SecondPhysicsSolutions

SecondPhysicsSolutions - Physics 2306 Spring 2007 Second...

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Physics 2306 Spring 2007 Second Exam 1-4) A solid conducting sphere (radius = r a ) has a total charge 3q. A hollow conducting sphere is concentric with the solid sphere and has an inner radius r b and an outer radius r c where r a < r b < r c . A positive charge q is on the hollow conducting sphere. 1) What is the charge density on the inner surface of the hollow conducting sphere? A. q/4 π r b 2 B. 2q/4 π r b 2 C. 3q/4 π r b 2 D. 4q/4 π r b 2 E. q/4 π r b 2 F. 2q/4 π r b 2 G. 3q/4 π r b 2 H. 4q/4 π r b 2 2) What is the charge density on the outer surface of the hollow conducting sphere? A. q/4 π r c 2 B. 2q/4 π r c 2 C. 3q/4 π r c 2 D. 4q/4 π r c 2 E. q/4 π r c 2 The total charge on the inner surface is +3q so that the electric field in the hollow conducting sphere is zero. This charge is spread uniformly over the inner surface (area = 4 π r b 2 ) of the hollow conducting sphere. The total charge on the hollow conducting sphere is +q. The charge on the inner surface is +3q and, therefore, the charge on the outer surface must be 2q. This charge is spread uniformly over the outer surface (area = 4 π r c 2 ) of the hollow conducting sphere.
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F. 2q/4 π r c 2 G. 3q/4 π r c 2 H. 4q/4 π r c 2 3) What is the electric potential difference V(r c ) V(r b )? A. kq/r a B. 2kq/r a C. kq/r c D. 2kq[1/r a 1/r b ] E. zero F. 2kq[1/r a 1/r b ] G. 2kq[1/r a 1/r c ] H. 2kq[1/r a 1/r c ] 4) What is the electric potential difference V(r b ) V(r a )? A. 3kq/r a B. 2kq[1/r b 1/r a ] C. kq[1/r b 1/r a ] D. 3 kq[1/r b 1/r a ] E. zero F. 3kq[1/r b 1/r a ] G. 2kq[1/r b 1/r a ] H. kq[1/r b 1/r a ] 5) Four Gaussian surfaces are enclosed within each other as shown in the figure below. Four charges are enclosed within the surfaces as shown in the figure. Which one of the following statements is true? When the statement refers to the strength of electric flux, assume positive electric flux is greater than negative electric flux. The electric field is zero between r b and r c since this is region of the hollow conducting sphere. Therefore, V(r c ) V(r b ) = rb rc E dl = zero The electric field in the region between r a and r b is the same as the electric field of a point charge, q = 3q, centered at r = 0. Therefore, V(r b ) V(r a ) = [V(r b ) V( )] [V(r a ) V( )] V(r b ) V(r a ) = 3kq/r b [ 3kq/r a ]
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A. The strength of the electric flux through S 2 is greater than the strength of the electric flux through S 1 . B. The strength of the electric flux through S 3 is greater than the strength of electric flux through S 1 . C.
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This note was uploaded on 04/02/2008 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.

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SecondPhysicsSolutions - Physics 2306 Spring 2007 Second...

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