1、CHAPTER 44.1. The value of E at P( = 2, = 40, z = 3) is given as E = 100a200a+ 300azV/m.Determine the incremental work required to move a 20C charge a distance of 6 m:a) in the direction of a: The incremental work is given by dW = qE dL, where in thiscase, dL = da= 6106a. ThusdW =(20106C)(100V/m)(61
2、06m) =12109J =12nJb) in the direction of a: In this case dL = 2da= 6106a, and sodW =(20106)(200)(6106) = 2.4108J = 24 nJc) in the direction of az: Here, dL = dzaz= 6106az, and sodW =(20106)(300)(6106) =3.6108J =36 nJd) in the direction of E: Here, dL = 6106aE, whereaE=100a200a+ 300az1002+ 2002+ 3002
3、1/2= 0.267a0.535a+ 0.802azThusdW =(20106)100a200a+ 300az 0.267a0.535a+ 0.802az(6106)=44.9nJe) In the direction of G = 2ax3ay+ 4az: In this case, dL = 6106aG, whereaG=2ax3ay+ 4az22+ 32+ 421/2= 0.371ax0.557ay+ 0.743azSo nowdW =(20106)100a200a+ 300az 0.371ax0.557ay+ 0.743az(6106)=(20106)37.1(a ax)55.7(
4、a ay)74.2(a ax) + 111.4(a ay)+ 222.9(6106)where, at P, (a ax) = (a ay) = cos(40) = 0.766, (a ay) = sin(40) = 0.643, and(a ax) =sin(40) =0.643. Substituting these results indW =(20106)28.435.8 + 47.7 + 85.3 + 222.9(6106) =41.8nJ404.2. A positive point charge of magnitude q1lies at the origin. Derive
5、an expression for theincremental work done in moving a second point charge q2through a distance dx from thestarting position (x,y,z), in the direction of ax: The incremental work is given bydW =q2E12 dLwhere E12is the electric field arising from q1evaluated at the location of q2, and wheredL =dxax.
6、Taking the location of q2at spherical coordinates (r,), we write:dW =q2q140r2ar (dx)axwhere r2= x2+y2+z2, and where ar ax= sincos. SodW =q2q140(x2+y2+z2)px2+y2px2+y2+z2| z sinxpx2+y2| z cosdx =q2q1xdx40(x2+y2+z2)3/24.3. If E = 120aV/m, find the incremental amount of work done in moving a 50m charge
7、adistance of 2 mm from:a) P(1,2,3) toward Q(2,1,4): The vector along this direction will be QP = (1,1,1)from which aPQ= axay+az/3. We now writedW =qE dL =(50106)120a(axay+az3(2103)=(50106)(120)(a ax)(a ay)13(2103)At P, = tan1(2/1) = 63.4. Thus (a ax) = cos(63.4) = 0.447 and (a ay) =sin(63.4) = 0.894
8、. Substituting these, we obtain dW = 3.1J.b) Q(2,1,4) toward P(1,2,3): A little thought is in order here: Note that the field has onlya radial component and does not depend on or z. Note also that P and Q are at thesame radius (5) from the z axis, but have dierent and z coordinates. We could justas
9、well position the two points at the same z location and the problem would not change.If this were so, then moving along a straight line between P and Q would thus involvemoving along a chord of a circle whose radius is5. Halfway along this line is a point ofsymmetry in the field (make a sketch to se
10、e this). This means that when starting fromeither point, the initial force will be the same. Thus the answer is dW = 3.1J as in parta. This is also found by going through the same procedure as in part a, but with thedirection (roles of P and Q) reversed.414.4. An electric field in free space is give
11、n by E = xax+ yay+ zazV/m. Find the work done inmoving a 1C charge through this fielda) from (1,1,1) to (0,0,0): The work will beW =qZE dL = 106Z01xdx +Z01ydy +Z01zdzJ = 1.5Jb) from ( = 2, = 0) to ( = 2, = 90): The path involves changing with and zfixed, and therefore dL = da. We set up the integral
12、 for the work asW =106Z/20(xax+yay+zaz) dawhere = 2, ax a= sin, ay a= cos, and az a= 0. Also, x = 2cos andy = 2sin. Substitute all of these to getW =106Z/20(2)2cossin + (2)2cossind = 0Given that the field is conservative (and so work is path-independent), can you see a mucheasier way to obtain this
