1 of 7Exam I Review.RI-1 Three equal mass charges are released from rest at the positions shown on the x-axis. Which mass has the largest initial acceleration?A: A B: B C: CD: 2 of the masses have the same acceleration. E: Some other answer. Answer: B. The particle which feels the largest E-field due to it neighbors will have the largest force. Since all the masses are equal, the largest force produces the largest acceleration. RI-2Four charges are arranged as shown, all a distance r from the origin. What is the magnitude of the E-field at the origin, in units if kQ/r2 ?A: 1 B: 2 C: 3 D: 4 E: None of these.Answer: None of these. The answer is 5. The magnitude of the E-field is 5kQ/r2.PHYS1120 Dubson, 1/14/2019 ©University of Colorado at Boulder+ - + x +Q -Q +Q A B C x y +Q +2Q -4Q +2Q2 of 7RI-3 A line of charge has length L and total charge Q. What is the relation between dx and dQ?A: dx = (Q/L) dQ B: dQ = (L/Q) dx C: dQ = (Q/L) dx D: NoneAt point A, what is the magnitude of the field dE due to the element of charge dQ?A: 2 2k dQh x+B: 2 2k dQh x+C: 2 2k dQh x+What is the correct expression for sin?A: 2 2xh x+B: 2 2hh x+C: 2 2hh x+D: 2 2xh x+What is the correct expression for dEx?A: 3/ 22 2k Q / L dxh xB: 2 2k h (Q / L)dxh xC: ( )( )3/ 22 2k x Q / L dxh x+Answers: dQ = (Q/L) dx2 2k dQdEh x=+2 2xsinh xq=+( )( )x3/ 22 2k x Q / L dxdE dE sinh x= q=+PHYS1120 Dubson, 1/14/2019 ©University of Colorado at BoulderL h dx, dQ dE point A x 0 3 of 7RI-4 A point charge +Q is brought near a large metal plate and the field lines are as shown. Does the plate have a non-zero net charge?A: Yes B: NoIs the magnitude of the charge on the plate greater, less, or the same as the charge Q?A: greater B: less C: same.Answers: Yes, the plate has a net negative charge, because field lines terminate on it, but no field lines begin on it. The plate has a smaller magnitude negative charge than the point charge . There are 8 fields lines coming out of the point charge, but only 5 field lines entering the plate. RI-5 If the electric field E is non-zero everywhere on a closed surface S, can the flux over a surface be zero E daS z0? A: Yes B: NoIf the electric field E is constant everywhere on a closed surface S, can you conclude that S SE da E da rr r� �? A: Yes B: NoAnswers: Yes. A simple example is a constant E-field �E E xur. The E-field is non-zero, yet E daS z0 for any closed surface in the field.NO!!! In order to "pull the E outside the surface integral" you must first have thatE daur rP everywhere on the surface so that E da E da ur r, and then you must have magnitude of the field = E = constant over the surface.PHYS1120 Dubson, 1/14/2019 ©University of Colorado at Boulder+4 of 7RI-6A long, straight line of charge has charge per length . Does Gauss's law allow you to compute the fluxSE darr�through the spherical surface of diameter D, centered on the line? A: Yes B: NoConsider now the open hemispherical surface (shaped like a cup) centered on the line as shown. Does Gauss's law allow you to compute the fluxSE da��rrthrough this surface? (Taking the outward normal direction as positive.) A: Yes B: NoAnswers: Yes. encoSQE da rr�, and Qenc = D, so oSDE da rr�.Yes. Apply Gauss's Law to the closed hemisphere. The surface integral over the flat portion is zero (since E-field perpendicular to surface there). So we can writeopencupDE da E da2l� = � =e� �r rr r�.PHYS1120 Dubson, 1/14/2019 ©University of Colorado at BoulderD D/25 of 7RI-7The electric field throughout a region of space is given by the formula� �E A y x B x y ur, where (x,y) are the coordinates of a point in space, and A, B are constants. What is ˆE x�ur?A: Ay B: Bx C: Ax D: By E: None of these.Answer: Ay. � � � � �� ��E x A y x B x y x A y x x B x y x Ay ur. Since �� ��x y 0 and x x 1 .RI-8 An insulating spherical shell with a uniform positive charge density on its surface is neara 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 inside is NOT zero. Etotal = Eshell + Epoint Inside, the shell, Eshell =0, but the field due to the point charge is still present. This is not a metal shell, so there is no polarization of the charge on the shell and no “screening”.RI-9 A spherical shell of charge has a non-uniform distribution of charge over its surface. Is the electric field everywhere within the shell zero?A: YesB: NoAnswer: No. The E-field inside a uniform, spherical shell of charge is zero. But if the charge is non-uniform, the E-field will not cancel inside the sphere.RI-10 Consider a point charge +Q off-center within a spherical metal shell. Does Gauss's Law allow you to compute the total charge on the inside surface of the shell? PHYS1120 Dubson, 1/14/2019 ©University of Colorado at BoulderE=0? + + + + + + + + + + + + + E=?6 of 7A: Yes B: NoDoes the total charge on the inside surface of the shell depend onthe total net charge of the whole shell?A: Yes B: NoConsider a dipole within a spherical metal shell.Is there a non-zero surface charge density on theinside surface of the shell?A: Yes B: NoAnswers: Yes, you can compute the total charge on theinside surface. Consider the flux SE darr� over a spherical surface within the metal. Since the E-field must be zero everywhere within the conductor, thenenc oSE da Q / 0. rr� Since Qenc = 0, there must be a charge -Q spread over the inside surface of the conductor. Since the point charge +Q is off-center, the surface -Q will not be spread uniformly over the inside surface.No, the total charge on the inside surface does not depend on the net charge of the shell. It only depends on the value of the charge within the cavity inside the shell. Yes, there must be a (non-zero, non-uniform) charge density on the inner surface. Since the E-field within the conductor must be zero, there must be some charges on the surfacesof the conductor which create an E-field which cancels the E-field due to the dipole.PHYS1120 Dubson, 1/14/2019 ©University of Colorado at Boulder+Q +Q -Q7 of 7RI-11Consider two very large, parallel metal plates in static equilibrium. Consider also the gaussian surface
View Full Document
Unlocking...