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Chapter 24

Gausss Law

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Electric Flux Electric fluxis the

product of the

magnitude of theelectric field and the

surface area,A,

perpendicular to the

field E = EA

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Electric Flux, General Area The electric flux is

proportional to the

number of electric fieldlines penetrating some

surface

The field lines may

make some angle

with the perpendicular

to the surface

Then E = EA cos

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Electric Flux, Interpreting the

Equation The flux is a maximum when the

surface is perpendicular to the field

The flux is zero when the surface is

parallel to the field

If the field varies over the surface, =

EA cos is valid for only a smallelement of the area

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Electric Flux, General In the more general

case, look at a small

area element

In general, this

becomes

cosE i i i i i

E A E A

0surface

limi

E i iA

E A d

E A

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Electric Flux, final The surface integral means the integral

must be evaluated over the surface in

question In general, the value of the flux will

depend both on the field pattern and on

the surface The units of electric flux will be N.m2/C2

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Electric Flux, Closed Surface Assume a closed

surface

The vectors Aipoint in differentdirections At each point, they

are perpendicular tothe surface By convention, they

point outward

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Active Figure 24.4

(SLIDESHOW MODE ONLY)

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Flux Through Closed Surface,

cont.

At (1), the field lines are crossing the surface from the

inside to the outside; < 90o

, is positive At (2), the field lines graze surface; = 90o, = 0 At (3), the field lines are crossing the surface from the

outside to the inside;180o > > 90o, is negative

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Flux Through Closed Surface,

final The netflux through the surface is

proportional to the net number of lines

leaving the surface This net number of lines is the number of

lines leaving the surface minus the number

entering the surface IfEn is the component ofE

perpendicular to the surface, then

E nd E dA E A

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Gausss Law, Introduction Gausss law is an expression of the

general relationship between the net

electric flux through a closed surfaceand the charge enclosed by the surface The closed surface is often called a

gaussian surface Gausss law is of fundamental

importance in the study of electric fields

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Gausss Law General A positive point

charge, q, is locatedat the center of asphere of radius r

The magnitude ofthe electric field

everywhere on thesurface of thesphere isE= keq / r

2

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Gausss Law General, cont. The field lines are directed radially outward

and are perpendicular to the surface at

every point

This will be the net flux through the

gaussian surface, the sphere of radius r We know E= keq/r

2 andAsphere = 4r2,

Ed E dA E A

4E eo

qk q

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Gausss Law General, notes The net flux through any closed surface

surrounding a point charge, q, is given by q/o

and is independent of the shape of that surface The net electric flux through a closed surface that

surrounds no charge is zero

Since the electric field due to many charges is

the vector sum of the electric fields produced bythe individual charges, the flux through any

closed surface can be expressed as

1 2d d

E A E E AK

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Gausss Law Final Gausss law states

qin is the net charge inside the surface

E represents the electric field at any point on

the surface

E is the total electric fieldand may have contributionsfrom charges both inside and outside of the surface

Although Gausss law can, in theory, be solved

to find E for any charge configuration, in

practice it is limited to symmetric situations

E A inEo

qd

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Active Figure 24.9

(SLIDESHOW MODE ONLY)

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Applying Gausss Law To use Gausss law, you want to

choose a gaussian surface over which

the surface integral can be simplifiedand the electric field determined

Take advantage of symmetry

Remember, the gaussian surface is asurface you choose, it does not have tocoincide with a real surface

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Conditions for a Gaussian

Surface Try to choose a surface that satisfies one or

more of these conditions:

The value of the electric field can be argued fromsymmetry to be constant over the surface

The dot product ofE.dA can be expressed as a

simple algebraic product EdA because E and dA

are parallel The dot product is 0 because E and dA are

perpendicular

The field can be argued to be zero over the surface

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Field Due to a Point Charge Choose a sphere as the

gaussian surface

E is parallel to dA at eachpoint on the surface

2

2 2

(4 )

4

E

o

e

o

qd EdA

E dA Er

q qE k

r r

E A

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Spherically Symmetric, cont. Select a sphere as

the gaussian

surface, r< a qin < Q

qin = r(4/3r3)

in

in2 34

E

o

e

o

qd EdA

q QE k r

r a

E A

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Spherically Symmetric

Distribution, final Inside the sphere, E

varies linearly with r

E 0 as r 0 The field outside the

sphere is equivalent

to that of a point

charge located atthe center of the

sphere

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Field Due to a Thin Spherical

