Q2S - b. In each case, Fnd the potential on both conductors...

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Quiz 2 Solutions Check the best answer. 1. With respect to the E-feld and potential in a region, which oF these is wrong? IF you move parallel to a uniForm E-feld a distance x the potential drops by the amount E x . IF the potential on a surFace is constant, it Follows that E = 0 at all points on that surFace. [ E could be perpendicular to the surFace.] IF two point charges oF like sign are moved closer together the potential energy oF the system increases. IF the potential does not depend on x , then E x = 0 . Check T or F depending on whether the statement is true or false. 2. As one moves away From a negative point charge the potential decreases. T ± [Increases] Physics 54 Summer 2011 1
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3. In the cases shown, a conducting sphere of radius a is surrounded by a larger concentric conducting shell of radius b and negligible thickness. Case 1. Charge + Q is placed on the inner sphere only. Case 2. Charge +Q is placed on the outer shell only. a. In each case, sketch lines of the E-Feld, in the gap between the conductors and in the space outside the shell.
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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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Q2S - b. In each case, Fnd the potential on both conductors...

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