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CHAPTER 4 4.1. The value of E at P( = 2, = 40 , z = 3) is given as E = 100a 200a + 300a z V/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 this case, dL = da = 610 6 a . Thus dW =(2010 6 C)(100V/m)(610 6 m) =1210 9 J =12nJ b) in the direction of a : In this case dL = 2da = 610 6 a , and so dW =(2010 6 )(200)(610 6 ) = 2.410 8 J = 24 nJ c) in the direction of a z : Here, dL = dza z = 610 6 a z , and so dW =(2010 6 )(300)(610 6 ) =3.610 8 J =36 nJ d) in the direction of E: Here, dL = 610 6 a E , where a E = 100a 200a + 300a z 100 2 + 200 2 + 300 2 1/2 = 0.267a 0.535a + 0.802a z Thus dW =(2010 6 )100a 200a + 300a z 0.267a 0.535a + 0.802a z (610 6 ) =44.9nJ e) In the direction of G = 2a x 3a y + 4a z : In this case, dL = 610 6 a G , where a G = 2a x 3a y + 4a z 2 2 + 3 2 + 4 2 1/2 = 0.371a x 0.557a y + 0.743a z So now dW =(2010 6 )100a 200a + 300a z 0.371a x 0.557a y + 0.743a z (610 6 ) =(2010 6 )37.1(a a x )55.7(a a y )74.2(a a x ) + 111.4(a a y ) + 222.9(610 6 ) where, at P, (a a x ) = (a a y ) = cos(40 ) = 0.766, (a a y ) = sin(40 ) = 0.643, and (a a x ) =sin(40 ) =0.643. Substituting these results in dW =(2010 6 )28.435.8 + 47.7 + 85.3 + 222.9(610 6 ) =41.8nJ 40 4.2. A positive point charge of magnitude q 1 lies at the origin. Derive an expression for the incremental work done in moving a second point charge q 2 through a distance dx from the starting position (x,y,z), in the direction of a x : The incremental work is given by dW =q 2 E 12 dL where E 12 is the electric field arising from q 1 evaluated at the location of q 2 , and where dL =dxa x . Taking the location of q 2 at spherical coordinates (r,), we write: dW = q 2 q 1 4 0 r 2 a r (dx)a x where r 2 = x 2 +y 2 +z 2 , and where a r a x = sincos. So dW = q 2 q 1 4 0 (x 2 +y 2 +z 2 ) p x 2 +y 2 p x 2 +y 2 +z 2 | z sin x p x 2 +y 2 | z cos dx = q 2 q 1 xdx 4 0 (x 2 +y 2 +z 2 ) 3/2 4.3. If E = 120a V/m, find the incremental amount of work done in moving a 50m charge a distance 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 a PQ = a x a y +a z / 3. We now write dW =qE dL =(5010 6 ) 120a (a x a y +a z 3 (210 3 ) =(5010 6 )(120)(a a x )(a a y ) 1 3 (210 3 ) At P, = tan 1 (2/1) = 63.4 . Thus (a a x ) = cos(63.4) = 0.447 and (a a y ) = sin(63.4) = 0.894. 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 only a radial component and does not depend on or z. Note also that P and Q are at the same radius ( 5) from the z axis, but have dierent and z coordinates. We could just as 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 involve moving along a chord of a circle whose radius is 5. Halfway along this line is a point of symmetry in the field (make a sketch to see this). This means that when starting from either point, the initial force will be the same. Thus the answer is dW = 3.1J as in part a. This is also found by going through the same procedure as in part a, but with the direction (roles of P and Q) reversed. 