Shell Use spheres as the gaussian surfaces

When r> a, the charge inside the surface is Q and

E= keQ / r2

When r< a, the charge inside the surface is 0 and E= 0

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Field at a Distance from a Line

of Charge Select a cylindrical

charge distribution The cylinder has a

radius ofrand a lengthof

E is constant inmagnitude and

perpendicular to thesurface at every pointon the curved part ofthe surface

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Field Due to a Line of Charge,

cont. The end view

confirms the field is

perpendicular to thecurved surface

The field through the

ends of the cylinder

is 0 since the field isparallel to these

surfaces

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Field Due to a Line of Charge,

final Use Gausss law to find the field

in

2

22

E

o

o

e

o

qd EdA

Er

E k r r

E Al

l

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Field Due to a Plane of

Charge E must be

perpendicular to theplane and must havethe same magnitude atall points equidistantfrom the plane

Choose a small cylinderwhose axis isperpendicular to theplane for the gaussiansurface

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Field Due to a Plane of

Charge, cont E is parallel to the curved surface and

there is no contribution to the surface

area from this curved part of thecylinder

The flux through each end of the

cylinder is EA and so the total flux is2EA

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Field Due to a Plane of

Charge, final The total charge in the surface is A

Applying Gausss law

Note, this does not depend on r Therefore, the field is uniform

everywhere

22

E

o o

A EA and E

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Electrostatic Equilibrium When there is no net motion of charge

within a conductor, the conductor is said

to be in electrostatic equilibrium

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Properties of a Conductor in

Electrostatic Equilibrium The electric field is zero everywhere inside

the conductor

If an isolated conductor carries a charge, thecharge resides on its surface The electric field just outside a charged

conductor is perpendicular to the surface and

has a magnitude of/o On an irregularly shaped conductor, the

surface charge density is greatest at locationswhere the radius of curvature is the smallest

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Property 1: Einside

= 0

Consider a conducting slab in

an external field E

If the field inside the conductor

were not zero, free electrons inthe conductor would

experience an electrical force

These electrons would

accelerate These electrons would not be

in equilibrium

Therefore, there cannot be a

field inside the conductor

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Property 1: Einside

= 0, cont.

Before the external field is applied, freeelectrons are distributed throughout theconductor

When the external field is applied, theelectrons redistribute until the magnitude ofthe internal field equals the magnitude of theexternal field

There is a net field of zero inside theconductor

This redistribution takes about 10-15s and can

be considered instantaneous

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Property 2: Charge Resides

on the Surface Choose a gaussian surface

inside but close to the actual

surface

The electric field inside iszero (prop. 1)

There is no net flux through

the gaussian surface

Because the gaussiansurface can be as close to

the actual surface as

desired, there can be no

charge inside the surface

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Property 2: Charge Resides

on the Surface, cont Since no net charge can be inside the

surface, any net charge must reside on

the surface Gausss law does not indicate the

distribution of these charges, only that it

must be on the surface of the conductor

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Property 3: Fields Magnitude

and Direction Choose a cylinder as

the gaussian surface

The field must beperpendicular to thesurface If there were a parallel

component to E,

charges wouldexperience a force andaccelerate along thesurface and it wouldnot be in equilibrium

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Property 3: Fields Magnitude

and Direction, cont. The net flux through the gaussian

surface is through only the flat face

outside the conductor The field here is perpendicular to the

surface

Applying Gausss law

E

o o

A EA and E

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Conductors in Equilibrium,

example The field lines are

perpendicular to

both conductors There are no field

lines inside the

cylinder

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Derivation of Gausss Law We will use a solid

angle,

A spherical surfaceof radius rcontains

an area element A

The solid angle

subtended at the

center of the sphere

is defined to be2

A

r

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Some Notes About Solid

Angles A and r2 have the same units, so is a

dimensionless ratio

We give the name steradian to thisdimensionless ratio

The total solid angle subtended by a

sphere is 4steradians

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Derivation of Gausss Law,

cont. Consider a point

charge, q, surroundedby a closed surface ofarbitrary shape

The total flux throughthis surface can befound by evaluatingE.A for each smallarea element andsumming over all theelements

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Derivation of Gausss Law,

final The flux through each element is

Relating to the solid angle

where this is the solid angle subtended byA

The total flux is

2cos

cosE eA

E A k qr

E A

2

cosA

r

2

cos

E e eo

dA q

k q k q d r

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Derivation of Gausss Law

Notes This result matches the earlier result for

the equation for Gausss law

This derivation was independent of theshape of the closed surface

This derivation was independent of the

position of the charge within the surface