Solution manal of algebral

2-1 Solutions for Chapter Problems Vectors in the Cartesian Coordinate System P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q APQ = ax + ay + az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az Qx-Px=Qx-4=2; Qx=6 Qy-Py=Qy-2=4; Qy=6 Qz-Pz=Qz-1=6; Qz=7 Ans: Q(6,6,7) P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, the vectors found in (a) through (f) Vector a Find the vector A AOP = ax + ay from the origin to P b Find the vector B BOQ = ax + ay from the origin to Q c Find the vector C CPQ = -3 ax + ay from P to Q d Find A + B A + B = ax + ay e Find C – A C - A = -7 ax + ay f Find B - A B - A = -3 ax + ay fill in the table and make a sketch of Mag 4.12 Unit Vector AOP = 0.97 ax + 0.24 ay 3.16 aOQ = 0.32 ax + 0.95 ay 3.61 aPQ = -0.83 ax + 0.55 ay 6.4 7.07 3.6 a = 0.78 ax + 0.62 ay a = -0.99 ax + 0.14 ay a = -0.83 ax + 0.55 ay a AOP = (4-0)ax + (1-0)ay + (0-0)az = ax + ay AOP = 42 + 12 = 17 = 4.12 ax + a y = 0.97ax + 0.24a y 17 17 (see Figure P2.2ab) aOP = b BOQ =(1-0)ax + (3-0)ay + (0-0)az = ax + ay BOQ = 12 + 32 = 10 = 3.16 Fig P2.2ab ax + a y = 0.32ax + 0.95a y 10 10 (see Figure P2.2ab) aOQ = c CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + ay C PQ = 32 + 2 = 13 = 3.61 −3 ax + a y = −0.83ax + 0.55a y 13 13 (see Figure P2.2cd) a PQ = Fig P2.2cd 2-2 d A + B = (4+1)ax + (1+3)ay + (0-0)az = ax + ay A + B = 52 + 42 = 41 = 6.4 ax + a y = 0.78ax + 0.62a y 41 41 (see Figure P2.2cd) a= e C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + ay C − A = 72 + 12 = 50 = 7.07 −7 ax + a y = −0.99ax + 0.14a y 50 50 (see Figure P2.2ef) a= FigP2.2ef f B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + ay B − A = 32 + 22 = 13 = 3.6 −3 ax + a y = −0.83ax + 0.55a y 13 13 (see Figure P2.2ef) a= P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary points in the Cartesian Coordinate System A program or function for this task is really overkill, as it is so easy to perform the task Enter points P and Q (for example, P=[1 3]; Q=[6 4]) Then, the vector from P toQ is simply given by Q-P As a function we could have: function PQ=vector(P,Q) % Given a pair of Cartesian points % P and Q, the program determines the % vector from P to Q PQ=Q-P; Running this function we have: >> P=[1 3]; >> Q=[6 4]; >> PQ=vector(P,Q) PQ = Alternatively, we could simply perform the math in the command line window: 2-3 >> PQ=Q-P PQ = >> Coulomb’s Law, Electric Field Intensity, and Field Lines P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m, 0.0, 0.0) = 1.0nC (a) Find the total force acting on the charge Q3 (b) Repeat the problem after changing the charge of Q2 to –4.0nC (c) Find the electric field intensity for parts (a) and (b) (a) F13 = so Q1Q2 a13 , where R13 = ax + ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay 4πε o R132 ( 