41 4.4. An electric field in free space is given by E = xa x + ya y + za z V/m. Find the work done in moving a 1C charge through this field a) from (1,1,1) to (0,0,0): The work will be W =q Z E dL = 10 6 Z 0 1 xdx + Z 0 1 ydy + Z 0 1 zdz J = 1.5J b) from ( = 2, = 0) to ( = 2, = 90 ): The path involves changing with and z fixed, and therefore dL = da . We set up the integral for the work as W =10 6 Z /2 0 (xa x +ya y +za z ) da where = 2, a x a = sin, a y a = cos, and a z a = 0. Also, x = 2cos and y = 2sin. Substitute all of these to get W =10 6 Z /2 0 (2) 2 cossin + (2) 2 cossin d = 0 Given that the field is conservative (and so work is path-independent), can you see a much easier way to obtain this result? c) from (r = 10, = 0 ) to (r = 10, = 0 + 180 ): In this case, we are moving only in the a direction. The work is set up as W =10 6 Z 0 + 0 (xa x +ya y +za z ) rda Now, substitute the following relations: r = 10, x = rsincos, y = rsinsin, z = rcos, a x a = coscos, a y a = cossin, and a z a =sin. Obtain W =10 6 Z 0 + 0 (10) 2 sincoscos 2 + sincossin 2 cossin d = 0 where we use cos 2 + sin 2 = 1. 4.5. Compute the value of R P A G dL for G = 2ya x with 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 have Z P A G dL = Z P A 2ydx The change in x occurs when moving between B and P, during which y = 1. Thus Z P A G dL = Z P B 2ydx = Z 2 1 2(1)dx = 2 b) straight-line segments A(1,1,2) to C(2,1,2) to P(2,1,2): In this case the change in x occurs when moving from A to C, during which y =1. Thus Z P A G dL = Z C A 2ydx = Z 2 1 2(1)dx =2 42 4.6. An electric field in free space is given as E = xa x + 4za y + 4ya z . Given V(1,1,1) = 10 V. Determine V(3,3,3). The potential dierence is expressed as V(3,3,3)V(1,1,1) = Z 3,3,3 1,1,1 (xa x + 4za y + 4ya z ) (dxa x +dya y +dza z ) = Z 3 1 xdx+ Z 3 1 4zdy + Z 3 1 4ydz We 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 from 1 to 3, holding x at 3 and y at 3. The integrals become: V(3,3,3)V(1,1,1) = Z 3 1 xdx+ Z 3 1 4(1)dy + Z 3 1 4(3)dz =36 So V(3,3,3) =36 +V(1,1,1) =36 + 10 =26 V 4.7. Let G = 3xy 3 a x + 2za y . Given an initial point P(2,1,1) and a final point Q(4,3,1), find R G dL using the path: a) straight line: y = x1, z = 1: We obtain: Z G dL = Z 4 2 3xy 2 dx+ Z 3 1 2zdy = Z 4 2 3x(x1) 2 dx+ Z 3 1 2(1)dy = 90 b) parabola: 6y = x 2 + 2, z = 1: We obtain: Z G dL = Z 4 2 3xy 2 dx+ Z 3 1 2zdy = Z 4 2 1 12 x(x 2 + 2) 2 dx+ Z 3 1 2(1)dy = 82 4.8. Given E =xa x +ya y , a) find the work involved in moving a unit positive charge on a circular arc, the circle centered at the origin, from x = a to x = y = a/ 2. In moving along the arc, we start at = 0 and move to = /4. The setup is W =q Z E dL = Z /4 0 E ada = Z /4 0 (x a x a | z sin +y a y a | z cos )ad = Z /4 0 2a 2 sincosd = Z /4 0 a 2 sin(2)d = a 2 2 cos(2) /4 0 = a 2 2 where q = 1, x = acos, and y = asin. Note that the field is conservative, so we would get the same result by integrating along a two-segment path over x and y as shown: W = Z E dL = Z a/ 2 a (x)dx+ Z a/ 2 0 ydy # =a 2 /2 43 4.8b) Verify that the work done in moving the charge around the full circle from x = a is zero: In this case, the setup is the same, but the integration limits change: W = Z 2 0 a 2 sin(2)d = a 2 2 cos(2) 2 0 = 0 4.9. A uniform surface charge density of 20 nC/m 2 is present on the spherical surface r = 0.6cm in free space. a) Find the absolute potential at P(r = 1cm, = 25 , = 50 ): Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter. The potential function for r 0.6 cm will be that of a point charge of Q = 4a 2 s , or V(r) = 4(0.610 2 ) 2 (2010 9 ) 4 0 r = 0.081 r V with r in meters At r = 1cm, this becomes V(r = 1cm) = 8.14 V b) Find V AB given 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 a result to obtain V AB = V A V B = 0.081 1 0.02 1 0.03 = 1.36 V 4.10. A sphere of radius a carries a surface charge density of s0 