y=1 1 J (a x ) dy dz dx dz 10 4 sin(2)e 2y dy dz sin(2)e 2y

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Todd McKinney
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1 Chapter 5 Odd-Numbered 5.. Given the current density J = 4 [sin(x)e y a x + cos(x)e y a y ]ka/m : a) Find the total current crossing the plane y = in the a y direction in the region <x<, <z<: This is found through y= I = J n da = J a y dx dz = 4 cos(x)e dx dz S S = 4 () sin(x) e =.3 MA b) Find the total current leaving the region <x,x<, <z<3by integrating J ds over the surface of the cube: Note first that current through the top and bottom surfaces will not exist, since J has no z component. Also note that there will be no current through the x = plane, since J x = there. Current will pass through the three remaining surfaces, and will be found through I = 3 J ( a y ) dx dz + y= 3 J (a y ) dx dz + y= 3 J (a x ) dy dz x= 3 = 4 [ cos(x)e cos(x)e ] 3 dx dz 4 sin()e y dy dz ( ) = 4 sin(x) (3 ) [ e ] ( ) + 4 sin()e y (3 ) = c) Repeat part b, but use the divergence theorem: We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume. We have J = J x x + J y y = 4 [ cos(x)e y cos(x)e y] = as expected 5.3. Let J = 4 sin θ r +4 a r A/m a) Find the total current flowing through that portion of the spherical surface r =.8, bounded by.π <θ<.3π, <φ<π: This will be π.3π 4 sin θ I = J n da = S.π (.8) +4 (.8) sin θdθdφ= 4(.8) π.3π sin dθ 4.64.π.3π = [ cos(θ)] dθ =77.4A.π b) Find the average value of J over the defined area. The area is Area = π.3π.π (.8) sin θdθdφ=.46 m The average current density is thus J avg = (77.4/.46) a r =53. a r A/m. 54

2 5.5. Let J = 5 ρ a ρ ρ +. a z A/m a) Find the total current crossing the plane z =. in the a z direction for ρ<.4: Use π.4 I = J n da = S z=. ρ +. ρdρdφ ( ) = ln(. + ρ ).4 (π) = π ln(7) = 78.A b) Calculate ρ v / t: This is found using the equation of continuity: ρ v t = J = ρ ρ (ρj ρ)+ J z z = ρ ρ (5) + ( ) z ρ = +. c) Find the outward current crossing the closed surface defined by ρ =., ρ =.4, z =, and z =.: This will be I = +. π π.4 5. a ρ ( a ρ )(.) dφ dz + ρ +. a z ( a z ) ρdρdφ+ since the integrals will cancel each other.. π π a ρ (a ρ )(.4) dφ dz ρ +. a z (a z ) ρdρdφ= d) Show that the divergence theorem is satisfied for J and the surface specified in part b. In part c, the net outward flux was found to be zero, and in part b, the divergence of J was found to be zero (as will be its volume integral). Therefore, the divergence theorem is satisfied Assuming that there is no transformation of mass to energy or vice-versa, it is possible to write a continuity equation for mass. a) If we use the continuity equation for charge as our model, what quantities correspond to J and ρ v? These would be, respectively, mass flux density in (kg/m s) and mass density in (kg/m 3 ). b) Given a cube cm on a side, experimental data show that the rates at which mass is leaving each of the six faces are.5, -9.85,.75, -., -4.5, and 4.45 mg/s. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center. We may write the continuity equation for mass as follows, also invoking the divergence theorem: ρ m t dv = J m dv = J m ds v v s where s J m ds = =.55 mg/s 55

3 5.7b. (continued) Treating our cm 3 volume as differential, we find ρ m t. =.55 3 g/s 6 m 3 = 55 g/m 3 s 5.9a. Using data tabulated in Appendix C, calculate the required diameter for a -m long nichrome wire that will dissipate an average power of 45 W when V rms at 6 Hz is applied to it: The required resistance will be R = V P = l σ(πa Thus the diameter will be lp d =a = σπv = (45) ( 6 )π() =.8 4 m=.8 mm b) Calculate the rms current density in the wire: The rms current will be I = 45/ = 3.75 A. Thus 3.75 J = π (.8 4 /) =6. 7 A/m 5.. Two perfectly-conducting cylindrical surfaces are located at ρ = 3 and ρ = 5 cm. The total current passing radially outward through the medium between the cylinders is 3 A dc. Assume the cylinders are both of length l. a) Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having σ =.5 S/m is present for 3 <ρ<5cm: Given the current, and knowing that it is radially-directed, we find the current density by dividing it by the area of a cylinder of radius ρ and length l: J = 3 πρl a ρ A/m Then the electric field is found by dividing this result by σ: E = 3 πσρl a ρ = 9.55 a ρ V/m ρl The voltage between cylinders is now: V = E dl = a ρ a ρ dρ = 9.55 ( ) 5 ln = 4.88 V ρl l 3 l Now, the resistance will be 5 3 R = V I = l =.63 l b) Show that integrating the power dissipated per unit volume over the volume gives the total dissipated power: We calculate l π.5 3 P = E J dv = (π) ρ (.5)l ρdρdφdz = 3 ( ) 5 π(.5)l ln = 4.64 W 3 l v.3 Ω 56

