QUIZ XV ELECTRIC FLUX and GAUSS LAW Consider a spherical insulating shell a R b with total charge Q uniformly distributed throughout the shell HINT Within the shell the volume charge density Q 4 3 b3 a3 a 2 b b 2 c 2 d 4 Determine the ELECTRIC FLUX through a Gaussian spherical surface R a Determine the ELECTRIC FLUX through a Gaussian spherical surface a R Determine the ELECTRIC FLUX through a Gaussian spherical surface R b Determine the magnitude of the ELECTRIC FIELD in the region a R b SHOW ALL WORK EXPLAIN LOGIC SOLUTION KEY CONCEPTS RELATIONSHIPS LAWS GAUSSIAN SURFACE ELECTRIC FLUX E GAUSS LAW E Qenclosed 0 VOLUME CHARGE DENSITY a b c d For any spherical Gaussian surface R a there is NO ENCLOSED CHARGE so by Gauss Law E 0 For a R b the enclosed charge within a spherical Gaussian surface Qenclosed dV dV since the charge density is UNIFORM and the volume enclosed is 4 R3 a3 3 So Qenclosed Q R3 a3 b3 a3 and E Q R3 a3 b3 a3 0 See LECTURE NOTES A spherical Gaussian surface R b encloses all of the charge Q so E Q 0 By Gauss Law for this spherically symmetric charge distribution through a spherical Gaussian surface E 4 R2 Q R3 a3 b3 a3 0 The ELECTRIC FIELD within the region a R b E Q 4 0 R3 a3 b3 a3 1 R2 The direction of the field due to the positive charge distribution is RADIALLY OUTWARD
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