Solutions to Problem Set 3, Physics 370, Spring 2014 1 TOTAL POINTS POSSIBLE: 70 points. ~ 1. In English, what is an electric field E? (5 points possible) In a nutshell, if we go by the definition of an elec~ in Griffiths, it is simply the electric force, as determined by tric field E Coulomb’s law, per unit charge being acted upon. It describes the effect (in terms of force per unit charge) a particular charge configuration will have on any new charge brought into the region. Fundamentally, what is it? Honestly, I am not sure. For now, I will accept this “operational” definition of the force per unit charge and state like Griffiths does on page 62 that “I can’t tell you what a field is – only how to calculate it.” 2. Recall Gauss’ Law is stated in two forms by Griffiths: The integral form of Gauss’ law as stated in equation (2.13) I ~ · d~a = Qenc E (1) ǫ0 S which is a variant of the form you may recall seeing in Physics II. There is also the differential form of Gauss’ law as stated in equation (2.14) ~ ·E ~ = ρ. ∇ ǫ0 (2) (a) Explain how these two expressions of Gauss’ law can be considered equivalent. For this problem, clearly cite any mathematical theorems you use to make your argument. (b) When is Gauss’ law valid? Does Gauss’ law hold even if we work ~ with a surface that is not perpendicular to the electric field E? Why would we typically choose such surfaces if Gauss’ law works equally well without such a restriction? Explain your reasoning. (c) What is “Qenc ”? Of course, it is the “enclosed charge,” but what is it enclosed by? Does the enclosing surface have to be a real surface? What makes a good choice of surface? Solutions to Problem Set 3, Physics 370, Spring 2014 2 (a) (4 points possible) Gauss’ law is basically a restatement of Coulomb’s law, expressed in terms of electric flux instead of electric force (if this doesn’t make sense, read Section 2.2 of Griffiths). It specifically states that the “electric flux through any surface enclosing a charge is Qenc /ǫ0 .” This is the form described by equation (2.13), where the left H hand side sums up the electric ~ · d~a over the entire surface. I flux through the surface ΦE ≡ S E can re-write this into equation (2.14) using the divergence theorem (equation 1.56 of the textbook) Z I ~ ~ ∇ · V dτ = V~ · d~a (3) V S and use it to re-write the integral form of Gauss’ law as the differential form I ~ · d~a = Qenc E (4a) ǫ0 S Z ~ ·E ~ dτ = Qenc (4b) ∇ ǫ0 V Z Z ρ ~ ~ dτ (4c) ∇ · E dτ = ǫ0 V V ~ ·E ~ = ρ ∴∇ (4d) ǫ0 where in step 4c we used the definition of the charge density. Those of you reading the textbook should realize this is basically outlined in pages 68 and 69 of Griffiths. (b) (3 points possible) Gauss’ Law is fundamentally just a restatement of Coulomb’s law. So whenever Coulomb’s law is valid, Gauss’ law should also be valid. W haven’t been too specific about this point yet, but Coulomb’s law is completely valid as the only description of the force in all electrostatic situations, that is where the charges are not moving. As such Gauss’ law works with any surface enclosing any static charge or charges (or even no charge)... regardless of the orientation of the surface to the electric field. Of course, for Gauss’ law to be useful, we typically choose a “Gaussian surface” that exploits the symmetry in the Solutions to Problem Set 3, Physics 370, Spring 2014 3 particular charge configuration so that the electric field is perpendicular (or parallel) to the Gaussian surface because it makes the mathematics much simpler. (c) (3 points possible) As noted in the answer to the last part of this problem, the Gaussian surface is an arbitrary choice we make. So we typically try to exploit the symmetries in the charge distribution (or rather, in the electric field we expect from such a charge distribution) by selecting a Gaussian surface that lies perpendicular (or parallel) to the electric field at that surface, which makes computation of the total electric flux through that surface easy. So the enclosed charge is sometimes up to us to choose, especially in the case of “infinite” distributions of charge. 