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Unformatted text preview: PHYS 2B Quiz 2 Solutions Aris January 24, 2010 1 Problem 1 From Gauss’s law, we know the electric field of a uniformly charged sphere with charge density ρ ρ 4 πr 3 3 = E 4 πr 2 (1) E = ρr 3 o (2) We are informed that E ( r = R 2 ) = 5 = ρR 6 o (3) Rewriting ρ as the total charge Q divided over the volume of a sphere, and dividing both sides by two, we obtain 2 . 5 = Q 4 π (2 R ) 2 (4) This gives us the Coulomb field at r = 2 R and the magnitude of the field (2.5). 2 Problem 2 The gain in kinetic energy from R to 2R is the loss in electrostatic potential energy: 1 2 mv 2 1 = kQq ( 1 R 1 2 R ) = kQq 2 R (5) Similarly for R to 3R, 1 2 mv 2 2 = kQq ( 1 R 1 3 R ) = 2 kQq 3 R (6) 1 Hence, v 2 2 v 2 1 = 2 3 1 2 (7) Taking the square root, we obtain v 2 = 1 . 15 v 1 . 3 Problem 3 Exploit the spherical symmetry of the problem so that we can use Gauss’ law: Q enclosed = E 4 πr 2 (8) For the electric field to be zero, the total charge enclosed must be zero. The total charge has two contributions from the inner and outer parts of the sphere:...
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 Winter '10
 Hirsch
 Electric charge

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