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Physics 7B Answers for E3 (rev. 3.0) Page 1 E-3. Gauss’s Law Questions for discussion 1. Using what you know about the electric field, write down some rules for electric field lines. • Electric field lines can only start (end) at positive (negative) charges, or at infinity. • Electric field lines can’t cross. • The number of electric field lines coming out of a charge is proportional to the charge. 2. Consider a pair of point charges ±Q, fixed in place near one another as shown. a) On the diagram above, sketch the field created by these two point charges. b) Now consider an imaginary spherical surface enclosing the +Q charge: Physics 7B Answers for E3 (rev. 3.0) Page 2 i) Reproduce here your drawing of the electric field lines from part (a), so you can get a sense of how the field lines pierce the imaginary spherical surface. ii) How much electric flux passes outward through the imaginary spherical surface? You should be able to arrive at the answer very quickly using Gauss’s Law. The imaginary sphere contains a net charge +Q. So according to Gauss's Law, the net flux passing outward through the sphere is ΦE = +Q/ε0. iii) By examining the field lines and how they pierce the imaginary spherical surface, try to explain why the flux turns out to be what Gauss’s Law said it was. (For example, try to explain why the net flux through the surface is outward.) There are plenty of field lines passing both in and out through the surface. But in the region where field lines are going out, the field is stronger. So the net flux is outward. The usual convention in drawing field lines is that the field lines are more closely spaced where the field is most intense. Provided we adhere to this convention in our drawing, then it suffices simply to count the number of field lines going in and going out. c) Next, consider an imaginary ellipsoidal surface enclosing both charges: i) Once again, reproduce your drawing of the electric field lines from part (a), so you can get a sense of how the field lines pierce the imaginary ellipsoidal surface. ii) How much electric flux passes outward through the imaginary ellipsoidal surface? Again, you should be able to arrive at the answer very quickly using Gauss’s Law. The net charge enclosed by the ellipsoid is +Q - Q = 0. So according to Gauss's Law, the electric flux passing outward through the ellipsoid is ΦE = 0. iii) By examining the field lines and how they pierce the ellipsoid, try to explain why the flux turns out to be what Gauss’s Law said it was. The drawing makes explicitly clear that many of the field lines passing outward through the ellipsoid eventually come back in. So these lines contribute nothing toward the net flux. Physics 7B Answers for E3 (rev. 3.0) Page 3 Since the region of the drawing is limited, it is not explicitly clear whether every field line passing outward eventually re-enters the surface. In fact this is the case: the net charge of the distribution is zero, so (almost) every field line beginning on the +Q must end on the -Q. (We can ignore the two exceptional fieldlines on the axis, because a single fieldline in itself doesn’t cover any finite area on the ellipsoidal surface.) But even if we did not realize that (almost) every field line leaving the surface eventually re- enters, we could still see that the net flux passing outward is zero. This is because the field strength and the Gaussian surface are both symmetric from right to left. So for every field line leaving on the left, there is a field line entering on the right that contributes an equal and opposite flux. From this point of view, it doesn't even matter that the field lines in question are actually the same line. d) Finally, consider an irregular imaginary closed surface that winds around between the charges as shown: i) Once again, reproduce your drawing of the electric field lines from part (a), so you can get a sense of how the field lines pierce the irregular imaginary closed surface. ii) How much electric flux passes outward through the irregular imaginary spherical surface? Again, you should be able to arrive at the answer very quickly using Gauss’s Law. The irregular imaginary surface encloses a net charge of zero. So according to Gauss's Law, the net electric flux passing outward through the surface is ΦE = 0. iii) By examining the field lines and how they pierce the irregular surface, try to explain why the flux turns out to be what Gauss’s Law said it was. In this case it is amply