C/m 2 . a) Find the absolute potential at the sphere surface: The setup for this is V 0 = Z a E dL where, from Gauss law: E = a 2 s0 0 r 2 a r V/m So V 0 = Z a a 2 s0 0 r 2 a r a r dr = a 2 s0 0 r a = a s0 0 V b) A grounded conducting shell of radius b where b a is now positioned around the charged sphere. What is the potential at the inner sphere surface in this case? With the outer sphere grounded, the field exists only between the surfaces, and is zero for r b. The potential is then V 0 = Z a b a 2 s0 0 r 2 a r a r dr = a 2 s0 0 r a b = a 2 s0 0 1 a 1 b V 44 4.11. Let a uniform surface charge density of 5nC/m 2 be present at the z = 0 plane, a uniform line charge density of 8nC/m be located at x = 0, z = 4, and a point charge of 2C be present at 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 function for the combined charges which is zero at M. That for the point charge we know to be V p (r) = Q 4 0 r Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have V l () = Z l 2 0 d +C 1 = l 2 0 ln() +C 1 For the sheet charge, we have V s (z) = Z s 2 0 dz +C 2 = s 2 0 z +C 2 The total potential function will be the sum of the three. Combining the integration constants, we obtain: V = Q 4 0 r l 2 0 ln() s 2 0 z +C The terms in this expression are not referenced to a common origin, since the charges are at dierent positions. The parameters r, , and z are scalar distances from the charges, and will be treated as such here. To evaluate the constant, C, we first look at point M, where V T = 0. At M, r = 2 2 + 5 2 = 29, = 1, and z = 5. We thus have 0 = 210 6 4 0 29 810 9 2 0 ln(1) 510 9 2 0 5 +C C =1.9310 3 V At point N, r = 1 + 4 + 9 = 14, = 2, and z = 3. The potential at N is thus V N = 210 6 4 0 14 810 9 2 0 ln( 2) 510 9 2 0 (3)1.9310 3 = 1.9810 3 V = 1.98kV 45 4.12. In spherical coordinates, E = 2r/(r 2 + a 2 ) 2 a r V/m. Find the potential at any point, using the reference a) V = 0 at infinity: We write in general V(r) = Z 2rdr (r 2 +a 2 ) 2 +C = 1 r 2 +a 2 +C With a zero reference at r , C = 0 and therefore V(r) = 1/(r 2 +a 2 ). b) V = 0 at r = 0: Using the general expression, we find V(0) = 1 a 2 +C = 0 C = 1 a 2 Therefore V(r) = 1 r 2 +a 2 1 a 2 = r 2 a 2 (r 2 +a 2 ) c) V = 100V at r = a: Here, we find V(a) = 1 2a 2 +C = 100 C = 100 1 2a 2 Therefore V(r) = 1 r 2 +a 2 1 2a 2 + 100 = a 2 r 2 2a 2 (r 2 +a 2 ) + 100 4.13. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.5 mm on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them? This will be the magnitude of the charge times the potential dierence between the finishing and starting positions, or W = (410 12 ) 2 2 0 1 2.5 1 5 10 4 = 5.7610 10 J = 576pJ 4.14. Given the electric field E = (y+1)a x +(x1)a y +2a z , find the potential dierence between the points a) (2,-2,-1) and (0,0,0): We choose a path along which motion occurs in one coordinate direction at a time. Starting at the origin, first move along x from 0 to 2, where y = 0; then along y from 0 to 2, where x is 2; then along z from 0 to 1. The setup is V b V a = Z 2 0 (y + 1) y=0 dx Z 2 0 (x1) x=2 dy Z 1 0 2dz = 2 b) (3,2,-1) and (-2,-3,4): Following similar reasoning, V b V a = Z 3 2 (y + 1) y=3 dx Z 2 3 (x1) x=3 dy Z 1 4 2dz = 10 46 4.15. Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = 1, y = 2 in free space. If the potential at the origin is 100 V, find V at P(4,1,3): The net potential function for the two charges would in general be: V = l 2 0 ln(R 1 ) l 2 0 ln(R 2 ) +C At the origin, R 1 = R 2 = 5, and V = 100 V. Thus, with l = 810 9 , 100 =2 (810 9 ) 2 0 ln( 5) +C C = 331.6 V At P(4,1,3), R 1 = |(4,1,3)(1,1,2)| = 10 and R 2 = |(4,1,3)(1,2,3)| = 26. Therefore V P = (810 9 ) 2 0 h ln( 10) + ln( 26) i + 331.6 =68.4 V 4.16. A spherically-symmetric charge distribution in free space (with a r ) note typo in problem statement, which says (0 r ) is known to have a potential function V(r) = V 0 a 2 /r 2 , where V 0 and a are constants. a) Find the electric field intensity: This is found through E =V = dV dr a r = 2V 0 a 2 r 3 a r V/m b) Find the volume charge density: Use Maxwells first equation: v = D = ( 0 E) = 1 r 2 d dr r 2 2 0 V 0 a 2 r 3 =2 0 V 0 a 2 r 4 C/m 3 c) Find the charge contained inside radius a: Here, we do not know the charge density inside radius a, but we do know the flux density at that radius. We use Gauss law to integrate D over the spherical surface at r = a to find the charge enclosed: Q encl = I r=a D dS = 4a 2 D| r=a = 4a 2 2 0 V 0 a 2 a 3 = 8 0 aV 0 C d) Find the total energy stored in the charge (or equivalently, in its electric field) in the region (a r 0. Assume a zero reference at infinity. Think of the line charge as an array of point charges, each of charge dq = L dx, and each having potential at the origin of dV = L dx/(4 0 x). The total potential at the origin is then the sum of all these potentials, or V = Z a L dx 4 0 x = Z a kdx 4 0 (x 2 +a 2 ) = k 4 0 a tan 1 x a a = k 4 0 a h 2 4 i = k 16 0 a 4.19. The annular surface, 1cm 3cm, z = 0, carries the nonuniform surface charge density s = 5nC/m 2 . Find V at P(0,0,2cm) if V = 0 at infinity: We use the superposition integral form: V P = Z Z s da 4 0 |rr 0 | where r = za z and r 0 = a . We integrate over the surface of the annular region, with da = dd. Substituting the given values, we find V P = Z 2 0 Z .03 .01 (510 9 ) 2 dd 4 0 p 2 +z 2 Substituting z = .02, and using tables, the integral evaluates as V P = (510 9 ) 2 0 2 p 2 + (.02) 2 (.02) 2 2 ln( + p 2 + (.02) 2 ) .03 .01 = .081V 48 4.20. In a certain medium, the electric potential is given by V(x) = 0 a 0 1e ax where 0 and a are constants. a) Find the electric field intensity, E: E =V = d dx 0 a 0 1e ax a x = 0 0 e ax a x V/m b) find the potential dierence between the points x = d and x = 0: V d0 = V(d)V(0) = 0 a 0 1e ad V c) if the medium permittivity is given by (x) = 0 e ax , find the electric flux density, D, and the volume charge density, v , in the region: D = E = 0 e ax 0 0 e ax a x = 0 a x C/m 2 Then v = D = 0. d) Find the stored energy in the region (0 x d), (0 y 1), (0 z 1): W e = Z v 1 2 D Edv = Z 1 0 Z 1 0 Z d 0 2 0 2 0 e ax dxdydz = 2 0 2 0 a e ax d 0 = 2 0 2 0 a 1e ad J 4.21. Let V = 2xy 2 z 3 +3ln(x 2 +2y 2 +3z 2 )V in free space. Evaluate each of the following quantities at P(3,2,1): a) V: Substitute P directly to obtain: V =15.0V b) |V |. This will be just 15.0V. c) E: We have E P =V P = 2y 2 z 3 + 6x x 2 + 2y 2 + 3z 2 a x + 4xyz 3 + 12y x 2 + 2y 2 + 3z 2 a y + 6xy 2 z 2 + 18z x 2 + 2y 2 + 3z 2 a z P = 7.1a x + 22.8a y 71.1a z V/m d) |E| P : taking the magnitude of the part c result, we find |E| P = 75.0V/m. e) a N : By definition, this will be a N P = E |E| =0.095a x 0.304a y + 0.948a z f) D: This is D P = 0 E P = 62.8a x + 202a y 629a z pC/m 2 . 