4 5.b. (continued). We also find the power by taking the product of voltage and current: P = VI = 4.88 (3) = 4.64 W l l which is in aggreement with the power density integration A hollow cylindrical tube with a rectangular cross-section has external dimensions of.5 in by in and a wall thickness of.5 in. Assume that the material is brass, for which σ =.5 7 S/m. A current of A dc is flowing down the tube. a) What voltage drop is present across a m length of the tube? Converting all measurements to meters, the tube resistance over a m length will be: R = (.5 7 )[(.54)(.54/) 4.54(.)(.54/)(.) 4 ] = Ω The voltage drop is now V = IR = ( =.47 V. b) Find the voltage drop if the interior of the tube is filled with a conducting material for which σ =.5 5 S/m: The resistance of the filling will be: R = (.5 5 )(/)(.54) 4 (.9)(.8) =.87 Ω The total resistance is now the parallel combination of R and R : R T = R R R + R =7.9 4 Ω and the voltage drop is now V = R T =.44 V Let V = (ρ +)z cos φ V in free space. a) Let the equipotential surface V = V define a conductor surface. Find the equation of the conductor surface: Set the given potential function equal to, to find: (ρ +)z cos φ = b) Find ρ and E at that point on the conductor surface where φ =.π and z =.5: At the given values of φ and z, we solve the equation of the surface found in part a for ρ, obtaining ρ =.. Then E = V = V ρ a ρ V ρ φ a φ V z a z = z cos φ a ρ + ρ + ρ z sin φ a φ (ρ +)zcos φ a z Then E(.,.π,.5) = 8. a ρ + 45 a φ 6.7 a z V/m 57

5 5.5c. Find ρ s at that point: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: E E ρ s = ɛ E n = ɛ = ɛ E E = ɛ (8.) + (45) surface E + (6.7) =.3 nc/m 5.7. Given the potential field V = xz x +4 V in free space: a) Find D at the surface z = : Use E = V = z ( ) x x x a x a y x +4 x +4 a z V/m At z =, we use this to find D(z =)=ɛ E(z =)= ɛ x x +4 a z C/m b) Show that the z = surface is an equipotential surface: There are two reasons for this: ) E at z = is everywhere z-directed, and so moving a charge around on the surface involves doing no work; ) When evaluating the given potential function at z =, the result is for all x and y. c) Assume that the z = surface is a conductor and find the total charge on that portion of the conductor defined by <x<, 3 <y<: We have ρ s = D a z z= = ɛ x x +4 C/m So Q = 3 ɛ x x +4 dx dy = (3)()ɛ ( ) ln(x +4) = 5ɛ ln =.9 nc 5.9. Let V =x yz z V in free space. a) Determine the equations of the equipotential surfaces on which V = and 6 V: Setting the given potential function equal to and 6 and simplifying results in: At V : x y z = At 6 V : x y z = 6 z 58

