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QuizSolnsThruQ16

# QuizSolnsThruQ16 - (Quiz 01 was simply attendance on the...

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(Quiz 01 was simply attendance on the first day.) First find |E|=kQ/r ² . This requires finding r = (x ² + y ² ) = (8 ² + 6 ² ) = 100 = 10 . So |E| = -kQ/100 . (Note that Q has been replaced with “ -Q ” as specified. Next, E x = |E|(x/r) = (-kQ/100)(-8/10) = +8kQ/1000 = kQ/125 , and E y = |E|(y/r) = (-kQ/100)(6/10) = -3kQ/500 . Finally, E = x ˆ E x + y ˆ E y , so E = x ˆ kQ/125 + y ˆ 3kQ/500 .

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The volumetric charge distribution is given by ρ = Q/ V . The volume of a sphere is given by V =(4/3) π r ³ = (4/3) π 2 ³ = (32/3) π . So ρ = 16 π /(32/3) π = 16·3/32 = 3/2 = 1.5C/m ³ .
Gauss’s law is E = Q in / A ε . The radius of the point of interest is less than the physical radius of the object, so the smaller value is used in getting Q in . Q in = ρ V in this case where V = π r ² l for a cylinder. So V = π 3 ² l = 9 π l , which means Q in = (-2)(9 π l ) = -18 π l . The area A always uses the radius for the point of interest ( Gaussian surface ). So A = 2 π r l = 2 π 3 l = 6 π l for a cylindrical surface. Putting it all together, Q in / A ε = -18 π l /6 π l ε = -3/ ε .

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