UT PHY 302L - Gauss Potential

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familiari (nf3795) – Gauss Potential HW – yeazell – (58395) 1This print-out should have 21 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsA cubic box of side a, oriented as shown, con-tains an unknown charge. The vertically di-rected electric field has a uniform magnitudeE at the top surface and 4 E at the bottomsurface.aE4 EHow much charge Q is inside the box?1. Qencl= 02. Qencl= 3 ǫ0E a2correct3. insufficient information4. Qencl= 2Eǫ0a25. Qencl= 2 ǫ0E a26. Qencl= 6 ǫ0E a27. Qencl= ǫ0E a28. Qencl=Eǫ0a29. Qencl=12ǫ0E a210. Qencl= 3Eǫ0a2Explanation:Electric flux through a surface S is, by con-vention, po si tive for electric field lines goingout of the surface S and negative for lines go-ing in. No flux passes through the verticalsides.The top receives Φtop= −E a2(inward isnegative) and the bottom Φbottom= 4 E a2,so the total electric flux isΦE= −E a2+ 4 E a2= 3 E a2.Using Gauss’s Law, the charge inside thebox isQencl= ǫ0ΦE= 3 ǫ0E a2.002 (part 1 of 6) 10.0 pointsA positive test charge is placed at the centerof a spherical Gaussian surface.What happens to the net flux through theGaussian surface when the surface is replacedby a cube of the same volume whose center i sat the same point?1. No change correct2. The net flux increases.3. The net flux decreases but is nonvanish-ing.4. The net flux is zero.Explanation:By Gauss’ Law,ΦS=IS~E · d~A =Qǫ0.As long as the charge remains the same, andit is enclosed by any Gaussian surface, the netflux will remain the same.003 (part 2 of 6) 10.0 pointsWhat happens to t he net flux through theGaussian surface when the sphere is replacedby a cube of one-third the volume centered atthe same point?1. The net flux is zero.2. The net flux increases.familiari (nf3795) – Gauss Potent ial HW – yeazell – (583 95) 23. No change correct4. The net flux decreases but is nonvanish-ing.Explanation:004 (part 3 of 6) 10.0 pointsWhat happens to the net flux through theGaussian surface when the charge is movedoff cent er in the original sphere, but remainswithin the sphere?1. The net flux is zero.2. The net flux increases.3. No change correct4. The net flux decreases but is nonvanish-ing.Explanation:005 (part 4 of 6) 10.0 pointsWhat happens to the net flux through theGaussian surface when the charge is movedjust outside the original sphere?1. No change2. The net flux increases.3. The net flux is zero. correct4. The net flux decreases but is nonvanish-ing.Explanation:If the charg e is moved outside the Gaussiansurface, then the amount of enclosed charg e iszero and the net flux is zero. One can imaginethe field lines going out from the charge. Anyline entering the closed Gaussian surface willalso leave, so the net flux is zero.006 (part 5 of 6) 10.0 pointsThe positive test charge i s again placed atthe center of a spherical Gaussian surface.What happens to the net flux throughthe Gaussian surface when a second positivecharge is placed near, but outside, the originalsphere?1. The net flux is zero.2. The net flux increases.3. No change correct4. The net flux decreases but is nonvanish-ing.Explanation:From Part 4 it is clear that an exteriorcharge does not contribute to the flux, so theflux remains the same.007 (part 6 of 6) 10.0 pointsWhat happens to t he net flux through theGaussian surface when a second posi tivecharge is placed inside the o riginal sphericalGaussian surface?1. The net flux decreases but is nonvanish-ing.2. The net flux is zero.3. The net flux increases. correct4. No changeExplanation:Since the net enclosed charge increases, theflux must increase.008 10.0 pointsA uniformly charged conducting plate witharea A has a total charge Q which is positive.Consider a cross-sectional view of the planeand the electric field lines due to the chargeon the plane.familiari (nf3795) – Gauss Potent ial HW – yeazell – (583 95) 3E E+Q+++++++++++PFind the magnitude of the field at poi nt P ,which is a distance a from the plate. Assumethat a is very small when compared t o thedimensions of the plate, such that edge effectscan be ignored.1. k~Ek =Q4 ǫ0A2. k~Ek = ǫ0Q A3. k~Ek =Qǫ0A4. k~Ek = 2 ǫ0Q A5. k~Ek = ǫ0Q a26. k~Ek =Q4 π ǫ0a27. k~Ek =Q2 ǫ0Acorrect8. k~Ek = 4 π ǫ0a2Q9. k~Ek = 4 π ǫ0a Q10. k~Ek =Q4 π ǫ0aExplanation:Consider the Gaussian surfaceE+Q+++++++++++ESDue to the symmetry of the problem, thereis an electric flux only through the right andleft surfaces and these two are equal. If thecross section of the surface is S, then Gauss’Law states thatΦTOTAL= 2 E S =1ǫ0QASE =Q2 ǫ0A.009 (part 1 of 2) 10.0 pointsA long coaxial cable consists of an inner cylin-drical conductor with radius R1and an outercylindrical conductor shell with inner radiusR2and outer radius R3as shown. The ca-ble extends out perpendicular to the planeshown. The charge on the inner conductorper unit length along the cable is λ and thecorresponding charge on the outer conductorper unit length is −λ (same magnitude butopposite in sign) and λ > 0.λR1R2R3b−λFind the magnitude of the electric field atthe point a distance r1from the axis of theinner conductor, where R1< r1< R2.1. E =λ2 π ǫ0r1correct2. E =λ R14 π ǫ0r123. E =2 λ√3 π ǫ0r14. E =λ2 π ǫ0R15. E =λ2R14 π ǫ0r126. E =λ R13 π ǫ0r127. E =λ√2 π ǫ0r1familiari (nf3795) – Gauss Potent ial HW – yeazell – (583 95) 48. E =λ√3 π ǫ0r19. E = 010. None o f theseExplanation:Pick a cylindrical Gaussian surface with theradius r1and apply Gauss’s law:E · ℓ · 2 π r1=λℓǫ0E =λ2 π ǫ0r1.010 (part 2 of 2) 10.0 pointsFind the magnitude of the electric field at thepoint a distance r2from the axis of the innerconductor, where R3< r2.1. E =2 λ√3 π ǫ0r22. E =λ√2 π ǫ0r23. E =λ√3 π ǫ0r24. E = 0 correct5. E =λ R14 π ǫ0r226. None of these7. E =λ R13 π ǫ0r228. E =λ2 π ǫ0r29. E =λ2R14 π ǫ0r2210. E =λ2 π ǫ0R1Explanation:Pick a cylindrical Gaussian surface with theradius r2and apply the Gauss’s l aw. Becausethere is no net charge inside the Gaussiansurface, the electric field E = 0 .011 10.0 pointsFour charges are placed at the corners of asquare of side a, with q1= q2= −q, q3= q4=+q, where q is positive. Initially there is nocharge at the center of the square.q1= −qq2= −qq4= +qq3= +qqFind the work required to bri ng the chargeq from infinity and place it at the center ofthe square.1. W =−2 k q2a22. W =8 k q2a23. W =4 k q2a4. W =−4 k q2a5. W =2 k

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# UT PHY 302L - Gauss Potential

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