ph1bA_08_hw5s

ph1bA_08_hw5s - Ph1b (analytic track): Solution to Homework...

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Unformatted text preview: 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 and radius a centered at A , and a smaller sphere of charge density 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 Gausss law, by invoking spherical symmetry: I ~ E d~a = 4 Z dV ~ E small ( A ) 4 a 2 2 = 4 ( ) 4 3 a 2 3 ~ E small ( A ) = 4 3 a 2 . The field must point towards the center of the negatively charged smaller sphere (see figure). Therefore: ~ E small ( A ) = 2 3 a y . By the principle of superposition ~ E ( A ) = ~ E big ( A ) + ~ E small ( A ) = 2 3 a y . 1 The field at B , on the surface of the larger sphere, is given by: ~ E ( B ) = ~ E big ( B ) + ~ E small ( B ) = 4 3 a y + 4 3 ( a 2 ) 3 ( a + a 2 ) 2 y = 4 3 a y 1 + 1 18 = 34 27 a y ....
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This note was uploaded on 04/28/2010 for the course PH 1b Anal taught by Professor Mckewn during the Winter '08 term at Caltech.

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ph1bA_08_hw5s - Ph1b (analytic track): Solution to Homework...

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