clear that every field line entering the surface also leaves. So the net outflow is zero. Physics 7B Answers for E3 (rev. 3.0) Page 4 3. The diagram below shows a single point charge Q. Sketch the field created by Q, and find the amount of electric flux passing through the imaginary infinite plane surface. A good way to look at this question is to ask, "How much flux does a point charge Q emit altogether?" The answer is, Q/ε0, in the sense that the outward flow through any closed surface containing Q is Q/ε0. Next we can ask, "Of this flux, how much eventually passes through the plane?" The answer here is pretty clearly "Half." So the amount of electric flux piercing the plane is Q/2e0. This problem shows how useful it can be to understand flux intuitively. (Evaluating the flux integral in this case would have been pretty difficult.) 4. A thin disk of radius R has uniform surface charge density σ. Let us imagine a “cubical” surface enclosing the disk. Physics 7B Answers for E3 (rev. 3.0) Page 5 a) What is the electric flux passing outward through the imaginary cubical surface? The net charge enclosed by the imaginary cubical surface is q = (charge per area)×(area) = σ πR2. So according to Gauss's Law, the net flux passing outward through the surface is πR2σ/ε0. b) Can you use this result to find the electric field created by the disk? Why or why not? You may want to sketch qualitative field lines on the diagram above. The only way Gauss's Law will help you to find the electric field of a charge distribution is if you will be able to pull the field strength out of a surface integral at some point. For this to happen, it has to be the case (for at least part of your surface) that 1. The field vectors have the same magnitude at all points of the surface; and 2. The field vectors make the same angle with all parts of the surface. The diagram above shows that the "angle" condition is not going to hold for any part of the cubical surface. Moreover, if we take the drawing literally as to the spacing of the field lines, then we can see that the "magnitude" condition is not going to hold either. 5. Use symmetry arguments to find the most general form for the electric field vector (magnitude and direction) for the following types of charge distributions. a) Spherically Symmetric - the charge distribution only depends on the radial distance r from the origin (for example, the point charge; a spherical shell of charge with a uniform surface charge density). The electric field will have the same symmetry as the charge distribution, so the electric field only depends on the radial distance from the origin and will only point in the radial direction, so ! r E = E r (r)! r . b) Cylindrically Symmetric - the charge distribution is infinitely long and only depends on the radial distance r from the axis of symmetry (for example, an infinitely long line of charge with a uniform linear charge density). The electric field will have the same symmetry as the charge distribution, so the electric field only depends on the radial distance from the axis of symmetry and will only point in the radial direction, so ! r E = E r (r)! r . c) Planar Symmetry - the charge distribution is infinite in two directions and only depends on the third direction in a cartesian coordinate system (for example, an infinite sheet of charge with uniform surface charge density). The electric field will have the same symmetry as the charge distribution, so the electric field only depends on the ‘third direction’, z, and will only point in the direction perpendicular to the plane of symmetry, so ! r E = E z (z)! z . Physics 7B Answers for E3 (rev. 3.0) Page 6 Answers to Problems 1. b) Yes c) 0° d) Yes e) Φlabel = E(2πrL) f) No, 90°, Yes, Φlid = 0 g) No, 90°, Yes, Φbottom = 0 h) Φ = E(2πrL) i) qencl = λL j) ! " 2#$ 0 r radially outward k) We needed the line to be very long so that we could assume enough symmetry to be able to exploit Gauss’ Law. 2. b) ! "R2 2# 0 r radially outward c) ! "r 2# 0 radially outward d) Yes, the values of the field just inside and just outside the surface are equal. 3. a) 0 b) ! " R B 3 # R A 3 ( ) 3$ 0 r 2 c) ! " r3 # R A 3 ( ) 3$ 0 r 2 d) Only the region r>RB will be affected, and the field will now be ! " R B 3 # R A 3 ( ) 3$ 0 r 2 + %R B 2 4$ 0 r 2 . 4. a) ! "#R4 b) ! "R4 4# 0 r 2 c) ! "r2 4# 0 5. a) Rightward d) ! E A = E B = "r 6# 0 e) ! "r 8# 0 6. a) The magnitudes are the same but the directions are opposite. b) Φlabel = 0 c) Φlid = Φbottom = EA d) E = σ/2ε0 away from the sheet.