13、result?c) from (r = 10, = 0) to (r = 10, = 0+ 180): In this case, we are moving only in theadirection. The work is set up asW =106Z0+0(xax+yay+zaz) rdaNow, substitute the following relations: r = 10, x = rsincos, y = rsinsin, z =rcos, ax a= coscos, ay a= cossin, and az a=sin. ObtainW =106Z0+0(10)2si
14、ncoscos2 + sincossin2cossind = 0where we use cos2 + sin2 = 1.4.5. Compute the value ofRPAG dL for G = 2yaxwith A(1,1,2) and P(2,1,2) using the path:a) straight-line segments A(1,1,2) to B(1,1,2) to P(2,1,2): In general we would haveZPAG dL =ZPA2ydxThe change in x occurs when moving between B and P,
15、during which y = 1. ThusZPAG dL =ZPB2ydx =Z212(1)dx = 2b) straight-line segments A(1,1,2) to C(2,1,2) to P(2,1,2): In this case the change inx occurs when moving from A to C, during which y =1. ThusZPAG dL =ZCA2ydx =Z212(1)dx =2424.6. An electric field in free space is given as E = xax+ 4zay+ 4yaz.
16、Given V(1,1,1) = 10 V.Determine V(3,3,3). The potential dierence is expressed asV(3,3,3)V(1,1,1) =Z3,3,31,1,1(xax+ 4zay+ 4yaz) (dxax+dyay+dzaz)=Z31xdx+Z314zdy +Z314ydzWe choose the following path: 1) move along x from 1 to 3; 2) move along y from 1 to 3,holding x at 3 and z at 1; 3) move along z fro
17、m 1 to 3, holding x at 3 and y at 3. The integralsbecome:V(3,3,3)V(1,1,1) =Z31xdx+Z314(1)dy +Z314(3)dz=36SoV(3,3,3) =36 +V(1,1,1) =36 + 10 =26 V4.7. Let G = 3xy3ax+ 2zay. Given an initial point P(2,1,1) and a final point Q(4,3,1), findRG dL using the path:a) straight line: y = x1, z = 1: We obtain:Z
18、G dL =Z423xy2dx+Z312zdy =Z423x(x1)2dx+Z312(1)dy = 90b) parabola: 6y = x2+ 2, z = 1: We obtain:ZG dL =Z423xy2dx+Z312zdy =Z42112x(x2+ 2)2dx+Z312(1)dy = 824.8. Given E =xax+yay, a) find the work involved in moving a unit positive charge on a circulararc, the circle centered at the origin, from x = a to
19、 x = y = a/2.In moving along the arc, we start at = 0 and move to = /4. The setup isW =qZE dL =Z/40E ada=Z/40(x ax a| z sin+y ay a| z cos)ad=Z/402a2sincosd =Z/40a2sin(2)d =a22cos(2)/40=a22where q = 1, x = acos, and y = asin.Note that the field is conservative, so we would get the same result by inte
20、grating alonga two-segment path over x and y as shown:W =ZE dL =“Za/2a(x)dx+Za/20ydy#=a2/2434.8b) Verify that the work done in moving the charge around the full circle from x = a is zero: Inthis case, the setup is the same, but the integration limits change:W =Z20a2sin(2)d =a22cos(2)20= 04.9. A unif
21、orm surface charge density of 20 nC/m2is present on the spherical surface r = 0.6cmin free space.a) Find the absolute potential at P(r = 1cm, = 25, = 50): Since the charge densityis uniform and is spherically-symmetric, the angular coordinates do not matter. Thepotential function for r 0.6 cm will b
22、e that of a point charge of Q = 4a2s, orV(r) =4(0.6102)2(20109)40r=0.081rV with r in metersAt r = 1cm, this becomes V(r = 1cm) = 8.14 Vb) Find VABgiven points A(r = 2cm, = 30, = 60) and B(r = 3cm, = 45, = 90):Again, the angles do not matter because of the spherical symmetry. We use the part aresult
23、to obtainVAB= VAVB= 0.08110.0210.03= 1.36 V4.10. A sphere of radius a carries a surface charge density of s0C/m2.a) Find the absolute potential at the sphere surface: The setup for this isV0=ZaE dLwhere, from Gauss law:E =a2s00r2arV/mSoV0=Zaa2s00r2ar ardr =a2s00ra=as00Vb) A grounded conducting shell
24、 of radius b where b a is now positioned around the chargedsphere. What is the potential at the inner sphere surface in this case? With the outersphere grounded, the field exists only between the surfaces, and is zero for r b. Thepotential is thenV0=Zaba2s00r2ar ardr =a2s00rab=a2s001a1bV444.11. Let
