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Unformatted text preview: Solution for Homework 6 Review for Exam 1 (Required but not Collected) Solution to Homework Problem 6.1(Solution to Problem 3.28) Problem: In charging by induction, the order you move things is very important. Imagine you have a metal sphere in the presence of the charged golf tube. You touch the sphere to ground the system. Discuss what happens if: (a)You remove your finger from the sphere first, then the golf tube. (b)You remove the golf tube first, then stop touching the sphere with your finger. Solution to Part(a) If an uncharged sphere is brought near a charged object, like a golf tube, the charge separates (I). If the object is then grounded while the tube is near, the separated negative charge that is not held in place by the golf tube escapes to the ground (II). If the ground is then removed, BEFORE (III) the golf tube is removed, the excess positive charge is trapped on the sphere. (IV). I: Charge Separates IV: Rod Removed III: Ground--net charge trapped II: Grounded + + + + + + + + + + + + + Neutral Net + Charge Net + Charge Net + Charge Grading Key: Part(a) 3 Points Solution to Part(b) If the rod is removed before the ground is, the positive charge on the object will draw an equal amount of negative charge from the groundleaving a neutral sphere. I: Charge Separates III: Rod Removed II: Grounded + + + + + + Neutral Net + Charge Neutral 1 Grading Key: Part (b) 2 Points 2 point(s) : Correct solution to part (b) no picture required. Total Points for Problem: 5 Points Solution to Homework Problem 6.2(Three Infinite Positive Planes) Problem: Three infinite parallel planes are evenly spaced along the x-axis. From left to right, they have charge densities , 3 , and where > . Compute and draw the electric field everywhere. Solution (a) Specialize Gauss Law for Planar Symmetry: Apply Gauss Law to a cylinder with end area A whose ends are parallel to the planes. Let the electric field at the right end of the Gaussian surface be vector E r = E r x and the electric field at the left side of the Gaussian surface, vector E l = E l x . The surface normals at the left and right of the Gaussian surface are, by observation, n l =- x and n r = x . The electric flux is e = integraldisplay S ( vector E n ) dA = integraldisplay left ( vector E l n l ) dA + integraldisplay right ( vector E r n r ) dA = A ( E r- E l ) . Applying Gauss Law gives e = A ( E r- E l ) = Q enclosed . (b) Compute the Electric Field in the Outermost Regions: In planar symmetry, the leftmost electric field is equal and opposite to the rightmost field. Therefore, vector E I =- vector E IV . Let vector E I = E I x and vector E IV = E IV x . Then, using the form of Gauss Law for planes on a Gaussian surface which encloses all charge, E r- E l = E IV- E I = Q enclosed A = E IV + E IV = 2 E IV , where I have used...
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