49 4.22. A line charge of infinite length lies along the z axis, and carries a uniform linear charge density of C/m. A perfectly-conducting cylindrical shell, whose axis is the z axis, surrounds the line charge. The cylinder (of radius b), is at ground potential. Under these conditions, the potential function inside the cylinder ( a, where a b: W e = Z v 1 2 D Edv = Z 1 0 Z 2 0 Z b a 2 8 2 0 2 dddz = 2 4 0 ln b a J 4.23. It is known that the potential is given as V = 80 .6 V. Assuming free space conditions, find: a) E: We find this through E =V = dV d a =48 .4 V/m b) the volume charge density at = .5m: Using D = 0 E, we find the charge density through v .5 = D .5 = 1 d d (D ) .5 =28.8 0 1.4 .5 =673pC/m 3 c) the total charge lying within the closed surface = .6, 0 z 1: The easiest way to do this calculation is to evaluate D at = .6 (noting that it is constant), and then multiply by the cylinder area: Using part a, we have D .6 = 48 0 (.6) .4 = 521pC/m 2 . Thus Q =2(.6)(1)52110 12 C =1.96nC. 50 4.24. A certain spherically-symmetric charge configuration in free space produces an electric field given in spherical coordinates by: E(r) = ( 0 r 2 )/(100 0 )a r V/m (r 10) (100 0 )/( 0 r 2 )a r V/m (r 10) where 0 is a constant. a) Find the charge density as a function of position: v (r 10) = ( 0 E 1 ) = 1 r 2 d dr r 2 0 r 2 100 = 0 r 25 C/m 3 v (r 10) = ( 0 E 2 ) = 1 r 2 d dr r 2 100 0 r 2 = 0 b) find the absolute potential as a function of position in the two regions, r 10 and r 10: V(r 10) = Z 10 100 0 0 r 2 a r a r dr Z r 10 0 (r 0 ) 2 100 0 a r a r dr 0 = 100 0 0 r 10 0 (r 0 ) 3 300 0 r 10 = 10 0 3 0 4(10 3 r 3 ) V V(r 10) = Z r 100 0 0 (r 0 ) 2 a r a r dr 0 = 100 0 0 r 0 r = 100 0 0 r V c) check your result of part b by using the gradient: E 1 =V(r 10) = d dr 10 0 3 0 4(10 3 r 3 ) a r = 10 0 3 0 (3r 2 )(10 3 )a r = 0 r 2 100 0 a r E 2 =V(r 10) = d dr 100 0 0 r a r = 100 0 0 r 2 a r d) find the stored energy in the charge by an integral of the form of Eq. (42) (not Eq. (43): W e = 1 2 Z v v Vdv = Z 2 0 Z 0 Z 10 0 1 2 0 r 25 10 0 3 0 4(10 3 r 3 ) r 2 sindrdd = 4 2 0 150 0 Z 10 0 40r 3 r 6 100 dr = 4 2 0 150 0 10r 4 r 7 700 10 0 = 7.1810 3 2 0 0 e) Find the stored energy in the field by an integral of the form of Eq. (44) (not Eq. (45). W e = Z (r10) 1 2 D 1 E 1 dv + Z (r10) 1 2 D 2 E 2 dv = Z 2 0 Z 0 Z 10 0 2 0 r 4 (210 4 ) 0 r 2 sindrdd + Z 2 0 Z 0 Z 10 10 4 2 0 2 0 r 4 r 2 sindrdd = 2 2 0 0 10 4 Z 10 0 r 6 dr + 10 4 Z 10 dr r 2 = 2 2 0 0 1 7 (10 3 ) + 10 3 = 7.1810 3 2 0 0 51 4.25. Within the cylinder = 2, 0 z 90 , positive work is done when moving (against the field) to the xy plane; if b 90 , negative work is done since we move with the field. 4.29. A dipole having a moment p = 3a x 5a y +10a z nC m is located at Q(1,2,4) in free space. Find V at P(2,3,4): We use the general expression for the potential in the far field: V = p (rr 0 ) 4 0 |rr 0 | 3 where rr 0 = P Q = (1,1,8). So V P = (3a x 5a y + 10a z ) (a x +a y + 8a z )10 9 4 0 1 2 + 1 2 + 8 2 1.5 = 1.31 V 53 4.30. A dipole for which p = 10 0 a z C m is located at the origin. What is the equation of the surface on which E z = 0 but E6= 0? First we find the z component: E z = E a z = 10 4r 3 2cos(a r a z ) + sin(a a z ) = 5 2r 3 2cos 2 sin 2 This will be zero when 2cos 2 sin 2 = 0. Using identities, we write 2cos 2 sin 2 = 1 2 1 + 3cos(2) The above becomes zero on the cone surfaces, = 54.7 and = 125.3 . 4.31. A potential field in free space is expressed as V = 20/(xyz)V. a) Find the tota
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