3. Griffiths Problem 2.7 (tweaked): Show that the electric field a distance z from the center of a spherical surface of radius R (see Figure 2.11), which carries a uniform charge density σ is zero for the case where z < R (inside) and Ez = 4πǫq0 z 2 for the case where z > R (outside) and q is the total charge on the sphere. [Hint: Use the law of cosines to write in terms of R and θ. Also, when you are working with your integral, you may have to rework your bounds of integration, thanks to a u-substitution you will need to apply to solve the problem. When reworking those limits, you may end up with a square root like the one shown below. Be sure to take the positive square root: √ (R − z), if R > z R2 + z 2 − 2Rz = (z − R), if R < z (10 points possible) Since the problem describes a surface charge Solutions to Problem Set 3, Physics 370, Spring 2014 4 density σ, we can read the problem as essentially stating “find the elec~ of a thin spherical shell with uniform charge density σ at a tric field E point on the z-axis.” The net electric field will be in the z-direction because of the symmetry of the charge distribution. The electric field due to the small patch of the surface indicated in Figure 2.11 is ~ = dE dq ˆ . 4πǫ0 2 (5) The magnitude of the separation vector, , is given by the law of cosines, 2 = R2 + z 2 − 2Rz cos θ. (6) Therefore: 1 σda ˆ. 2 2 4πǫ0 (R + z − 2Rz cos θ) (7) ~ · zˆ = dE ˆ · zˆ = dE cos ψ dEz = dE (8) ~ = dE Now, thanks to symmetry, we know that any component of the electric field in the xy-pane due to some piece of charge on the sphere is cancelled by another piece of charge on the sphere. As such, the total ~ can be found by just integrating over the z-component electric field E contributions tot he electric field by all the charge elements. The z~ is just component of dE where angle ψ between ˆ and zˆ. We can solve for cos ψ using the law of cosines R2 = 2 + z 2 − 2 z cos ψ 2 + z 2 − R2 cos ψ = 2 z (R2 + z 2 − 2Rz cos θ) + z 2 − R2 = 2 z z − R cos θ . = (9a) (9b) (9c) (9d) Combining equations 6, 7, 8, and 9, we can write the z-component of 5 Solutions to Problem Set 3, Physics 370, Spring 2014 the electric field as dEz = dE ˆ · zˆ = dE cos ψ 1 z − R cos θ σda = 2 2 4πǫ0 (R + z − 2Rz cos θ) σda z − R cos θ = . 2 4πǫ0 (R + z 2 − 2Rz cos θ)3/2 (10a) (10b) (10c) The differential area element for a sphere is da = R2 sin θdθdφ, so the z-component of the electric field is Z Z z − R cos θ σR2 2π π Ez = sin θdθdφ (11a) 2 4πǫ0 φ=0 θ=0 (R + z 2 − 2Rz cos θ)3/2 Z z − R cos θ σR2 π sin θdθ. (11b) = 2ǫ0 θ=0 (R2 + z 2 − 2Rz cos θ)3/2 There are several√ways to do the integral over θ. One is to make the substitution u = R2 + z 2 − 2zR cos θ, such that zR sin θ dθ R2 + z 2 − 2zR cos θ sin θ du =√ dθ. 2 zR R + z 2 − 2zR cos θ du = √ (12a) (12b) Making become √ √ the corresponding change to the limits of integration 2 2 2 2 u = R + z − 2Rz = |z − R| for θ = 0 and u = R + z + 2Rz = z + R for θ = π. Using this, we re-write equation 11 Z σR2 z+R z − R cos θ du Ez = (13a) 2ǫ0 u=|z−R| u2 zR Z σR z+R z − R cos θ du (13b) = 2ǫ0 z u=|z−R| u2 I can eliminate the R cos θ by re-writing it as R cos θ = R2 +z 2 −u2 . 2z The 6 Solutions to Problem Set 3, Physics 370, Spring 2014 integral for the electric field is σR Ez = 2ǫ0 z Z z+R σR = 2ǫ0 z Z z+R u=|z−R| u=|z−R| Z z+R R2 +z 2 −u2 2z u2 du (14a) 2z 2 −R2 −z 2 +u2 2z u2 du (14b) z− z 2 − R 2 + u2 du u2 u=|z−R| Z z+R σR z 2 − R2 = + 1 du 4ǫ0 z 2 u=|z−R| u2 = σR 4ǫ0 z 2 (14c) (14d) Now doing the actual integrals 2 z+R z − R2 σR − +u Ez = 4ǫ0 z 2 u |z−R| 2 2 2 σR z −R z − R2 = − + + (z + R − |z − R|) 4ǫ0 z 2 z+R |z − R| z 2 − R2 1 σR −(z − R) + + (z + R − |z − R|) = 4ǫ0 z 2 |z − R| 2 (15a) (15b) (15c) At this point it is convenient to consider separately the two cases z > R and z < R. For points outside the sphere z > R and |z − R| = z − R, and Ez is σR [−(z − R) + (z + R) + (z + R − (z − R))] 4ǫ0 z 2 σR = (2R + 2R) 4ǫ0 z 2 σR2 . = ǫ0 z 2 Ez = (16a) (16b) (16c) The total charge of the sphere is q = 4πR2 σ so that we can write the field as q Ez = , (17) 4πǫ0 z 2 which is just the field of a point charge at the origin! Solutions to Problem Set 3, Physics 370, Spring 2014 7 For points inside the sphere z < R and |z − R| = R − z so that σR [−(z − R) − (z + R) + (z + R − (R − z))] 4ǫ0 z 2 σR = (−2z + 2z) 4ǫ0 z 2 = 0, Ez = (18a) (18b) (18c) so the field is zero everywhere inside the spherical shell. (Thanks to Dr. Craig for providing a pre-LATEXed version of an early draft of this solution.) 