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PracticeTest1solutions

# PracticeTest1solutions - Section A 1 The electric field...

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Section A 1. The electric field must be zero inside an isolated conductor. So a spherical Gaussian surface within each conductive shell must have zero flux since E=0, which implies that by Gauss’ Law the enclosed charge must be zero. Thus, the inside surface of each shell must have an electric charge that exactly balances the charge from the inner conductors. Therefore, the inside surface of the outermost shell must balance a charge of (–3Q+Q), which is thus +2Q 2. (b). As demonstrated in Example 23.8, the electric field on the axis of the ring is due to the charge elements of the ring, all a distance r from point P . In order to visualize the situation, one could approximate the infinite number of infinitesimal charge elements as a finite number of charge elements that add up to the total charge Q . For example, if there were 1000 charge elements, each element would have the same charge of Q /1000. If one were to flatten the ring into a disk and in essence fill up the hole in the ring, except for

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