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EE1 Quiz Solution

EE1 Quiz Solution - -a d Q and d = a 2 d To solve for the...

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EE1 Spring 2008 - Quiz Solution A positive point charge Q is located at a distance d from the center of a grounded condictiong sphere of radius a ( a < d ). What is the potential in space for R > a and R < a ? For R < a , since the metallic sphere is grounded, V = 0. For R > a , we can simplify this problem by using the method of images to replace the conducting sphere with a point charge. To do so, we can predict a point charge within the spherical region that is positioned along a line from the origin to the original point charge. Following the procedure in Cheng’s Field and Waves Electronics (pp. 170-171) we find:

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Unformatted text preview: -a d Q and d = a 2 d To solve for the total voltage for any arbitrary point P, we can add the contributions of the potential from each point charge. For the original point charge, the potential is: V Q = Q 4 π± R Q where, by the law of cosines, R Q = [ R 2 + d 2-2 Rdcosθ ] 1 / 2 1 and θ is the angle between the line from the origin to the charge and the vector describing the observation point. For the image charge, the potential is: V Q = Q 4 π± R Q where R Q = [ R 2 + ( a 2 d ) 2-2 R ( a 2 d ) cosθ ] 1 / 2 The total potential is: V tot = V Q + V Q 2...
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