708_Physics ProblemsTechnical Physics

# 708_Physics ProblemsTechnical Physics - 48 Gausss Law...

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48 Gauss’s Law P24.69 EA ⋅= = zz d q a r rd r r in 00 2 0 1 4 π Er a rdr a r E a r 4 44 2 2 2 0 0 0 2 0 ππ = = = = z constant magnitude (The direction is radially outward from center for positive a ; radially inward for negative a .) P24.70 In this case the charge density is not uniform , and Gauss’s law is written as dd V 1 0 ρ . We use a gaussian surface which is a cylinder of radius r , length A , and is coaxial with the charge distribution. (a) When rR < , this becomes a r b dV r 2 0 0 0 A bg = F H G I K J z . The element of volume is a cylindrical shell of radius r , length A , and thickness dr so that dV r dr = 2 A . r ar b 2 2 23 2 0 0 πρ A A = F H G I K J F H G I K J so inside the cylinder, E r a r b = F H G I K J 0 0 2 2 3 . (b) When > , Gauss’s law becomes a r b r R 22 0 0 0 AA b g = F H G I K J z or outside the cylinder, E R r a R b = F H G I K J 0 2 0 2 2 3 . P24.71 (a) Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x : Use Gauss’s law:
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