Gauss’s LawGauss’s Law

The electric flux through a The electric flux through a closed surface is proportional closed surface is proportional to the charge enclosedto the charge enclosed

Definition of FluxDefinition of Flux• The amount of field, material or The amount of field, material or

other physical entity passing other physical entity passing through a surface.through a surface.

• Surface area can be represented as Surface area can be represented as vector defined normal to the vector defined normal to the surface it is describingsurface it is describing

• Electric Flux is defined by the Electric Flux is defined by the equation:equation:

surface

Ad

E

Electric FluxElectric Flux

• The amount of The amount of electric field electric field passing through a passing through a surface areasurface area

• The units of electric The units of electric flux are N-mflux are N-m22/C/C

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Carl Friedrich Gauss 1777-1855

Yes it’s the same Yes it’s the same guy that gave you guy that gave you the Gaussian the Gaussian distribution and … distribution and …

To give you some To give you some perspective he was born 50 perspective he was born 50 years after Newton died years after Newton died (1642-1727). Predicted the (1642-1727). Predicted the time and place of the first time and place of the first asteroid asteroid CERESCERES (Dec. 31, (Dec. 31, 1801). Had the unit of 1801). Had the unit of magnetic field named after magnetic field named after him and of course had him and of course had much to do with the much to do with the development of development of mathematicsmathematics

Ceres Ceres year , d=4.6 Auyear , d=4.6 Au

Gauss’ Law in E& M

• Uses symmetry to determine E-field due to a charge distribution

• Method: Considers a hypothetical surface enclosing some charge and calculates the E-field

• The shape of that surface is “EVERYTHING”

KEY TO USING Gauss’s LawKEY TO USING Gauss’s Law

• The shape of the surrounding surface is one that MIMICS the symmetry of the charge distribution …..

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The electric flux passing through a spherical The electric flux passing through a spherical surface surrounding a point chargesurface surrounding a point charge

24A r

22

0 0

44E

q qE d A E dA EA r

r

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Use a spherical surface with areaUse a spherical surface with area

Electric field for a point charge at Electric field for a point charge at rr

204

qE

r

Due to symmetry, Due to symmetry, EE and and dAdA are parallel and are parallel and EE is constant is constant over all of over all of AA. So…. So…

Gauss’s LawGauss’s LawThe previous example is true in general:The previous example is true in general:

The total flux passing through a closed surface is The total flux passing through a closed surface is proportional to the charge enclosed within that proportional to the charge enclosed within that surface.surface.

Eoclosed

surface

qE dA

Note: The area vector points outwardNote: The area vector points outward

The Gaussian SurfaceThe Gaussian SurfaceAn imaginary closed surface created to enable An imaginary closed surface created to enable the application of Gauss’s Lawthe application of Gauss’s Law

What is the total flux through each surface?What is the total flux through each surface?

Solving problems with Gauss’s LawSolving problems with Gauss’s Law

1. Charge densities –

It is convenient to define charge densities for linear, surface and volume charge distributions

2. Symmetry and coordinate systems –

Choose that coordinate system that most nearly matches the symmetry of the charge distribution. For example, we chose spherical coordinates to determine the flux due to a point charge because of spherical symmetry.

Solving problems with Gauss’s LawSolving problems with Gauss’s Law

3. 3. Determining the Electric Field from Gauss’s Determining the Electric Field from Gauss’s LawLaw – Symmetry must be present in the – Symmetry must be present in the charge distribution so that over the Gaussian charge distribution so that over the Gaussian surface surface EE and and AA vectors are parallel and vectors are parallel and EE is is constant over the surface. constant over the surface.

Consider three examples: (1) the long straight Consider three examples: (1) the long straight line of charge, (2) the infinite plane sheet of line of charge, (2) the infinite plane sheet of charge, and (3) a charged sphere.charge, and (3) a charged sphere.

