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PHYS 110B - HW #4 Fall 2005, Solutions by David Pace Equations Referenced As ”EQ

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PHYS 110B - HW #4 Fall 2005, Solutions by David Pace Equations Referenced As ”EQ PHYS 110B - HW #4 Fall 2005, Solutions by David Pace Equations referenced as ”EQ. #” are from Griffiths Problem statements are paraphrased [1.] Problem 8.2 from Griffiths Reference problem 7.31 (figure 7.43). (a) Let the charge on the ends of the wire be zero at t = 0. Find the electric and magnetic fields in the gap, E~ (s, t) and B~ (s, t). (b) Find the energy density and Poynting vector in the gap. Verify equation 8.14. (c) Solve for the total energy in the gap; it will be time-dependent. Find the total power flowing into the gap by integrating the Poynting vector over the relevant surface. Verify that the input power is equivalent to the rate of increasing energy in the gap. (Griffiths Hint: This verification amounts to proving the validity of equation 8.9 in the case where W = 0.) Solution (a) The electric field between the plates of a parallel plate capacitor is known to be (see example 2.5 in Griffiths), σ E~ = zˆ (1) o where I define zˆ as the direction in which the current is flowing. We may assume that the charge is always spread uniformly over the surfaces of the wire. The resultant charge density on each “plate” is then time-dependent because the flowing current causes charge to pile up. At time zero there is no charge on the plates, so we know that the charge density increases linearly as time progresses. It σ(t) = (2) πa2 where a is the radius of the wire and It is the total charge on the plates at any instant (current is in units of Coulombs/second, so the total charge on the end plate is the current multiplied by the length of time over which the current has been flowing). The electric field between the ends of the wire is, ~ It E = 2 zˆ (3) oπa The magnetic field is found from Ampere’s law, ! I Z ∂E~ B~ · d~l = µ I + µ · d~a Eq. 7.38 (4) o enc o o ∂t where the enclosed current is zero, Ienc = 0, and the magnetic field is generated entirely by the displacement current. 1 The displacement current, Id (the non-zero term in Ampere’s law above), through the gap must be in the zˆ direction (the only proper current flow is through the wire), Bφ(2πs) = Id (5) Solving for the displacement current in the gap gives, ! Z ∂E~ I = µ · d~a (6) d o o ∂t Z ∂ It = µoo 2 zˆ · d~a (7) ∂t oπa Z s Z 2π µoI = 2 (ˆz) · s ds dφ zˆ (8) πa 0 0 2 s µoI s = 2 (2π) (9) πa 2 0 2 µoIs = (10) a2 Now the magnetic field is found from (5), µoIs B~ = φˆ (11) 2πa2 where it should be noted that the field increases with s. This is only for the region in the gap and inside the wire radius of a. The linear increase of the magnetic field in this region agrees with that we would find in a solid wire with a uniform current density. For s > a the magnetic field is a fringing field that goes to zero as s → ∞. (b) The energy density is given in terms of the fields in the gap, 2 1 2 B uem = oE + Eq. 8.13 (12) 2 µo 2 2 2 2 2 1 I t 1 µoI s = o 2 2 4 + 2 4 (13) 2 oπ a µo 4π a 2 2 2 1 I t µos = 2 4 + (14) 2 π a o 4 noting that E2 = E~ · E~ . The Poynting vector in the gap is, 1 S~ = E~ × B~ Eq. 8.10 (15) µo 1 It µoIs ˆ = 2 zˆ × 2 φ (16) µo oπa 2πa I2ts = − 2 4 sˆ (17) 2oπ a 2 The Poynting vector represents energy flow. Taking special note of the direction found above we see that the energy is flowing into the gap. Verification of equation 8.14 follows (umech = 0 because there are no charges to move in the gap and the energy within the gap is therefore due entirely to that of the fields), ∂ (u + u ) = −∇~ · S~ Eq. 8.14 (18) ∂t mech em 2 2 2 2 ∂ I t µos 1 ∂ −I ts 2 4 + = − s 2 4 (19) ∂t 2π a o 4 s ∂s 2oπ a I2t 1 I2ts 2 4 = − − 2 4 (20) oπ a s oπ a I2t = 2 4 (21) oπ a (c) Since we have the energy density in the gap we are ready to determine the total energy. Z 2 1 2 B Uem = oE + dτ Eq. 8.5 (22) 2 µo Z a Z 2π Z w 2 2 2 1 I t µos = 2 4 + s ds dφ dz (23) 2 0 0 0 π a o 4 2 Z a 2 3 I t s µos = 2 4 (2πw) + ds (24) 2π a 0 o 4 2 2 2 4 I