4a + 3a ) x10 C )(1x10 C ) ( = 4π (10 F 36π m ) ( 5m ) −9 F13 −9 x y −9 FV NM C VC = 1.15 x10−9 ax + 0.86 x10−9 a y N Similarly, F23 = 1.15 x10−9 ax − 0.86 x10−9 a y N , so FTOT = 2.3a x nN (b) with Q2 = -4 nC, F13 is unchanged but F23 = −1.15x10−9 ax + 0.86 x10−9 a y N , so FTOT = 1.7a y nN −9 FTOT ( 2.3x10 ax N ) VC V (c) Ea = = = 2.3ax -9 Q3 m (1x10 C ) Nm Likewise, Eb = 1.7a y V m Fig P2.4 P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 µC on Q2(6.0m, 9.0m, 3.0m) = 10 nC Q1Q2 a12 , where 4πε o R122 R12 = (6-3)ax + (9-3)ay + (3-3)az = ax + ay m 3a + 6a y R12 = 32 + 62 = 45m, a12 = x , and 45 F12 = 2-4 (1x10 C )(10 x10 C ) 3a = 4π (10 F 45m ) 36π m ) ( −6 F12 −9 −9 x + 6a y FV NM , so F12 = 0.89ax + 1.8a y µ N C VC 45 Fig P2.5 P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0 cm Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P We arbitrarily label the charges as shown in Figure P2.6 Then ROP = 0.1 ax + 0.1 ay ROP = 0.141 m aOP = 0.707 ax + 0.707 ay FOP = (10nC )(10nC )( 0.707 ) ( ax + a y ) −9 ( 4π ) 10 F 36π m ( 0.141m ) ( ) = 32 ( a x + a y ) µ N FTP = FSP = (10nC )(10nC ) ay −9 ( 4π ) 10 F 36π m ( 0.1m ) = 90a y µ N (10nC )(10nC ) ax −9 ( 4π ) 10 F 36π m ( 0.1m ) = 90a x µ N ( ( ) ) Fig P2.6 and then the total (adjusting to significant digits) is: FTOT = 120 ( a x + a y ) µ N 2-5 P2.7: 1.00 nC point charges are located at (0.00, -2.00, 0.00)m, (0.00, 2.00, 0.00)m, (0.00, 0.00, -2.00)m and (0.00, 0.00, +2.00)m Find the total force acting on a 1.00 nC charge located at (2.00, 0.00, 0.00)m Figure P2.7a shows the situation, but we need only find the x-directed force from one of the charges on Qt (Figure P2.7b) and multiply this result by Because of the problem’s symmetry, the rest of the components cancel 2a + 2a y QQt F1t = a R , R = 2a x + 2a y , R = m, a R = x , 4πε o R (1x10−9 C )(1x10−9 C ) ⎛⎜⎝ 2a x + 2a y ⎞⎟⎠ so F1t = = 796 x10−12 ( a x + a y ) N −9 4π 10 F 8m ) 36π m ( The force from all charges is then FTOT = ( ) ( 796 x10−12 a x ) nN = 3.2a x nN ) ( Fig P2.7a Fig P2.7b P2.8: A 20.0 nC point charge exists at P(0.00,0.00,-3.00m) Where must a 10.0 nC charge be located such that the total field is zero at the origin? For zero field at the origin, we must cancel the +az directed field from QP by placing Q at the point Q(0,0,z) (see Figure P2.8) Then we have Etot = EP + EQ = 20 x10−9 C ) a z ( QP FV V So, E P = a = = 20a z R −9 m 4πε o R 4π 10 F 3m C 36π m ( ) ( ) 2-6 and EQ = Q 4πε o R aR (10 x10 C ) (−a ) −9 = ( z −9 4π 10 F 36π m ) ( z ( m) ) = −90 az z2 So then 90 20a z − a z = 0, z 90 z = , z = 2.12 20 Thus, Q(0,0,2.12m) Fig P2.8 The Spherical Coordinate System P2.9: Convert the following points from Cartesian to Spherical coordinates: a P(6.0, 2.0, 6.0) b P(0.0, -4.0, 3.0) c P(-5.0,-1.0, -4.0) ⎛ ⎞ o −1 ⎛ ⎞ o (a) r = 62 + 22 + 62 = 8.7, θ = cos −1 ⎜ ⎟ = 47 , φ = tan ⎜ ⎟ = 18 ⎝ 8.7 ⎠ ⎝6⎠ ⎛3⎞ ⎛ −4 ⎞ (b) r = 02 + 42 + 32 = 5, θ = cos −1 ⎜ ⎟ = 53o , φ = tan −1 ⎜ ⎟ = −90o ⎝5⎠ ⎝ ⎠ ⎛ −4.0 ⎞ o −1 ⎛ − ⎞ o (c) r = 52 + 12 + 42 = 6.5, θ = cos −1 ⎜ ⎟ = 130 , φ = tan ⎜ ⎟ = 190 ⎝ 6.5 ⎠ ⎝ −5 ⎠ P2.10: a b c Convert the following points from Spherical to Cartesian coordinates: P(3.0, 30.