4. Griffiths Problem 2.8 (tweaked): Use the electric field inside and outside a uniformly charged spherical shell as determined in Problem 2.7 to find the field inside and outside a sphere of radius R, which carries a uniform volume charge density ρ. Express your answers in terms of the total charge of the sphere, q. What does the field outside ~ as a function of the the sphere look similar to? Draw a graph of |E| distance from the center. NOTE: If you were not able to do Problem 2.7, you can still tackle this problem starting with the stated solutions for the electric field inside and outside the uniformly charged spherical shell. (10 points possible) The key here is to think of the sphere as being a collection of spherical shells. From the previous problem we know that the electric field of a spherical shell with charge dq is given in my rephrasing of Problem 2.7 as dE = dq 4πǫ0 z 2 (19) for points outside the sphere and 0 for points inside. Furthermore, since there is spherical symmetry, the z axis is somewhat arbitrary, so I’ll just called it the radial coordinate, r ′ . The sphere in this problem has charge density ρ. For points outside the sphere, a distance r ′ from the center, the electric field is Z q Z q 1 q dq ′ = , (20) dq ′ = Er = ′2 ′2 4πǫ0 r q′ =0 4πǫ0 r ′2 q ′ =0 4πǫ0 r Solutions to Problem Set 3, Physics 370, Spring 2014 8 where q is the total charge of the sphere. Outside the sphere the field is just that of a point charge at the origin. Inside the sphere we need to be a bit more careful. In preparation for doing the integral note that the charge in a spherical shell of thickness dr ′ should be dq = ρAdr ′ where A = 4πr ′2 is the surface area of a shell, such that dq = 4πρr ′2 dr ′ . Here the integral for the electric field goes from r ′ = 0 to r ′ = r. The electric field inside the sphere is Z r Z q ρ ρr ρ r ′3 dq ′ ′2 ′ . (21) = = r dr = Er = ′2 ′2 ′2 ǫ0 r r′ =0 ǫ0 r 3 3ǫ0 q=0 4πǫ0 r Note that the boundary between the solutions inside and outside the sphere are equal at r ′ = R (the surface of the sphere) implies q ρR = 2 4πǫ0 R 3ǫ0 4 q = πR3 ρ 3 3q . ρ= 4πR3 (22a) (22b) (22c) So the complete solution is: Er = or ~ = E q , 4πǫ0 r 2 q r 4πǫ0 R3 if r ≥ R if r ≤ R q rˆ, 4πǫ0 r 2 q rˆ r 4πǫ0 R3 if r ≥ R if r ≤ R (23) (24) A graph of the electric field (done using Maple) is below. Solutions to Problem Set 3, Physics 370, Spring 2014 9 (Thanks to Dr. Craig for providing me with a draft pre-LATEXed solution.) Solutions to Problem Set 3, Physics 370, Spring 2014 10 5. Griffiths Problem 2.10: A charge q sits at the back corner of a cube, ~ through the shaded area? as shown in Figure 2.17. What is the flux of E (HINT: For simplicity, treat the cube drawn in Figure 2.17 as one octant of a cube with a charge at the center. If you start doing complicated calculus, you are on the wrong path to E&M enlightenment.) (5 points possible) This problem is computationally very nasty. This is because across the shaded face of the cube shown, the distance from that surface to the surface varies, which means the electric field strength varies, which means the electric flux per unit area varies across the surface. As such, solvingH for the electric flux through that surface by direct ~ · d~a would be very non-trivial. integration of ΦE = E Instead, we use the hint provided. Envision the charge at the “center” of the surface of a larger cube consisting of eight cubes of the size of those in in Figure 2.17, but with the charge at the center, as shown here. Since the charge is centered now, by symmetry it should be clear each of the six faces of the cube should have 1/6 of the total flux. Furthermore, Solutions to Problem Set 3, Physics 370, Spring 2014 11 if we view one of these six faces face-on, the charge is centered, which means that while the flux per unit area varies across the face, it does so symmetrically across each quadrant of the face. In other words, each of the 24 squares (each a quadrant of a face of the cube) that make up the