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Gauss s LawCopyright 2009 Pearson Education Inc Outline of Gauss s Law Electric Flux Gauss s Law.

Applications ofGauss s Law Experimental Basis ofGauss s Coulomb s LawsCopyright 2009 Pearson Education Inc .

Gauss s Law Gauss s Law can be used asan alternative procedure forcalculating electric fields It is based on the inverse.

behavior of the electric forcebetween point charges Copyright 2009 Pearson Education Inc Gauss s Law Gauss s Law is convenient in.

calculations of the electric field ofHighly Symmetric ChargeDistributions Gauss s Law is important inunderstanding verifying the.

properties of conductors inelectrostatic equilibrium Copyright 2009 Pearson Education Inc Johann Carl Friedrich Gauss 1736 1806 Germany .

Mathematician Astronomer Physicist Sometimes called the Prince of Mathematics Johann Carl Friedrich Gauss.

1736 1806 Germany Mathematician Astronomer Physicist Sometimes called the Prince of Mathematics .

A child prodigy in math Do you have trouble believing some of the following I do Johann Carl Friedrich Gauss 1736 1806 Germany Mathematician Astronomer .

Physicist Sometimes called the Prince of Mathematics A child prodigy in math Do you havehis.

Age 3 He informed trouble believingfather of asome of the following mistake I do in a payrollcalculation gave the correct answer .

Johann Carl Friedrich Gauss 1736 1806 Germany Mathematician Astronomer Physicist Sometimes called the.

Prince of Mathematics A child prodigy in math Do you havehis Age 3 He informed trouble believingfather of asome of the following .

mistake I do in a payrollcalculation gave the correct answer Age 7 His teacher gave the problem of summing allintegers 1to 100 to his class to keep them busy Gauss.

quickly wrote the correct answer 5050 on his slate Johann Carl Friedrich Gauss 1736 1806 Germany Mathematician Astronomer Physicist .

Sometimes called the Prince of Mathematics A child prodigy in math Do you havehis Age 3 He informed trouble believing.

father of asome of the following mistake I do in a payrollcalculation gave the correct answer Age 7 His teacher gave the problem of summing all.

integers 1to 100 to his class to keep them busy Gaussquickly wrote the correct answer 5050 on his slate Whether or not you believe all of this it is 100 true thatMade a HUGE number of contributions to Johann Carl Friedrich Gauss.

Genius Made a HUGE number ofcontributions to Mathematics Physics Astronomy Johann Carl Friedrich GaussGenius Made a HUGE number of.

contributions to Mathematics Physics Astronomy Some are 1 Proved The Fundamental Theorem of Algebra that every polynomial has a root of the form Johann Carl Friedrich Gauss.

Genius Made a HUGE number ofcontributions to Mathematics Physics Astronomy Some are 1 Proved The Fundamental Theorem of Algebra that every polynomial has a root of the form.

2 Proved The fundamental Theorem ofArithmetic that every natural number can berepresented as a product of primes in only one way Johann Carl Friedrich GaussGenius Made a HUGE number of.

contributions to Mathematics Physics Astronomy Some are 1 Proved The Fundamental Theorem of Algebra that every polynomial has a root of the form2 Proved The fundamental Theorem of.

Arithmetic that every natural number can be3 Proved that every number is theassumrepresented of at mosta product 3 triangularof primes in onlynumbers .

4 Developed the method of least squares fitting many other methodsin statistics probability 5 Proved many theorems of integral calculus including the divergencetheorem when applied to the E field it is what is called Gauss s Law Johann Carl Friedrich Gauss.

Genius Made a HUGE number ofcontributions to Mathematics Physics Astronomy Some are 1 Proved The Fundamental Theorem of Algebra that every polynomial has a root of the form.

2 Proved The fundamental Theorem ofArithmetic that every natural number can be3 Proved that every number is theassumrepresented of at mosta product 3 triangular.

of primes in onlynumbers 4 Developed the method of least squares fitting many other methodsin statistics probability 5 Proved many theorems of integral calculus including the divergencetheorem when applied to the E field it is what is called Gauss s Law .

6 Proved many theorems of number theory 7 Made many contributions to the orbital mechanics of the solar system 8 Made many contributions to Non Euclidean geometry9 One of the first to rigorously study the Earth s magnetic field Electric Flux.

The Electric Flux Ea cross sectional area A isproportional to the total numberof field lines crossing the area isDefined as.

constant E only Copyright 2009 Pearson Education Inc The Electric Flux is defined as the productof the magnitude of the electric field E thesurface area A perpendicular to the field .

E EAFlux Units Example Electric flux Calculate the electric flux through the rectangle shown The rectangle is 10 cm by 20 cm E 200 N C 30 .

Copyright 2009 Pearson Education Inc Example Electric flux Calculate the electric flux through the rectangle shown The rectangle is 10 cm by 20 cm E 200 N C 30 E EAcos 30 A 0 02 m2.

E 200 0 02 cos 30 3 46 N m2 CCopyright 2009 Pearson Education Inc Electric Flux General Area The electric flux isproportional to the.

number of electric fieldlines penetrating some The field lines may makesome angle with theperpendicular to the.

surface Then E EA cos Copyright 2009 Pearson Education Inc Electric Flux Interpreting Its Meaning E EA cos .

E is a maximum when the surfaceis perpendicular to the field 0 E is zero when the surface isparallel to the field 90 If the field varies over the surface .

E EA cos is valid for only asmall element of the area Copyright 2009 Pearson Education Inc Flux Through a Cube Example The field lines pass.

2 surfaces perpendicularly are parallel to the other 4For face 1 E E 2 For face 2 E E 2 For the other sides .