Example - Long straight line of chargeExample - Long straight line of chargeLooking at the diagram (b), we can determine that the Looking at the diagram (b), we can determine that the problem has a problem has a cylindrical symmetrycylindrical symmetry..Therefore the Gaussian surface is a cylinder of radius Therefore the Gaussian surface is a cylinder of radius r r and length and length ll..There are three surfaces to consider. The There are three surfaces to consider. The upper and upper and lower circular surfaceslower circular surfaces have normals parallel to the z have normals parallel to the z axis which are perpendicular to the electric field, thus axis which are perpendicular to the electric field, thus contribute zero to the fluxcontribute zero to the flux..

The area to be evaluated for the integral is that of The area to be evaluated for the integral is that of the curved sides of the cylinder, the curved sides of the cylinder, 22rr. The . The charge charge enclosed is enclosed is ..

Example - Long straight line of Example - Long straight line of chargecharge

0 02 2enclosed

o

q lE

A rl r

Since the field has radial symmetry, it is also Since the field has radial symmetry, it is also constant at a fixed distance of r.constant at a fixed distance of r.

0

enclosedqE dA EA

Gauss’ Law: Determining the E-field near the surface of a nonconducting (insulating) sheet

o

o

o

o

o

E

E

AEA

qEA

qAdE

2

so, sideother from also and

Results of other geometriesResults of other geometriesUniformly charged dielectric Uniformly charged dielectric (insulating) sphere(insulating) sphere

E rr ao

3

Ea

rr a

o

3

23

Uniformly charged dielectric Uniformly charged dielectric infinite plane sheetinfinite plane sheet

Eo

2

Can you derive these results?Can you derive these results?

Conclusions Gauss’ LAW

1.1. Only the charge enclosed within a volume defined Only the charge enclosed within a volume defined by a closed surface contributes to the net electric by a closed surface contributes to the net electric flux through the surface. flux through the surface.

2.2. That net flux through the surface is proportional to That net flux through the surface is proportional to the charge enclosed within the volume.the charge enclosed within the volume.

Conclusions3. 3. Gaussian surface is an imaginary closed surface Gaussian surface is an imaginary closed surface

necessary to solve a problem using Gauss’s Lawnecessary to solve a problem using Gauss’s Law

4. 4. Gauss’s Law can be used to determine the electric field Gauss’s Law can be used to determine the electric field of a charge distribution if there is a high degree of of a charge distribution if there is a high degree of symmetrysymmetry

5. 5. Applying Gauss’s Law to the Applying Gauss’s Law to the interior of an interior of an electrostaticallyelectrostatically charged conductor we charged conductor we conclude that the electric field conclude that the electric field within the conductor is zerowithin the conductor is zero

6. Any Net charge on a conductor 6. Any Net charge on a conductor must reside on its surfacemust reside on its surface

Conclusion: Gauss’ Law and Conclusion: Gauss’ Law and ConductorsConductors

Shell Theorems: Conductors

1)1) A shell of uniform charge attracts A shell of uniform charge attracts or repels a charge particle that is or repels a charge particle that is outside the shell as though all outside the shell as though all charge is concentrated at the charge is concentrated at the center.center.

2)2) If a charged particle is located If a charged particle is located inside such a shell, there is no inside such a shell, there is no electrostatic force on the particle electrostatic force on the particle from the shellfrom the shell

r̂r

q

oE 24

1

0E

Electric Field at surface of conductor is perpendicular to the surface and proportional to the charge density at the

surface

Shown that the excess charge resides Shown that the excess charge resides on the outer surface of conductor on the outer surface of conductor surface.surface.

Unless the surface is spherical the Unless the surface is spherical the charge density charge density (chg. per unit area) (chg. per unit area) variesvaries

However the E-field just outside of a However the E-field just outside of a conducting surface is easy to conducting surface is easy to determine using Gauss’ Lawdetermine using Gauss’ Law

o

o

o

o

o

E

AEA

qEA

qdAE

qAdE

Gauss’ Law can be used to determine Gauss’ Law can be used to determine E-field in cases where we have:E-field in cases where we have:

• Spherical symmetry

• Cylindrical symmetry

• Planar symmetry