1 of 12CTGAUSS-1Two vectors A and B are perpendicular to each other. The dot product AB is equal to..A) AB B) zero C) –ABD) Answer depends on the locations of the two vectorsE) Some other answerAnswer: zero CTGAUSS-2A prism-shaped closed surface is in a constant, uniform electric field E, filling all space, pointingright. 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.)Which face has the largest magnitude electric flux through it?A) A B) B C) C D) A and B have the same magnitude fluxAnswer: A and B have the same magnitude flux 1/14/2019 PHYS1120 Dubson Fa09 ©University of ColoradoECBAA B2 of 12CTGAUSS-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) AB) BC) Both surfaces have the same flux.Answer: Both surfaces have the same flux.CTGAUSS-4 Two open surfaces are in anelectric field as shown. Surface A is a flatcircular disk of radius R. Surface B is ahollow-cup hemisphere of the same radius R.Which surface has the greater flux through it?A) AB) BC) Both surfaces have the same flux.Answer: A has greater flux1/14/2019 PHYS1120 Dubson Fa09 ©University of ColoradoABAB3 of 12CTGAUSS-5 The flux thru an area ˆA A x=r where the electricfield is x yˆ ˆE E x E y= +r (Ex and Ey are constants) is ...A) ()( )2 2x yE E A+B) ( )( )x yE E A+C) xE AD) yE AE) None of these.Answer: xE ACTGAUSS-6 The net electric flux flowing through the closed cylindrical surface shown is: A) Zero B)Positive C) NegativeAnswer: net flux is zero1/14/2019 PHYS1120 Dubson Fa09 ©University of ColoradoxzyAE4 of 12CTGAUSS-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 cubeB) smaller sphereC) larger sphereD) Impossible to tell without moreinformationE) 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) a student draws the cylindrical gaussian surface of radius r and length L. What is the LHS of Gauss's Law ( E da��vv�) for this surface?A) 2E 2 rpB) E 2 r LpC) 2E (2 r 2 r L)p + pD) 2E rpE) None of theseAnswer: E 2 r Lp1/14/2019 PHYS1120 Dubson Fa09 ©University of Colorador + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +L+ green yellow pink5 of 12CTGAUSS-9The non-zero electric field everywhere on a closed surface is constant: constantE =r (meaning vector Eris everywhere constant in magnitude and direction). Is the following calculation correct?E da E da EA� = =� �rr� �A) Definitely correct B) Definitely incorrectC) 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 insideB) E 0 insideC) Not enough info to answer. Answer: E 0 inside The total E-field is the field due to the spherical shell of charge + the fielddue 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.1/14/2019 PHYS1120 Dubson Fa09 ©University of Colorado+ + + + + + + + + + + + + E=?6 of 12CTGAUSS-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..A) –EA B) +2EA C) (–2A+Lr2) ED) Lr2E E) None of theseAnswer: None of these. The answer is –2EA (the flux is negative)1/14/2019 PHYS1120 Dubson Fa09 ©University of ColoradoE E L cap area A radius r+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 7 of 12CTGAUSS-12 A dipole sits near the origin. We drawan imaginary Gaussian sphere (radius r) around it.Gauss’ law says:enclosed0qE da� =e�vv�Do we conclude that E=0 everywhere on that sphere?A) Yes, E=0 everywhereB) 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, inother places, it is negative. The total flux sums to zero.CTGAUSS-13 Using Gauss's Law, we determined thatE=0 inside an insulating sphere with a uniform chargedistribution on its surface. What if we change the sphere toa 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 E da E A� =�vv�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.1/14/2019 PHYS1120 Dubson Fa09 ©University of Colorador–+8 of 12CTGAUSS-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/r2B) E = kq/A2C) E = kq/(r–A)2D) E = 0E) None of theseAnswer: 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 shell. The conducting metal shell has no net charge. What is the total charge on the inner surface of the shell?A) –QB) +QC) +2QD) zeroE) Some other answer.Hint: Consider the gaussian surface shown.What is the total charge on the exterior surface of the shell?A) –Q B) +Q C) +2Q D) zero E) Some other answer.Answers: Total charge on the inside surface is +Q. The total
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