6 5.9b) Assume these are conducting surfaces and find the surface charge density at that point on the V = 6 V surface where x = and z =. It is known that V 6 V is the field-containing region: First, on the 6 V surface, we have Now Then, at the given point, we have x y z 6 z = () y() 6= y = 7 8 E = V = 4xyz a x x z a y [xy z] a z D(, 7/8, ) = ɛ E(, 7/8, ) = ɛ [7 a x +8a y +5a z ]C/m We know that since this is the higher potential surface, D must be directed away from it, and so the charge density would be positive. Thus ρ s = D D =ɛ =.4 nc/m c) Give the unit vector at this point that is normal to the conducting surface and directed toward the V = surface: This will be in the direction of E and D as found in part b, or [ ] 7ax +8a y +5a z a n = = [.6a x +.68a y +.43a z ] Let the surface y = be a perfect conductor in free space. Two uniform infinite line charges of 3 nc/m each are located at x =,y =, and x =,y =. a) Let V = at the plane y =, and find V at P (,, ): The line charges will image across the plane, producing image line charges of -3 nc/m each at x =,y =, and x =, y =. We find the potential at P by evaluating the work done in moving a unit positive charge from the y = plane (we choose the origin) to P : For each line charge, this will be: V P V,, = ρ [ ] l final distance from charge ln πɛ initial distance from charge where V,, =. Considering the four charges, we thus have [ V P = ρ ( ) ( ) ( ) ( )] l 7 ln +ln ln ln πɛ [ = ρ ( ) l ( ) ( )] [ ] ln () + ln +ln +ln = ln πɛ πɛ =. kv 59

7 b) Find E at P : Use E P = ρ [ l (,, ) (,, ) (,, ) (,, ) πɛ (,, ) + (,, ) ] (,, ) (,, ) (,, ) (,, ) (, 3, ) (, 4, ) = ρ [ ] l (,, ) (,, ) (, 3, ) (, 4, ) + = 73 a x 8.9 a y V/m πɛ A dipole with p =.a z µc m is located at A(,, ) in free space, and the x = plane is perfectly-conducting. a) Find V at P (,, ). We use the far-field potential for a z-directed dipole: V = p cos θ 4πɛ r = p 4πɛ z [x + y + z ].5 The dipole at x = will image in the plane to produce a second dipole of the opposite orientation at x =. The potential at any point is now: V = p [ ] z 4πɛ [(x ) + y + z ].5 z [(x +) + y + z ].5 Substituting P (,, ), we find V =. 6 4πɛ [ ] = 89.5 V b) Find the equation of the -V equipotential surface in cartesian coordinates: We just set the potential exression of part a equal to V to obtain: [ ] z [(x ) + y + z ].5 z [(x +) + y + z ].5 = Electron and hole concentrations increase with temperature. For pure silicon, suitable expressions are ρ h = ρ e = 6T.5 e 7/T C/m 3. The functional dependence of the mobilities on temperature is given by µ h =.3 5 T.7 m /V s and µ e =. 5 T.5 m /V s, where the temperature, T, is in degrees Kelvin. The conductivity will thus be σ = ρ e µ e + ρ h µ h = 6T.5 e 7/T [. 5 T T.7].3 9 = e 7/T [ +.95T.] S/m T Find σ at: a) C: With T = 73 K, the expression evaluates as σ()=4.7 5 S/m. b) 4 C: With T = = 33, we obtain σ(4) =. 3 S/m. c) 8 C: With T = = 353, we obtain σ(8) =. S/m. 6

8 5.7. Atomic hydrogen contains atoms/m 3 at a certain temperature and pressure. When an electric field of 4 kv/m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7. 9 m. a) Find P: With all identical dipoles, we have P = Nqd =(5.5 5 )(.6 9 )(7. 9 )=6.6 C/m =6.6 pc/m b) Find ɛ R : We use P = ɛ χ e E, and so χ e = P ɛ E = Then ɛ R =+χ e = (8.85 )(4 3 =.76 4 ) 5.9. A coaxial conductor has radii a =.8 mmandb = 3 mm and a polystyrene dielectric for which ɛ R =.56. If P =(/ρ)a ρ nc/m in the dielectric, find: a) D and E as functions of ρ: Use E = P ɛ (ɛ R ) = (/ρ) 9 a ρ (8.85 )(.56) = 44.9 ρ a ρ V/m Then D = ɛ E + P = 9 a ρ ρ [ ].56 + = a ρ ρ C/m = 3.8a ρ ρ nc/m b) Find V ab and χ e : Use χ e = ɛ r =.56, as found in part a..8 ( ) V ab = dρ = 44.9ln = 9 V 3 ρ.8 c) If there are 4 9 molecules per cubic meter in the dielectric, find p(ρ): Use p = P N = ( 9 /ρ) a ρ = a ρ C m ρ 5.3. The surface x = separates two perfect dielectrics. For x>, let ɛ R = ɛ R = 3, while ɛ R =5 where x<. If E =8a x 6a y 3a z V/m, find: a) E N : This will be E a x =8V/m. b) E T. This will consist of the components of E not normal to the surface, or E T = 6a y 3a z V/m. c) E T = (6) + (3) =67.V/m. d) E = (8) + (6) + (3) = 4.4V/m. 6

