HW3Solns - element will be parallel and we have E d A = | E...

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HW 3 Solutions :
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For the next 2 questions we make use of the following result. For a positive spherically symmetric charge distribution the electric ±eld must have the same spherical symmetry, thus the electric ±eld points radially outward at all points. If we suspect the electric ±eld is zero in some region, we can take it to be nonzero and radially outward and show that this leads to a contradiction unless the electric ±eld is indeed zero. The Gaussian surfaces for all three regions shown below in 53 and the one region in 59 are spheres concentric with the shell. Thus at each point on the Gaussian surface the electric ±eld and the area
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Unformatted text preview: element will be parallel and we have E d A = | E || d A | cos(0) = | E || d A | = EdA . Moreover, the electric eld is also constant at a given radial distance from the center of the sphere, which is again from the spherical symmetry of the charge distribution. Thus the eld is constant over an area integral and we have E d A = E dA = EA = 4 r 2 where is the radius of our Gaussian surface. We make use of the above result in 53 a, b, c and 59....
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HW3Solns - element will be parallel and we have E d A = | E...

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