PHY2049: Chapter 231Chapter 23: Gauss’ LawPHY2049: Chapter 232Conductors with No CurrentÎE is zero everywhere inside Why? Conductors are full of mobile charges (e.g., conduction electrons in a background formed by immobile positive ions). If there were E, then the charges must be moving around due to force F=qE. This would contradict “no current.”Note: even if there is an externally imposed E, it cannot go insideÎAll excess charge must be on outer surface. Why? Since E=0 everywhere inside, qencenclosed by any Gaussian surface is also zero everywhere inside. Note: distribution of surface charge must be such to make E=0 everywhere inside ÎE is always normal to surface on conductor Why? E component parallel to surface would cause surface charge to move. This would contradict “no current.”PHY2049: Chapter 233Spherical Symmetry (1)ÎInsulator or conductor? Conducting sphere cannot be uniformly chargedÎInside By symmetry, E must be radially symmetric E field has constant mag., ⊥ to Gaussian surfaceÎOutsideGaussian surface(sphere)Gauss’ LawSolve for EUniformly Charged Sphere0enc2)4(εqrπEdS==⋅∫AE+ + + ++ + + + + ++ + + + + + + ++ + + + + + + ++ + + + + + + ++ + + + + + + ++ + + + + ++ + + +334RπQρ =Q333)34(RrQrπρ =Q302033414/RQrπεrπεRQrE ==2041rQπεE =PHY2049: Chapter 234Spherical Symmetry (2)ÎInside conductor E must be 0 Charge can be only on surfacesÎOutside By symmetry, E must be radially symmetric E field has constant mag., ⊥ to Gaussian surfaceGaussian surface(sphere)Gauss’ LawMust be 0Concentric Conducting Spherical ShellQ−-Q++++++–0enc0εqdS==⋅∫AE+Q uniformly distributed on inner wall0enc2)4(εqrπEdS==⋅∫AEGauss’ Law2041rQπεE −=– Q uniformly distributed on outer surfacePoints toward shell–––––––––––++++++PHY2049: Chapter 235Spherical Symmetry (2)ÎNote: Problem 23-4 on page 614 In this problem, point charge –Q in cavity is not at center. This makes the problem harder. In particular, you do not need to understand explanation given in book for why charge on outer surfaceis uniform. Rigorous explanation in fact requires a tool, so-called uniqueness theorem, which is beyond introductory physics.Concentric Conducting Spherical ShellPHY2049: Chapter 236Axial Symmetry (1): Line Charge ÎInfinitely long line, uniformly charged By symmetry, E must be axially symmetric On curved surface, E field has constant mag., ⊥ to Gaussian surface Through top and bottom surfaces, no ΦEsince E is || Solve for E0enc00)2(εqrπhEdS=++=⋅∫AEGauss’ LawhλrλπεE021=PHY2049: Chapter 237Axial Symmetry (2): Uniformly Charged CylinderÎInfinitely tall cylinder, uniformly charged By symmetry, E must be axially symmetric On curved surface, E field has constant mag., ⊥ to Gaussian surface Through top and bottom surfaces, no ΦEsinceE is || )(2rπhρSolve for E0enc00)2(εqrπhEdS=++=⋅∫AEGaussian surface(cylindrical)ρrRhGauss’ lawrρεE021=PHY2049: Chapter 238Rectangular Symmetry (1): Uniformly Charged SheetÎInfinitely wide and tall By symmetry, E must be ⊥ , same on both sidesÎParallel conducting plates: read Section 23-8Solve for EGauss’ law)(2rπσ02εσE =Constant!0enc2200)()(εqrπErπEdS=+++=⋅∫AEPHY2049: Chapter 239λπεE021=Some Comparisons2rtotal041qπεE =r02εσE =no distance dependenceÎSpherically symmetric charge distributionÎUniformly charged, infinitely long (i.e., very long) lineÎUniformly charged planePHY2049: Chapter 2310ÎNumerically integrate dEproduced by all charge elements dqÎUsually easier to numerically compute electric potential V first and then EElectrical potential is subject of Chapter 24How to calculate E field, if there is no
View Full Document
Unlocking...