closed surface get the same electric field flux as every other one. By Gauss’ law, the total electric flux through any closed surface H ~ · d~a = Qenc , therefore the flux through the shaded surface is ΦE = E H ǫ0 1 ~ · d~a = 1 Qenc = 1 q . If you are concerned about is ΦE,shaded = 24 E 24 ǫ0 24 ǫ0 the fact that this is a cube and not a sphere, note that the symmetry of the situation means that even though the flux through a given ’square’ varies as you move across the square, all 24 squares will see the same 1 total flux, so the flux through any one square must be 24 the total flux through the closed surface. Solutions to Problem Set 3, Physics 370, Spring 2014 12 6. Griffiths Problem 2.11: Use Gauss’s law to find the electric field inside and outside a spherical shell of radius R, which carries a uniform surface charge density of σ. Compare your answer to Problem 2.7. (10 points possible) Clearly the Gaussian surface here should exploit the spherical symmetry in the problem, so let’s use a spherical surfaces illustrated below. Outs ide Guassian S urfa ce Inside Guassian Su rf e ac r For a spherical surface, the area element d~a = (rdθ)(r sin θdφ)ˆ r = r 2 sinθdθdφˆ r (25) But we don’t even need to concern ourselves with this, since there is ~ is radial and has the same magnispherical symmetry, we know the E H ~ · d~a = E · A = E(4πr 2 ). tude over a co-centric spherical surface, so E For the Inside Gaussian surface, I ~ · d~a = E(4πr 2 ) = Qenc = 0 E ǫ0 (26) ~ inside = 0. therefore E For the Outside Gaussian surface, I 2 ~ · d~a = E(4πr 2 ) = Qenc = 4πR σ E ǫ0 ǫ0 therefore σR2 ~ rˆ. Eoutside = ǫ0 r 2 (27) (28) These answers are the same as what we arrived at through much more painful means in Problem 2.7 as shown in equations 18a and 17. Solutions to Problem Set 3, Physics 370, Spring 2014 13 7. Griffiths Problem 2.12: Use Gauss’s law to find the electric field inside a uniformly charged sphere (charge density ρ). Compare your answer to Problem 2.8. (10 points possible) Just as Problem 2.8 relied on the results of Problem 2.7, we can use our experience from Problem 2.11 and use the same Guassian surfaces as in that problem to solve the problem. Its just that in this case, we will have to consider a charge volume density ρ for radius r < R and from symmetry we again know that for a co-centric spherical surface the E field is radial and of equal magnitude everywhere on that surface. For the Inside Gaussian surface (r < R), the enclosed charge is 43 πr 3 ρ, thus I ~ · d~a = E(4πr 2 ) = Qenc = 1 4 πr 3 ρ (29) E ǫ0 ǫ0 3 therefore ~ inside = 1 ρrˆ r. (30) E 3ǫ0 . You can also express this in terms of the total charge on the sphere since Qtot = 34 πR3 ρ, thus ~ inside = E or Qtot rˆ r 4πǫ0 R3 (31) Qtot ~r (32) 4πǫ0 R3 This is the same as equation 23 for r < R in our solution of Problem 2.8. ~ inside = E Solutions to Problem Set 3, Physics 370, Spring 2014 14 8. Griffiths Problem 2.18: Two spheres, each of radius R and carrying uniform charge densities +ρ and −ρ, respectively, are placed so that they partially overlap (Fig. 2.28). Call the vector from the positive cen~ Show that the field in the region of overlap ter to the negative center d. is constant, and find its value. [Hint: Use the answer to Problem 2.12.] (10 points possible) From Problem 2.12, the field inside the positive sphere is ~ + = ρ ~r+ (33) E 3ǫ0 where r+ is the vector from center of the positive charge distribution to the point in question (as shown in the figure below). The negative sphere should have a field ~ − = − ρ ~r− E 3ǫ0 (34) where r− is the vector from center of the negative charge distribution to the point in question (as shown below). rr+ d + -rr+ r+ – r- =d Solutions to Problem Set 3, Physics 370, Spring 2014 15 This means the total field should be the sum of equations 33 and 34: ~ =E ~+ + E ~− E ρ = [~r+ −~r− ] . 3ǫ0 (35a) (35b) And by definition the dipole vector, d~ = ~r+ −~r− (see lower right hand corner of diagram above), thus: ~ ~ = ρ d. E 3ǫ0 (36) Since d~ is just the seperation between the charges and is constant, the ~ field must be constant in the overlap region (which is inside both E ~ fields). spheres, such that equation 32 holds for both sphere’s E