pre1ans - sphere, 5.0 10-7 C. Therefore, V/m 8000 ) m 75 ....

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PHYS-1200 PHYSICS II SPRING 2006 PREVIEW QUIZ 1- ANSWERS PART A. 1. C 4. E 2. B 5. D 3. B 6. B 7. E 8. B PART B. F 5 . 1 F 0 . 3 F 0 . 3 F 0 . 3 F) 0 . 3 F)( 0 . 3 ( F 0 . 3 C 1. μ + = + + = C = 4.5 F = + = + + = 0 . 18 9 0 . 6 1 0 . 3 1 0 . 3 0 . 3 1 0 . 3 1 R 1 2. R = 2.0 3. a) COUNTERCLOCKWISE b) THE NET FORCE IS ZERO PART C. 1. a) so and V 20 A i r R r R i R i r = + + = = - = - E E E , , . . 60 010 R = 0.20 b) so )( m m 2 R L A L RA = = = × × - - ρ , ( . . ) . 0 20 34 10 17 10 6 8 L = 40.0 m c) INTO THE PAGE d) For one wire, = (4 10 T m/A)(20.0 A) 2 (0.050 m) T -7 B i r 1 0 5 2 8 0 10 = × = × - π . The fields of the two wires add, so the result is double that for one wire. B = 1.6 ×10 -4 T ¯¯¯¯¯¯¯¯¯¯¯¯
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2. Since there is spherical symmetry, Gauss' law can be used. Select a spherical Gaussian surface, centered about the center of the shells. a) . ² 4 1 , Then . ² 4 or 0 0 0 r q E q r E q A d E πε ε π = = = In this expression, q is the charge enclosed in the Gaussian surface, and r is the radius of a spherical Gaussian surface with the point A on its surface. Then, the Gaussian surface encloses the charge on the inner
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Unformatted text preview: sphere, 5.0 10-7 C. Therefore, V/m 8000 ) m 75 . ( C 10 . 5 ) F / m 10 . 9 ( 4 1 7 9 = = =-r q E A E A = 8000 V/m b) If the Gaussian surface is in the metal of the outer shell, E = 0 everywhere on the Gaussian surface, so the net charge enclose equals zero. Since there is a charge of 5.0 10-7 C on the inner sphere, there must be a charge of 5.0 10-7 C on the inside of the outer sphere to cancel it. There is also a charge of 5.0 10-7 C on the outside of the outer shell to keep the outer shell neutral, but the question does not ask about that. q i = 5.0 10-7 C q i is: NEGATIVE...
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pre1ans - sphere, 5.0 10-7 C. Therefore, V/m 8000 ) m 75 ....

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