E 0 Therefore Copyright 2009 Pearson Education Inc Gauss s LawFlux through a closed surface Copyright 2009 Pearson Education Inc .

The Electric Flux E through a closed surfacedefined as the closed surface sum integral of thescalar dot product of the electric field E thesurface area A Copyright 2009 Pearson Education Inc .

Gauss s Law The net number of field lines throughthe surface is proportional to the chargeenclosed also to the flux givingGauss s Law .

This can be used to find thefield in situations with a high degreeCopyright 2009 Pearson Education Inc Gauss s Law The net number of E field lines through a.

surface is proportional to the charge enclosed to the flux Gauss swhich gives This is a VERYfind the Ewhich can be usedintosituations where there is.

field especiallya high degree of symmetry It can be shown that ofcourse the E field calculated this way is identical tothat obtained by Coulomb s Law Often however insuch situations it is often MUCH EASIER to use.

Gauss s Law than to use Coulomb s Law Copyright 2009 Pearson Education Inc For a Point Charge E EA E 4 r2 E Qencl o .

Of course solving for Egives the same result asCoulomb s Law Copyright 2009 Pearson Education Inc Using Coulomb s Law to.

evaluate the flux of the field due toa point charge over the surface ofa sphere of surface area A1surrounding the charge gives Now consider a point charge surrounded by an.

Arbitrarily Shaped closed surface of area A2 It can beseen that the same flux passes through A2 as passesthrough the spherical surface A1 So This Result isValid for any Arbitrarily Shaped Closed Surface Copyright 2009 Pearson Education Inc .

The power of this is that the problemsolver can choose the closed surface called a Gaussian Surface at your convenience In cases where there is a large.

of symmetry in the problem this will simplify thecalculation considerably Copyright 2009 Pearson Education Inc Now consider a Gaussian.

Surface enclosing several pointcharges We can use thesuperposition principle to showCopyright 2009 Pearson Education Inc So Gauss s Law is valid for ANY.

Charge Distribution Note though that it only refers to thefield due to charges within theGaussian surface Charges outside thesurface will also create fields .

Copyright 2009 Pearson Education Inc Conceptual Example Flux from Gauss s law Consider the 2 Gaussian surfaces A1 A2 shown The only charge present is the charge Qat the center of surface A1 .

Calculate the net flux through each surface A1 Copyright 2009 Pearson Education Inc Applications of Gauss s LawCopyright 2009 Pearson Education Inc By symmetry clearly a.

Spherical Conductor SPHERICAL A thin spherical shell of Gaussian surfaceis needed radius r0 possesses a totalnet charge Q that is.

uniformly distributed on it Calculate the E field at a Outside the shell r r0 b Inside the shell r r0 c What if the conductor.

were a solid sphere Copyright 2009 Pearson Education Inc Example By symmetry clearly aSolid Sphere of ChargeGaussian surface.

An electric charge Q isis needed distributed uniformlythroughout a nonconducting sphere radius r0 .

Calculate the electric field a Outside the sphere r r0 b Inside the sphere r r0 Copyright 2009 Pearson Education Inc Example By symmetry clearly a.

Solid Sphere of ChargeGaussian surface An electric charge Q isis needed distributed uniformly.

throughout a nonconducting sphere radius r0 Calculate the electric field a Outside the sphere r r0 Outside r r0 .

E Q 4 0r2 Copyright 2009 Pearson Education Inc Example By symmetry clearly aSolid Sphere of ChargeGaussian surface.

An electric charge Q isis needed distributed uniformlythroughout a nonconducting.

sphere radius r0 Calculate the electric field b Inside the sphere r r0 Qencl Vr Vr0 Q Results 4 r3 3 4 r03 3 Q.

Inside r r0 E Qr 4 0r03 Copyright 2009 Pearson Education Inc Summary By symmetry clearly aGaussian surface is needed .

Outside r r0 E Q 4 0r2 Inside r r0 E Qr 4 0r03 E Qr 4 0r03 Note E inside.

has E Q 4 0r2 a very different rdependence than ECopyright 2009 Pearson Education Inc Example Long Uniform Line of Charge.

A very long straight effectively wire of radius R hasa uniform positive charge per unit length Calculate theE field at points near outside the wire far from the ends By symmetry clearly aCylindrical Gaussian.

Surface is needed Gauss s Law E EA Qencl 0 For a cylinder A 2 R and Qencl 0 So Gauss s Law is E 2 R 0 So this gives.

Copyright 2009 Pearson Education Inc E 2 R Summary Long Uniform Line of Charge A very long straight effectively wire of radius R hasa uniform positive charge per unit length Calculate theE field at points near outside the wire far from the ends .

By symmetry clearly aCylindrical GaussianSurface is needed Results E 2 0R E for the wire has a very different R.

dependenceCopyright 2009 Pearson Education Inc Example Infinite Plane of Charge A charge Q is distributed uniformly on aplane with a surface charge density.

charge per unit area Q A over a very large but very thinnon conducting flat planesurface Calculate the electricfield at points near the plane .

Use Gauss s Law Choose a cylindricalGaussian surface as shownin the figure Copyright 2009 Pearson Education Inc .

A CylindricalInfinite Plane of Chargesurface was chosen but The charge is distributed here the shape of theuniformly with a surface Gaussian surface.

charge density doesn t matter The charge per unit area Q A result is independent ofGaussâ€™s law states that the product EÂ·dA, ... Physics, & Astronomy. Some are: ... The power of this is that the problem. solver can choose the closed surface (called a . Gaussian Surface) at your convenience! In cases where there is a large amount . of symmetry in the problem,

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