°, 45.°) P(5.0, π/4, 3π/2) P(10., 135°, 180°) (a) x = r sin θ cos φ = 3sin 30o cos 45o = 1.06 y = r sin θ sin φ = 3sin 30o sin 45o = 1.06 z = r cos θ = 3cos 30o = 2.6 so P (1.1,1.1, 2.6) (b) 2-7 x = r sin θ cos φ = 5sin 45o cos 270o = y = r sin θ sin φ = 5sin 45o sin 270o = −3.5 z = r cos θ = 5cos 45o = 3.5 so P (0, −3.5,3.5) (c) x = r sin θ cos φ = 10sin135o cos180o = −7.1 y = r sin θ sin φ = 10sin135o sin180o = z = r cos θ = 10 cos135o = −7.1 so P (−7.1, 0, −7.1) P2.11: Given a volume defined by 1.0m ≤ r ≤ 3.0m, ≤ θ ≤ 0°, 90° ≤ φ ≤ 90°, (a) sketch the volume, (b) perform the integration to find the volume, and (c) perform the necessary integrations to find the total surface area (a) Fig P2.11 (b) 90o π 0 V = ∫∫∫ r sin θ drdθ dφ = ∫ r dr ∫ sin θ dθ 2 ∫ dφ = 13π = 13.6m3 So volume V = 14 m (c) There are surfaces: an inner, an outer, and identical sides S side = ∫∫ rdrdφ = ∫ rdr π ∫ dφ = 2π m ; Souter = ∫∫ r sin θ dθ dφ = π 90o π 0 ∫ sin θ dθ ∫ dφ = m ; STOT = 11π m = 34.6m 2 So Stotal = 35 m2 Sinner = S sides = 6π m 2 9π m 2-8 Line Charges and the Cylindrical Coordinate System P2.12: Convert the following points from Cartesian to cylindrical coordinates: a P(0.0, 4.0, 3.0) b P(-2.0, 3.0, 2.0) c P(4.0, -3.0, -4.0) ⎛4⎞ (a) ρ = 02 + 42 = 4, φ = tan −1 ⎜ ⎟ = 90o , z = 3, so P (4.0,90o ,3.0) ⎝0⎠ ⎛ ⎞ (b) ρ = 22 + 32 = 3.6, φ = tan −1 ⎜ ⎟ = 124o , z = 2, so P (3.6,120o , 2.0) ⎝ −2 ⎠ ⎛ −3 ⎞ (c) ρ = 42 + 32 = 5, φ = tan −1 ⎜ ⎟ = −37 o , z = −4, so P (5.0, −37 o , −4.0) ⎝ ⎠ P2.13: a b c Convert the following points from cylindrical to Cartesian coordinates: P(2.83, 45.0°, 2.00) P(6.00, 120.°, -3.00) P(10.0, -90.0°, 6.00) (a) x = ρ cos φ = 2.83cos 45o = 2.00 y = ρ sin φ = 2.83sin 45o = 2.00 z = z = 2.00 so P(2.00, 2.00, 2.00) (b) x = ρ cos φ = 6.00 cos120o = −3.00 y = ρ sin φ = 6.00sin120o = 5.20 z = z = −3.00 so P(−3.00,5.20, −3.00) (c) x = ρ cos φ = 10.0 cos(−90.0o ) = y = ρ sin φ = 10.0sin(−90.0o ) = −10.0 z = z = 6.00 so P(0, −10.0, 6.00) P2.14: A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diameter of 6.00 cm a Sketch the pipe conveniently overlaying the cylindrical coordinate system, lining up the length direction with the z-axis b Determine the total surface area (this could actually be useful if, say, you needed to an electroplating step on this piece of pipe) c Determine the weight of the pipe given the density of copper is 8.96 g/cm3 2-9 (a) See Figure P2.14 (b) The top area, Stop, is equal to the bottom area We must