Gauss Law.pd

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Chapter 23 Gauss' Law

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Instead of considering the electric fields of charge elementsin a given charge distribution, Gauss' law considers ahypothetical closed surface enclosing the charge distrib-ution. This Gaussian surface can have any shape.

find the electric field on the surface by usingthe fact thatGauss' law relates the electric fields at points ona (closed) Gaussian surface to the net chargeenclosed by that surface.

If we know the electric field on a Gaussian surface,we can find the net charge enclosed by the surface.

23-2 Flux


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Let represent the volume flow rate (volume per unit time) at which air flows

through the loop.

the rate of volume flow through the loop is

where is an area vector whose magnitude isequal to an area and whose direction is normalto the plane of the area.

This rate of flow through an area is an exampleof a flux a volume flux in this situation.

In a more abstract way, the above equation isregarded as the flux of the velocity field throughthe loop. Thus flux means the product of an areaand the field across that area.

A provisional definition for the flux of the electric fieldfor the Gaussian surface is

A

= E A .

23-3 Flux of an Electric Field

= v cos A = vA cos = vA ,


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For the area vectors approaching a differential limit, the sum becomes anintegral and the definition of electric flux is

The flux of the electric field is a scalar, and its SI unit is the newton square-meter per coulomb (N m2/C).

the magnitude E is proportional to the number of electric field lines per unitarea. Thus, the scalar product is proportional to the number of electric fieldlines passing through the area dA. Thus

The electric flux through a Gaussian surface is proportional to the net numberof electric field lines passing through that surface.

problem 23-1 problem 23-2

=Ed A electric flux through a Gaussian surface .


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Gauss' law relates the net flux of an electric field through a closed surface (a

Gaussian surface) to the netcharge qenc that is enclosedby that surface,

If qenc is positive, the net flux is outward; if qenc is

negative, the net flux is inward.

The only things that matter are the magnitude andsign of the net enclosed charge. Charge outside the

surface, and the exact form and location of the chargesinside the Gaussian surface are of no concern.

problem 23-3

23-4 Gauss' Law

0 = 0

Ed A = qenc Gauss' Law ,


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From the symmetry of the situation, at anypoint the electric field is perpendicular to the

spherical Gaussian surface and directed outwardfrom the interior, then

Although E varies radially with distance from q, ithas the same value everywhere on the spherical surface,

thus

This is exactly Coulomb's law.

Gauss' law permits us to prove an important theorem about conductors:

If an excess charge is placed on an isolated conductor, that amount of charge willmove entirely to the surface of the conductor. None of the excess charge will befound within the body of the conductor.

E =1

4 0

q

r2

.

23-6 A Charged Isolated Conductor

0

E

dA =

0

E 4 r2 = q.

0 EdA 0 Ed A = qenc q.

23-5 Gauss' Law and Coulomb's Law


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The electric field inside this conductor must be zero. If this werenot so, the field would exert forces on the conduction (free)electrons, which are always present in a conductor, and thus currentwould always exist within a conductor.

Since there is no such perpetual current in an isolatedconductor, and so the internal electric field is zero.

the charges are then in electrostatic equilibrium if thenet force on each charge is zero.

If the electric field is zero everywhere inside the copperconductor, it must be zero for all points on the Gaussiansurface because that surface is inside the conductor. AndGauss' law tells us that the net charge inside the Gaussiansurface must also be zero.

the excess charge is not inside the Gaussian surface,

it must be outside that surface, which means it must lieon the actual surface of the conductor.

An Isolated Conductor with a CavityBecause the electric field vanishes inside the conductor, there

can be no flux through this new Gaussian surface. From Gauss' law, that surfacecan enclose no net charge. We conclude that there is no net charge on the cavity

walls;


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The Conductor Removedthe electric field is set up by the charges and not by the conductor. The

conductor simply provides an initial pathway for the charges to take up theirpositions.

The External Electric Fieldthe surface charge density varies over the surface

of any nonspherical conductor. And this variationmakes the determination of the electric field set upby the surface charges very difficult.

However, the electric field just outside the surface

of a conductor is easy to determine using Gauss' law.

The electric field at and just outside the conductor'ssurface must be perpendicular to that surface. Other-wise, then it exert forces on the surface charges,causing them to move. Thus

then

the magnitude of the electric field just outside aconductor is proportional to the surface charge density

on the conductor.

E =

0

conducting surface .

0 EA 0 = qenc A.


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problem 23-4

at every point on the cylindrical part of the

Gaussian surface, the electric field must havethe same magnitude E and (for a positivelycharged rod) must be directed radially outward.

The flux through this cylindrical surface is

23-7Applying Gauss' Law: CylindricalSymmetry

=Ed A = E 2 rh .


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with Gauss' law,

yields

This is the electric field due to an infinitely long, straight line of charge, at apoint that is a radial distance r from the line. It also approximates the field of afinite line of charge at points that are not too near the ends (compared with thedistance from the line).

problem 23-5

Nonconducting SheetFrom symmetry, the electric field must be perpendicular

to the sheet and hence to the end caps.

0

E 2 rh 0

= qenc

h ,

E =

2 0

rline of charge .

23-8Applying Gauss' Law: Planar Symmetry


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With Gauss' law

yields

Since we are considering an infinite sheet with uniform charge density, thisresult holds for any point at a finite distance from the sheet.

Two Conducting Plates

Since the plates are conductors, when we bring them into this arrangement, theexcess charge on one plate attracts the excess charge on the other plate, and allthe excess charge moves onto the inner faces of the plates. Thus, the electricfield at any point between the plates has the magnitude

0

E EA EA 0 Ed A = qenc A ,

E =

0.

E =

2 0

sheet of charge .


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Since no excess charge is left on the outer faces, the electric field to the leftand right of the plates is zero.

the charge distribution of the two-plate system is not merely the sum of thecharge distributions of the individual plates.

problem 23-6

Two shell theorems:

A shell of uniform charge attracts or repels a charged particle that is outside theshell as if all the shell s charge were concentrated at the center of the shell.

If a charged particle is located inside a shell of uniform charge, there is no

electrostatic force on the particle from the shell. 11

23-9Applying Gauss' Law: Spherical Symmetry


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E = 0 spherical shell, field at r R .

applying Gauss' law to surface S2 , for which

r R, we would find that

Applying Gauss' law to surface S1, for

which r < R, leads directly to

Any spherically symmetric charge distribution,

can be constructed with a nest of concentricspherical shells.

For purposes of applying the two shell theorems, the volume charge densityshould have a single value for each shell but need not be the same from shell toshell. Thus, for the charge distribution as a whole, can vary, but only with r,

the radial distance from the center.

For r > R,

If the charge density is uniform, then

E =1

4 0 r

2

4

3

R3 = R3

3 0 r

2.

E =1

4 0

q

r2

spherical shell, field at r R .

E =1

4 0

q

r2

=1

4 0 r2 0

R

4 r '2

dr ' .


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Let q' represent that enclosed charge, then

If the full charge q enclosed within radius R is

uniform (=const), then q' enclosed within radius rin Fig (b) is proportional to q:

thus

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charge enclosed bysphere of radius r

volume enclosed by

sphere of radius r

=full charge

full volume.

q '

4 r3 / 3=

q

4 R3 / 3

q ' = qr

3

R3

.

E = q4 0 R3 r= r

3 0uniform charge, field at r R .

E =1

4 0

q '

r2

spherical shell, field at r R .

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