9 5.3. (continued) e) The angle θ between E and a normal to the surface: Use cos θ = E a x E = θ =4. f) D N = D N = ɛ R ɛ E N = 3(8.85 )(8) =. nc/m. g) D T = ɛ R ɛ E T = 5(8.85 )(67.) =.97 nc/m. h) D = ɛ R ɛ E N a x + ɛ R ɛ E T =.a x.66a y.33a z nc/m. i) P = D ɛ E = D [ (/ɛ R )]=(4/5)D =.7a x.3a y.6a z nc/m. j) the angle θ between E and a normal to the surface: Use cos θ = E a x E = D a x D =. (.) =(.66) +(.33) =.58 Thus θ = cos (.58) = Two perfect dielectrics have relative permittivities ɛ R = and ɛ R = 8. The planar interface between them is the surface x y+z = 5. The origin lies in region. If E = a x +a y 5a z V/m, find E : We need to find the components of E that are normal and tangent to the boundary, and then apply the appropriate boundary conditions. The normal component will be E N = E n. Taking f = x y +z, the unit vector that is normal to the surface is n = f f = [a x a y +a z ] 6 This normal will point in the direction of increasing f, which will be away from the origin, or into region (you can visualize a portion of the surface as a triangle whose vertices are on the three coordinate axes at x =5,y = 5, and z =.5). So E N =(/ 6)[ ] = 8.7V/m. Since the magnitude is negative, the normal component points into region from the surface. Then E N = 8.65 Now, the tangential component will be ( 6 ) [a x a y +a z ]= 33.33a x a y 66.67a z V/m E T = E E N = 33.3a x a y +6.67a z Our boundary conditions state that E T = E T and E N =(ɛ R /ɛ R )E N =(/4)E N.Thus E = E T + E N = E T + 4 E N = 33.3a x a y +6.67a z 8.3a x +8.3a y 6.67a z = 5a x + 75a y V/m Let the cylindrical surfaces ρ = 4 cm and ρ = 9 cm enclose two wedges of perfect dielectrics, ɛ R = for <φ<π/, and ɛ R = 5 for π/ <φ<π. IfE = (/ρ)a ρ V/m, find: a) E : The interfaces between the two media will lie on planes of constant φ, to which E is parallel. Thus the field is the same on either side of the boundaries, and so E = E. 6

10 5.35. (continued) b) the total electrostatic energy stored in a m length of each region: In general we have w E =(/)ɛ R ɛ E. So in region : W E = π/ 9 In region, we have W E = 4 π 9 π/ 4 ()ɛ () ρdρdφdz = π ρ ɛ () ln (5)ɛ () ρdρdφdz = 5π ρ 4 ɛ () ln ( ) 9 =45. µj 4 ( ) 9 = 338 µj Capacitors tend to be more expensive as their capacitance and maximum voltage, V max, increase. The voltage V max is limited by the field strength at which the dielectric breaks down, E BD. Which of these dielectrics will give the largest CV max product for equal plate areas: (a) air: ɛ R =,E BD = 3 MV/m; (b) barium titanate: ɛ R =, E BD = 3 MV/m; (c) silicon dioxide: ɛ R =3.78, E BD = 6 MV/m; (d) polyethylene: ɛ R =.6, E BD =4.7 MV/m? Note that V max = EBDd, where d is the plate separation. Also, C = ɛ R ɛ A/d, and so V max C = ɛ R ɛ AE BD, where A is the plate area. The maximum CV max product is found through the maximum ɛ R E BD product. Trying this with the given materials yields the winner, which is barium titanate A parallel plate capacitor is filled with a nonuniform dielectric characterized by ɛ R =+ 6 x, where x is the distance from one plate. If S =. m, and d = mm, find C: Start by assuming charge density ρ s on the top plate. D will, as usual, be x-directed, originating at the top plate and terminating on the bottom plate. The key here is that D will be constant over the distance between plates. This can be understood by considering the x-varying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x. The approximation becomes exact as the layer thicknesses approach zero. We know that D, which is normal to the layers, will be continuous across each boundary, and so D is constant over the plate separation distance, and will be given in magnitude by ρ s. The electric field magnitude is now E = D ɛ ɛ R = ρ s ɛ (+ 6 x ) The voltage beween plates is then V = 3 ( ρ s dx ɛ (+ 6 x ) = ρ s x ) ɛ 4 6 tan = ρ s ( π ) ɛ 3 4 Now Q = ρ s (.), and so C = Q V = ρ s(.)ɛ ( 3 )(4) ρ s π =4.5 F= 45 pf 63