25、a uniform surface charge density of 5nC/m2be present at the z = 0 plane, a uniform linecharge density of 8nC/m be located at x = 0, z = 4, and a point charge of 2C be presentat P(2,0,0). If V = 0 at M(0,0,5), find V at N(1,2,3): We need to find a potential functionfor the combined charges which is z
26、ero at M. That for the point charge we know to beVp(r) =Q40rPotential functions for the sheet and line charges can be found by taking indefinite integralsof the electric fields for those distributions. For the line charge, we haveVl() =Zl20d +C1=l20ln() +C1For the sheet charge, we haveVs(z) =Zs20dz
27、+C2=s20z +C2The total potential function will be the sum of the three. Combining the integration constants,we obtain:V =Q40rl20ln()s20z +CThe terms in this expression are not referenced to a common origin, since the charges are atdierent positions. The parameters r, , and z are scalar distances from
28、 the charges, and willbe treated as such here. To evaluate the constant, C, we first look at point M, where VT= 0.At M, r =22+ 52=29, = 1, and z = 5. We thus have0 =21064029810920ln(1)5109205 +C C =1.93103VAt point N, r =1 + 4 + 9 =14, =2, and z = 3. The potential at N is thusVN=21064014810920ln(2)5
29、10920(3)1.93103= 1.98103V = 1.98kV454.12. In spherical coordinates, E = 2r/(r2+ a2)2arV/m. Find the potential at any point, usingthe referencea) V = 0 at infinity: We write in generalV(r) =Z2rdr(r2+a2)2+C =1r2+a2+CWith a zero reference at r , C = 0 and therefore V(r) = 1/(r2+a2).b) V = 0 at r = 0: U
30、sing the general expression, we findV(0) =1a2+C = 0 C =1a2ThereforeV(r) =1r2+a21a2=r2a2(r2+a2)c) V = 100V at r = a: Here, we findV(a) =12a2+C = 100 C = 10012a2ThereforeV(r) =1r2+a212a2+ 100 =a2r22a2(r2+a2)+ 1004.13. Three identical point charges of 4 pC each are located at the corners of an equilate
31、ral triangle0.5 mm on a side in free space. How much work must be done to move one charge to a pointequidistant from the other two and on the line joining them? This will be the magnitude ofthe charge times the potential dierence between the finishing and starting positions, orW =(41012)22012.515104
32、= 5.761010J = 576pJ4.14. Given the electric field E = (y+1)ax+(x1)ay+2az, find the potential dierence betweenthe pointsa) (2,-2,-1) and (0,0,0): We choose a path along which motion occurs in one coordinatedirection at a time. Starting at the origin, first move along x from 0 to 2, where y = 0;then a
33、long y from 0 to 2, where x is 2; then along z from 0 to 1. The setup isVbVa=Z20(y + 1)y=0dx Z20(x1)x=2dy Z102dz = 2b) (3,2,-1) and (-2,-3,4): Following similar reasoning,VbVa=Z32(y + 1)y=3dx Z23(x1)x=3dy Z142dz = 10464.15. Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x
34、 = 1, y = 2in free space. If the potential at the origin is 100 V, find V at P(4,1,3): The net potentialfunction for the two charges would in general be:V =l20ln(R1)l20ln(R2) +CAt the origin, R1= R2=5, and V = 100 V. Thus, with l= 8109,100 =2(8109)20ln(5) +C C = 331.6 VAt P(4,1,3), R1= |(4,1,3)(1,1,
35、2)| =10 and R2= |(4,1,3)(1,2,3)| =26. ThereforeVP=(8109)20hln(10) + ln(26)i+ 331.6 =68.4 V4.16. A spherically-symmetric charge distribution in free space (with a 0. Assume a zero reference at infinity.Think of the line charge as an array of point charges, each of charge dq = Ldx, and eachhaving pote
