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CU-Boulder PHYS 1120 - Questions

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1 of 12 CTGAUSS-1. Two vectors A and B are perpendicular to each other. The dot product A⋅B is equal to.. A B A) AB B) zero C) –AB D) Answer depends on the locations of the two vectors E) Some other answer Answer: zero CTGAUSS-2. A prism-shaped closed surface is in a constant, uniform electric field E, filling all space, pointing right. The 3 rectangular faces of the prism are labeled A, B, and C. Face A is perpendicular to the E-field. The bottom face C is parallel to E. Face B is the leaning face. (The two triangular side faces are not labeled.) B AC E Which face has the largest magnitude electric flux through it? A) A B) B C) C D) A and B have the same magnitude flux Answer: A and B have the same magnitude flux 9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado2 of 12 CTGAUSS-3. Two open surfaces are in an electric field as shown. Surface A is a flat circular disk of radius R, which squarely faces the charge. Surface B is a hollow-cup hemisphere of the same radius R. The flat rim of the hemisphere is the same distance from the charge as the rim of the flat disk. Which surface has the greater flux through it? A) A A B) B B C) Both surfaces have the same flux. Answer: Both surfaces have the same flux. CTGAUSS-4. Two open surfaces are in an electric field as shown. Surface A is a flat circular disk of radius R. Surface B is a hollow-cup hemisphere of the same radius R. Which surface has the greater flux through it? A B A) A B) B C) Both surfaces have the same flux. Answer: A has greater flux9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado3 of 12 CTGAUSS-5. The flux thru an area where the electric field is ˆAAx=GxyˆˆEExEy=+yGE(Ex and Ey are constants) is ... AA) ()()22xyEEA+ x B) ()()xyEEA+z C) D) xEAyEAE) None of these. Answer: xEACTGAUSS-6. The net electric flux flowing through the closed cylindrical surface shown is: A) Zero B)Positive C) Negative Answer: net flux is zero9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado4 of 12 CTGAUSS-7. Three closed surfaces enclose a point charge. The three surfaces are a small cube, a small sphere, and a larger sphere – all centered on the charge. Which surface has the largest flux through it? A) Small cube + green yellow pink B) smaller sphere C) larger sphere D) Impossible to tell without more information E) All three have the same flux. Answer: All three have the same flux. CTGAUSS-8. To compute the E-field around an infinite line of charge (with charge per length λ) a student draws the cylindrical gaussian surface of radius r and length L. r + + + + + + + + ++ ++ ++ ++ ++ ++ ++ ++ ++ + What is the LHS of Gauss's Law ( Eda⋅∫KKv) for this surface? L A) B) E2 C) 2E2 rπrLπ2E(2r 2rL)π+πD) E) None of these 2ErπAnswer: E2 rLπ 9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado5 of 12 CTGAUSS-9. The non-zero electric field everywhere on a closed surface is constant: constantE =G (meaning vector EGis everywhere constant in magnitude and direction). Is the following calculation correct? Eda E da EA⋅ ==∫∫GGvv A) Definitely correct B) Definitely incorrect C) Possibly correct, possibly incorrect – depends on details of the surface and E. Answer: This is a tough one. I am pretty sure the answer is B) Definitely incorrect. CTGAUSS-10. A spherical shell with a uniform positive charge density on its surface is near a positive point charge. Is the electric field inside the sphere zero? A) E=0 inside + + + + + + + + + + + + + E=? B) E ≠ 0 inside C) Not enough info to answer. Answer: E ≠ 0 inside The total E-field is the field due to the spherical shell of charge + the field due to the point charge. The field due to the spherical shell is zero (inside the shell). But the field due to the point charge is present inside the shell.9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado6 of 12 CTGAUSS-11. A uniform, infinite plane of negative charge creates a uniform E-field of magnitude E perpendicular to the plane and pointing toward the plane as shown. An imaginary gaussian surface in the shape of a right cylinder is shown. (This shape is sometimes called a "pillbox".) The flux through surface is.. E E L cap area A radius r A) –EA B) +2EA C) (–2A+Lπr2) E D) Lπr2E E) None of these Answer: None of these. The answer is –2EA (the flux is negative)9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado7 of 12 CTGAUSS-12. A dipole sits near the origin. We draw an imaginary Gaussian sphere (radius r) around it. Gauss’ law says: CTGAUSS-12. A dipole sits near the origin. We draw an imaginary Gaussian sphere (radius r) around it. Gauss’ law says: ++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + enclosed0qEda⋅=ε∫KKv Do we conclude that E=0 everywhere on that sphere? A) Yes, E=0 everywhere B) No, E is not 0 at all points on that sphere. Answer: No, E is not 0 at all points on that sphere. The total flux through the sphere is zero, but the E is non-zero everywhere on the sphere. In some places on the sphere, the flux is positive, in other places, it is negative. The total flux sums to zero. r CTGAUSS-13. Using Gauss's Law, we determined that E=0 inside an insulating sphere with a uniform charge distribution on its surface. What if we change the sphere to a cube? Is E=0 inside? A) Yes, E=0 everywhere inside. B) No, E is not zero everywhere inside. Answer: No, E is not zero everywhere inside. With the sphere, we were able to argue from symmetry that for the imaginary spherical surface. In the case of a cube, we cannot make the same symmetry argument regardless of which imaginary Gaussian surface we pick.Eda EA⋅=∫KvK–+9/11/2009 PHYS1120 Dubson Fa09 ©University of Colorado8 of 12 CTGAUSS-14. A point charge +q sits outside a solid neutral copper sphere of radius A. What is the magnitude of the E-field at the center of the sphere? A) E = kq/r2 B) E = kq/A2 +qC) E = kq/(r–A)2 D) E = 0 E) None of these Answer: E = 0 everywhere inside a metal in equilibrium. Polarization charges form on the surface of the copper sphere which produce a field inside the copper which exactly cancels the field due to the point charge q. CTGAUSS-15. A negative point charge with charge -Q sits in the interior of a spherical metal


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CU-Boulder PHYS 1120 - Questions

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