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UW-Madison PHYSICS 208 - Chapter 28 - Gauss’s Law

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 28. Gauss’s Law Using Gauss’s law, we can deduce electric fields, particularly those with a high degree of symmetry, simply from the shape of the charge distribution. The nearly spherical shape of the girl’s head determines the electric field that causes her hair to stream outward. Chapter Goal: To understand and apply Gauss’s law.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The electric field inside a conductor in electrostatic equilibrium is A. uniform. B. zero. C. radial. D. symmetric.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The electric field inside a conductor in electrostatic equilibrium is A. uniform. B. zero. C. radial. D. symmetric.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Electric Flux The electric flux measures the flow of electric field. The flux through a small surface of area A whose normal to the surface is tilted at angle θ from the field is: Or using the dot-product: We have assumed A is so small that E is uniform within it.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 FluxCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EXAMPLE 28.1 The electric flux inside a parallel-plate capacitor QUESTION:Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EXAMPLE 28.1 The electric flux inside a parallel-plate capacitorCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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.  δΦEi= EiδAicosθ= E i•δ A iCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Electric Flux through a Closed Surface The electric flux through a closed surface is The electric flux is still the summation of the fluxes through a vast number of tiny pieces, pieces that now cover a closed surface. NOTE: A closed surface has a distinct inside and outside. The area vector dA is defined to always point toward the outside.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s Law For any closed surface enclosing total charge Qin,the net electric flux through the surface is This result for the electric flux is known as Gauss’s Law. Gauss’s law follows from Coulomb’s law but is more general. It applies when charges moves and Coulomb’s law is invalid.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Using Gauss’s Law 1. Gauss’s law applies only to a closed surface, called a Gaussian surface. 2. A Gaussian surface is not a physical surface. It is an imaginary, mathematical surface in the space surrounding one or more charges. 3. We can’t generally find the electric field from Gauss’s law alone. We can apply Gauss’s law when from symmetry and superposition, we already can guess the shape of the field.Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s 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 same electric field:  E =14πεoqR2R  ΦE= E • d A ∫= EdA∫= E dA∫= EA =14πεoqR2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4πR2( )= q /εoCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s Law for a point charge 13 13 q Two spheres of radius R and 2R are centered on the positive 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 2RCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s Law for a point charge 14 q Two spheres of the same size enclose different positive charges, 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 BCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. General Gauss’s Law  ΦE= E • d A ∫=Qenclosedεo Works for any closed surface, not only sphere q  Works for any charge distribution (by superposition)Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Field of a sphere of chargeCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EXAMPLE 28.3 Outside a sphere of chargeCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EXAMPLE 28.3 Outside a sphere of chargeCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 : 2πrL   E • d A  E λL E 2πrL Gauss’ law:  E 2πrL =λL /εo⇒ E =12πεoλrCopyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 : 2πr2   E • d A  E ηπr2 E 2πr2 Gauss’ law:  E 2πr2=ηπr2/εo⇒ E =η2εoSurface charge C/m2Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conductors


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UW-Madison PHYSICS 208 - Chapter 28 - Gauss’s Law

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