chp26_explained - Week 4 Chapter 26 Problem 15 Two...

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Week 4, Chapter 26 Problem 15 Two conducting spheres of radius a are separated by a distance l>>a; since the distance is large, neither sphere appects the other’s electric field significantly, and the fields remain spherically symmetric. (a) If the spheres carry equal but opposite charges ±q, show that the potential difference between them is 2kq/a. (b) Write an expression for the work dW involved in moving an infinitesimal charge dq from the negative to the positive sphere. (c) Integrate your expression to find the work involved in transferring a charge Q from one sphere to the other, assuming both are initially uncharged.. (a) Since the electric fields of each sphere are not affected substantially by each other’s presence, their voltage will not be either, so we can use the standard formula for the potential of a sphere or point charge with r = infinity as the reference point. The potential of each conducting sphere is going to be equal to the point charge potential at a the radius a. The potential will not increase higher than that because the electric field inside the conductor is zero, so the potential cannot increase once inside that radius a. a kq V a V a V V a kq a V a kq a V 2 ) ( ) ( ) ( , ) ( 2 1 2 1 = ! " = ! " = = (b) If we move a differential amount of charge onto the sphere, it will not change the voltage, so taking the differential of W=qV, we’ll get dW=Vdq. dq a kq dW Vdq dW 2 = = (c) If both spheres are initially uncharged, then the lower bound of our integration is 0, and since we’re charging them up to a voltage Q, that’s our upper bound. Notice that while the potential doesn’t really change much for any little dq added, it does charge over the course of the integration, starting with V = 0 and ending with V = 2kQ/a. ( ) a kQ W Q a k W q a k W dq a kq W Q Q 2 2 0 2 0 0 2 2 2 = ! = " = = #
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Problem 36 A capacitor consists of a conducting shell of radius b. Show that its capacitance is ( ) a b k b a C !
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