36、ntial at the origin of dV = Ldx/(40x). The total potential at the origin isthen the sum of all these potentials, orV =ZaLdx40x=Zakdx40(x2+a2)=k40atan1xaa=k40ah24i=k160a4.19. The annular surface, 1cm b: Eout= 0 because the cylinder is at groundpotential.d) Find the stored energy in the electric field
37、 per unit length in the z direction within thevolume defined by a, where a 90, positive work is done when moving (againstthe field) to the xy plane; if ba:We start withE(r,) =qd40r32cosar+ sinaThen the energy will beWe=Zvol120E Edv =Z20Z0Za(qd)23220r64cos2 + sin2| z 3cos2+1r2sindrdd=2(qd)2322013r3aZ
38、03cos2 + 1sind =(qd)24820a3cos3cos0| z 4=(qd)2120a3Jb) Why can we not let a approach zero as a limit? From the above result, a singularity in theenergy occurs as a0. More importantly, a cannot be too small, or the original far-fieldassumption used to derive Eq. (36) (a d) will not hold, and so the f
39、ield expressionwill not be valid.4.33. A copper sphere of radius 4 cm carries a uniformly-distributed total charge of 5C in freespace.a) Use Gauss law to find D external to the sphere: with a spherical Gaussian surface atradius r, D will be the total charge divided by the area of this sphere, and wi
40、ll be ar-directed. ThusD =Q4r2ar=51064r2arC/m2b) Calculate the total energy stored in the electrostatic field: UseWE=Zvol12D Edv =Z20Z0Z.0412(5106)21620r4r2sindrdd= (4)12(5106)21620Z.04drr2=251012801.04= 2.81Jc) Use WE= Q2/(2C) to calculate the capacitance of the isolated sphere: We haveC =Q22WE=(51
41、06)22(2.81)= 4.451012F = 4.45pF554.34. A sphere of radius a contains volume charge of uniform density 0C/m3. Find the total storedenergy by applyinga) Eq. (43): We first need the potential everywhere inside the sphere. The electric fieldinside and outside is readily found from Gausss law:E1=0r30arr
42、a and E2=0a330r2arr aThe potential at position r inside the sphere is now the work done in moving a unitpositive point charge from infinity to position r:V(r) =ZaE2 ardrZraE1 ardr0=Za0a330r2drZra0r030dr0=0603a2r2Now, using this result in (43) leads to the energy associated with the charge in the sph
43、ere:We=12Z20Z0Za020603a2r2r2sindrdd =030Za03a2r2r4dr =4a520150b) Eq. (45): Using the given fields we find the energy densitieswe1=120E1 E1=20r2180r a and we2=120E2 E2=20a6180r4r aWe now integrate these over their respective volumes to find the total energy:We=Z20Z0Za020r2180r2sindrdd +Z20Z0Za20a6180
44、r4r2sindrdd =4a520150564.35. Four 0.8nC point charges are located in free space at the corners of a square 4 cm on a side.a) Find the total potential energy stored: This will be given byWE=124Xn=1qnVnwhere Vnin this case is the potential at the location of any one of the point charges thatarises fro
45、m the other three. This will be (for charge 1)V1= V21+V31+V41=q401.04+1.04+1.042Taking the summation produces a factor of 4, since the situation is the same at all fourpoints. Consequently,WE=12(4)q1V1=(.8109)220(.04)2 +12= 7.79107J = 0.779Jb) A fifth 0.8nC charge is installed at the center of the s
46、quare. Again find the total storedenergy: This will be the energy found in part a plus the amount of work done in movingthe fifth charge into position from infinity. The latter is just the potential at the squarecenter arising from the original four charges, times the new charge value, orWE=4(.8109)
47、240(.042/2)= .813JThe total energy is nowWEnet= WE(parta) +WE= .779 +.813 = 1.59J4.36 Surface charge of uniform density slies on a spherical shell of radius b, centered at the originin free space.a) Find the absolute potential everywhere, with zero reference at infinity: First, the electricfield, found from Gauss la