w t a µoa = 4 + (25) πa 2o 16 2 2 2 I w t µoa = 2 + (26) 2πa o 8 where this result is for the volume of the cylindrical gap (length w and radius a). The problem tells us to determine the total power flowing into the gap by integrating the Poynting vector over the surface enclosing it. This is the cylindrical surface occurring at s = a. Technically, this also includes the circular surfaces at each of the plates, but for these surfaces the product S~ ·d~a = 0 so they do not contribute to the solution. Z 2π Z w Power in = S~ · s dφ dz sˆ (27) 0 0 Z 2π Z w I2ts = − 2 4 s dφ dz (28) 0 0 2oπ a I2ta2 = − 2 4 (2wπ) (29) 2oπ a I2tw = − 2 (30) oπa 3 Finally, we essentially want to verify equation 8.9, Z I dW d 1 2 1 2 1 = − oE + B dτ − E~ × B~ · d~a (31) dt dt V 2 µo µo S where this represents the rate at which work is done on a collection of charges in the volume V, that is enclosed by the surface S. Since there are no charges in the gap, W = 0 and dW/dt = 0. The equation becomes, Z I d 1 2 1 2 1 oE + B dτ = − E~ × B~ · d~a (32) dt V 2 µo µo S d I Uem = − S~ · d~a (33) dt S we have solved for Uem in (26), and the integral on the right hand side of the above equation is the power flowing into the gap, given by (30). Continuing with the verification of equation 8.9, 2 2 2 2 d I w t µoa I tw 2 + = − − 2 (34) dt 2πa o 8 oπa I2tw I2tw 2 = 2 (35) oπa oπa this equation is verified. [2.] Professor Carter Problem Two charged shells, both of radius R and with surface charge density σ, are placed so that there is a distance d > 2R between their centers. Calculate the force of one shell on the other by integrating Maxwell’s stress tensor. (Hint: a good surface to choose includes the plane which passes directly between the two and then closes out at infinity). Solution Generally, the force on charges within a volume V is, I ↔ d Z F~ = T · d~a − oµo S~ dτ Eq. 8.22 (36) S dt V In this problem we will determine the force on one of the spheres due to the other sphere using the equation above. Enclose one of the spheres (call it number 2 for no specific reason) in a volume. In this problem there is no time dependence. The term involving the time derivative of the Poynting ↔ vector is zero. We will find the expression for the Maxwell stress tensor, T , which is determined by the total field in the system (we sum the contributions from sphere 1 and sphere 2, recalling that the stress tensor includes self-fields). There are no magnetic fields in this problem so the Maxwell stress tensor is, 2 E2 Ex − 2 ExEy ExEz ↔ = 2 E2 T o EyEx Ey − 2 EyEz (37) 2 E2 EzEx EzEy Ez − 2 4 The expression in (37) was given directly in lecture on 10/21/2005, but it may also be derived from the general form of the stress tensor. 1 2 1 1 2 Tij ≡ o EiEj − δijE + BiBj − δijB Eq. 8.19 (38) 2 µo 2 I will draw the enclosed region around sphere 2 and determine the force that 1 exerts on it. The electric fields outside of the spheres are determined from Gauss’ law. To best simplify the geometry of the surface integral that must be done, I use a surface that includes the plane exactly between the centers of the spheres. This surface closes at infinity, where the electric field is zero, so the integration needs only be completed along the plane directly between the spheres (see figure 1). Figure 1: Geometry of problem 2. The force is determined by integrating the Maxwell stress tensor over a surface including the plane directly between the spheres. Setting the origin of this system at the center of sphere 1, the electric field due to sphere 1 is given by, ~ q E1 = 2 rˆ (39) 4πor σR2 = 2 rˆ (40) or where q is the total charge on the sphere (therefore, q = σ(4πR2)). Writing this in terms of Cartesian coordinates, and setting x = d/2 since we are only interested in the electric field along the plane between the charges gives, 2 ! ~ σR 1 (d/2)ˆx + yyˆ + zzˆ E1 = 2 2 2 p (41) o (d/2) + y + z (d/2)2 + y2 + z2 ! σR2 1 (d/2)ˆx + yyˆ + zzˆ = 2 2 2 p (42) o (d /4) + y + z (d2/4) + y2 + z2 5 p using r = (d/2)2 + y2 + z2 as the distance from the center of sphere 1 to the plane.
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