Physics 2102 Jonathan Dowling Flux Capacitor Operationa Physics 2102 Lecture 05 TUE 02 FEB Gauss Law II QuickTime and a decompressor are needed to see this Carl Friedrich Gauss 1777 1855 QuickTime and a decompressor are needed to see this picture What Dat Flux A QuickTime and a decompressor are needed to see this picture QuickTime and a decompressor are needed to see this picture Constant Area Flux r r EgA EA cos dA E Changing Area Flux r r EgdA E cos dA What Dat is a Registered Trademark of Dowling Productions Inc All Rights Reserved Gauss Law General Case Consider any ARBITRARY CLOSED surface S NOTE this Gaussian Surface does NOT have to be a real physical object The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED S E q r r q E dA 0 Surface The results of a complicated integral is a very simple formula it avoids long One of Maxwell s 4 equations calculations Examples r r q E dA 0 Surface What is Flux Through Surfaces q 0 S1 q 0 S2 0 S3 0 S4 Faraday s Cage Electric Field Inside Hollow Conductor is Zero Safe in the Plane Safe in the Car QuickTime and a decompressor are needed to see this picture E 0 Choose any arbitrary surface inside the metal Since E 0 flux 0 Hence total charge enclosed 0 Inside cavity is shielded from all external electric fields Faraday Cage effect YUV420 codec decompressor are needed to see this picture Gauss Law Cylindrical Symmetry Charge of q 10 C is uniformly spread over a line of length L 1 m Use Gauss Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line Approximate as infinitely long line E radiates outwards Choose cylindrical surface of radius R length L co axial E R 1 mm 1m Line of Charge q L Units C m Gauss Law Cylindrical Symmetry Approximate as infinitely long line E radiates outwards Choose cylindrical surface of radius R length L co axial with line of charge E R 1 mm 1m E A E 2 RL q L 0 0 dA E so cos 1 L E 2k 2 0RL 2 0R R Recall Finite Line Example L 2 L 2 dx x E y k a 2 2 3 2 k a 2 2 2 a x a x a L 2 L 2 2k L 2 2 a 4a L the Line Is Infinitely Long L a 2k L 2k Ey 2 a a L r E QuickTime and a TIFF Uncompressed decompressor are needed to see this picture a R Gauss Law Insulating Plate Infinite INSULATING plane with uniform charge density s E is NORMAL to plane Construct Gaussian box as shown q A Applying Gauss law we have 2 AE 0 0 Solving for the electric field we get E 2 0 Surface Charge q A Units C m2 For an insulator E 2 0 and for a conductor E 0 Insulating and Conducting Planes Q E 2 0 2 A 0 Q Insulating Plate Charge Distributed Homogeneously Q 2 Q E 0 2 A 0 Conducting Plate Charge Distributed on the Outer Surfaces Electric Field Inside a Conductor is ZERO Gauss Law Spherical Symmetry Consider a POINT charge q pretend that you don t know Coulomb s Law Use Gauss Law to compute the electric field at a distance r from the charge Use symmetry place spherical surface of radius R centered around the charge q E has same magnitude anywhere on surface E normal to surface q 0 r q E E A E 4 r q kq E 2 2 4 0 r r 2 Electric Fields With Spherical Symmetry Shell Theorem 10 C A spherical shell has a charge of 10C and a point charge of 15C at the center What is the electric field produced OUTSIDE the shell 15C If the shell is conducting Field Inside a Conductor is ZERO And if the shell is insulating Charged Shells Behave Like a Point Charge of Total Charge Q at the Center E E k 15C r2 E 0 E k 5C r2 r Conducting Electric Fields With Insulating Sphere r R Venclosed 4 r 3 3 r3 qenclosed Q Q Q 3 3 R 4 R 3 V total r R qenclosed Q QuickTime and a TIFF Uncompressed decompressor are needed to see this picture EA qenclosed 0 3 r r R E4 r 2 Q 3 0 R r R E4 r 2 Q 0 Summary Electric Flux a Surface Integral Vector Calculus Useful Visualization Electric Flux Lines Like Wind Through a Window Gauss Law Provides a Very Direct Way to Compute the Electric Flux QuickTime and a decompressor are needed to see this picture
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