11 5.4. Let ɛ R =.5 for <y< mm, ɛ R = 4 for <y<3 mm, and ɛ R3 for 3 <y<5 mm. Conducting surfaces are present at y = and y = 5 mm. Calculate the capacitance per square meter of surface area if: a) ɛ R3 isthatofair; b)ɛ R3 = ɛ R ;c)ɛ R3 = ɛ R ;d)ɛ R3 is silver: The combination will be three capacitors in series, for which C = C + C + C 3 = d ɛ R ɛ () + d ɛ R ɛ () + d 3 ɛ R3 ɛ () = 3 ɛ [ ] ɛ R3 So that C = (5 3 )ɛ ɛ R ɛ R3 Evaluating this for the four cases, we find a) C =3.5 nf for ɛ R3 =,b)c =5. nf for ɛ R3 =.5, c) C =6.3 nf for ɛ R3 = 4, and d) C =9.83 nf if silver (taken as a perfect conductor) forms region 3; this has the effect of removing the term involving ɛ R3 from the original formula (first equation line), or equivalently, allowing ɛ R3 to approach infinity Two coaxial conducting cylinders of radius cm and 4 cm have a length of m. The region between the cylinders contains a layer of dielectric from ρ = c to ρ = d with ɛ R = 4. Find the capacitance if a) c = cm, d = 3 cm: This is two capacitors in series, and so C = + = C C πɛ [ 4 ln ( 3 ) +ln ( 4 3 )] C = 43 pf b) d = 4 cm, and the volume of the dielectric is the same as in part a: Having equal volumes requires that 3 =4 c, from which c =3.3 cm. Now C = + = [ ( ) 3.3 ln + ( )] 4 C C πɛ 4 ln C = pf Two conducting spherical shells have radii a = 3 cm and b = 6 cm. The interior is a perfect dielectric for which ɛ R =8. a) Find C: For a spherical capacitor, we know that: C = 4πɛ Rɛ a b = 4π(8)ɛ ( 3 6) () =.9πɛ =53.3pF b) A portion of the dielectric is now removed so that ɛ R =., <φ<π/, and ɛ R =8, π/ <φ<π. Again, find C: We recognize here that removing that portion leaves us with two capacitors in parallel (whose C s will add). We use the fact that with the dielectric completely removed, the capacitance would be C(ɛ R =)=53.3/8 =6.67 pf. With one-fourth the dielectric removed, the total capacitance will be C = 4 (6.67) + 3 (53.4) = 4.7pF With reference to Fig. 5.7, let b =6m,h = 5 m, and the conductor potential be 5 V. Take ɛ = ɛ. Find values for K, ρ L, a, and C: We have [ h + ] [ h K = + b 5 + ] (5) = + (6) =3. b 6 64

12 5.47. (continued) We then have ρ L = 4πɛ V = 4πɛ (5) = 8.87 nc/m ln K ln(3) Next, a = h b = (5) (6) =3.8m. Finally, C = πɛ cosh (h/b) = πɛ cosh (5/6) =35.5pF A cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the potential of the cylinder be V and that of the plane be V. Find the surface charge density on the: a) cylinder at a point nearest the plane: The cylinder will image across the plane, producing an equivalent two-cylinder problem, with the second one at location 5 cm below the plane. We will take the plane as the zy plane, with the cylinder positions at x = ±5. Now b = cm, h = 5 cm, and V = V. Thus a = h b =4.9 cm. Then K =[(h + a)/b] =98., and ρ L =(4πɛ V )/ ln K =.43 nc/m. Now D = ɛ E = ρ [ L (x + a)ax + ya y π (x + a) + y (x a)a ] x + ya y (x a) + y and ρ s, max = D ( a x ) = ρ L x=h b,y= π [ h b + a (h b + a) h b a ] (h b a) = 473 nc/m b) plane at a point nearest the cylinder: At x = y =, D(, ) = ρ [ L aax π a aa ] x a = ρ L π a a x from which ρ s = D(, ) a x = ρ L πa = 5.8nC/m 65

Chapter 1 The Electric Force

Chapter 1 The Electric Force 1. Properties of the Electric Charges 1- There are two kinds of the electric charges in the nature, which are positive and negative charges. - The charges of opposite sign