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Unformatted text preview: b. In each case, Fnd the potential on both conductors (taking V ( ) = ). c. Suppose a thin conducting wire is connected between the sphere and the shell. What difference would that make to your answers? a. See drawing above. In Case 2 the charge is only on the outer surface of the shell. b. or r > b the situation is like a point charge Q at the center in both cases, so V ( b ) = kQ / b . In Case 2 the potential is constant for r < b at this value. or Case 1 we have V ( a ) V ( b ) = kQ r 2 dr b a = kQ 1 a 1 b . Since V ( b ) = kQ b we Fnd V ( a ) = kQ a . [The negligible thickness of the shell makes it this simple.] c. No difference in Case 2, because the two conductors are already at the same potential. In Case 1 the charge on the inner sphere would cancel with the charge on the inner surface of the shell, making this case like Case 2. Case 1 Case 2 Physics 54 Summer 2011 2...
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- Summer '09