Nm2/C.E.0 Nm2/C.


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A.6 Nm2/C.B.4 Nm2/C.C.2 Nm2/C. D.1 Nm2/C.E.0 Nm2/C.

The total electric flux through this box is

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