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ph1bA_08_hw5s - Ph1b(analytic track Solution to Homework 5...

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Ph1b (analytic track): Solution to Homework 5 Chefung Chan, Winter 2008 Problem 1-16 (5 points) As far as electrostatics is concerned, we can think of the given configuration as being equivalent to that made by superposing two sources: the original uncarved sphere of charge density ρ and radius a centered at A , and a smaller sphere of charge density ρ and radius a/ 2, in the location of the cavity. The electric field at A (and at B ) is then the sum of the electric fields due to the big sphere and the small sphere, by the principle of superposition. At A , the electric field due to the big sphere, E big ( A ), is zero because A is at the center of the charge distribution. The electric field due to the small sphere, E small ( A ), can be calculated using Gauss’s law, by invoking spherical symmetry: E · da = 4 π ρ dV E small ( A ) 4 π a 2 2 = 4 π ( ρ ) 4 3 π a 2 3 E small ( A ) = 4 3 π a 2 ρ . The field must point towards the center of the negatively charged smaller sphere (see figure). Therefore: E small ( A ) = 2 3 πaρ ˆ y . By the principle of superposition E ( A ) = E big ( A ) + E small ( A ) = 2 3 πaρ ˆ y . 1
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The field at B , on the surface of the larger sphere, is given by: E ( B ) = E big ( B ) + E small ( B ) = 4 3 πρa ˆ y + 4 3 πρ ( a 2 ) 3 ( a + a 2 ) 2 ˆ y = 4 3 πρa ˆ y 1 + 1 18 = 34 27 πρa ˆ y .
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