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Unformatted text preview: HW-5 1.For a point charge q at a distance afrom a spherical conductor of radius R (R<a), we obtained in the lecture the result of the image charge inside the conductor. Starting from that result, calculate the following quantities. (a)The surface charge density and the force per unit area on the sphere. (b) The total induced charge and the total force on the sphere. Is the result what you expect? (c) The potential energy of the system. (d) If the potential of the conductor is Vinstead of zero, how many image charges are needed and where they should be placed? Solution:(a)From the lecture, we know that an image charge of should qaRq'be placed at aRb2. Then the potential at rR should be cos24cos24cos24cos244'4),,(2222222221brbraRqararqbrbraRqararqrqrqrVThe electric field at the surface of the sphere is 2/3222/322)cos2()cos()cos2(cos4),,(bRbRabRRaRaRaRqrrVERrNote 2/3222/3324222/322)cos2()cos()cos/2/()cos/()cos2()cos(aRRaRRaaaRaRRaaRRRbRbRabRR. We can simplify the result of E to. RRaaRaRqRaRaRaRqaRRaRRaaaRaRaRqE)()cos2(4)cos2(4)cos2()cos()cos2(cos4222/32222/3222/3222/322The surface charge density is RRaaRaRqE)()cos2(4222/322. The force per unit area is rRRaaRaRqnEf222232222)()cos2()4(22(b)The total induced surface charge is '22)(112)(112)()2(1)2(12)()cos2(12)()cos2(sin2)()cos2(2sin)(sin2sin222222222/1222/...
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- Fall '09