also find the inner area, Sinner, and the outer area, Souter 2π Stop = ∫∫ ρ d ρ dφ = ∫ ρ d ρ ∫ dφ = 5π cm2 Sbottom = Stop 2π 20 0 Souter = ∫∫ ρ dφ dz = ∫ dφ ∫ dz = 120π cm 2π 20 0 Sinner == ∫∫ ρ dφ dz = ∫ dφ ∫ dz = 80π cm The total area, then, is 210π cm2, or Stot = 660 cm2 (c) Determining the weight of the pipe requires the volume: V = ∫∫∫ ρ d ρ dφ dz 2π 20 0 = ∫ ρ d ρ ∫ dφ ∫ dz = 100π cm3 g ⎞ ⎛ M pipe = ⎜ 8.96 ⎟ (100π cm3 ) cm ⎠ ⎝ = 2815 g So Mpipe = 2820g Fig P2.14 P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00 (a) A charge Q = 8.00 nC exists somewhere along the y-axis Where must you locate Q so that the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of part (a) that you locate a charge Q at (0.00, 6.00m, 0.00) What value of Q will result in a total electric field intensity of zero at the origin? (a) The contributions to E from the line and point charge must cancel, or E = E L + EQ For the line: E L = ( 2nC / m ) ρL V aρ = a y = 18 a y −9 2πε o ρ m 2π 10 F 2m 36π m ( ) ( ) 2-10 and for the point charge, where the point is located a distance y along the y-axis, we (8nC ) ( −a y ) Q 72 have: EQ = − = = ( −a y ) a ( ) y −9 4πε o y 4π 10 F y2 y 36π m Therefore: 72 72 = 18, or y = = 2m y 18 ) ( So Q ( 0,2.0m,0 ) (b) Q 4πε o ( ) Q= = 18, (18)( 36 ) = 72nC Fig P2.15 P2.16: You are given two z-directed line charges of charge density +1 nC/m at x = 0, y = -1.0 m, and charge density –1.0 nC/m at x = 0, y = 1.0 m Find E at P(1.0m,0,0) The situation is represented by Figure P2.16a A better 2-dimensional view in Figure P2.16b is useful for solving the problem ⎛ a + ay ⎞ ρ L1 E1 = a ρ , and ρ a ρ = ⎜ x ⎟ 2πε o ρ ⎠ ⎝ (1x10 C ) −9 E1 = ( −9 2π 10 F 36π m So ETOT = 18 ay V/m Fig P2.16a ) ( 2m ) (a x + a y ) FV V V = ( a x + a y ) , and E = ( -a x + a y ) C m m Fig P2.16b 2-39 ρLa zdz ∫ 2ε o a + z ( ) 2 Letting u = a + z , du = 2z dz, we have ρ a −3 ρ a Vho = − L ∫ u du = L 4ε o 2ε o u Replacing u and evaluating from to h, ρ a⎡ 1⎤ Vho = L ⎢ − ⎥ 2ε o ⎣ a + h a ⎦ h Vho = − ∫ EidL = − = −36π V = −113V Fig P2.49 10 Conductivity and Current P2.50: A columnular beam of electrons from ≤ ρ ≤ mm has a charge density ρv =-0.1 cos(πρ/2) nC/mm3 (where ρ is in mm) and a velocity of x 106 m/sec in the +az direction Find the current ⎛ πρ ⎞ Let’s let ρ v = ρ o cos ⎜ ⎟ , where ρo = -0.1 nC/mm Then we’ll let u = uoaz, where uo = ⎝ ⎠ 6x10 mm/s Notice we convert the units to mm Now, ⎛ πρ ⎞ J = ρ v u = ρ ouo cos ⎜ ⎟ az , ⎝ ⎠ and with dS = ρ dρ dφ az we then have a 2π ⎛ πρ ⎞ I = ∫ J idS = ρ ouo ∫ ρ cos ⎜ ⎟ d ρ ∫ dφ ⎝ ⎠ 0 This becomes a ⎛ πρ ⎞ I = A∫ ρ cos ⎜ ⎟ d ρ , where A = 2πρouo ⎝ ⎠ Now we can integrate by parts, or ∫ udv = uv − ∫ vdu, where u = Aρ, du = Adρ, ⎛ πρ ⎞ ⎛ πρ ⎞ sin ⎜ ⎟ , and dv = cos ⎜ ⎟ d ρ π ⎝ ⎠ ⎝ ⎠ We then have Aa ⎛ π a ⎞ 4A ⎡ ⎛ π a ⎞ ⎤ I= sin ⎜ ⎟ + ⎢ cos ⎜ ⎟ − 1⎥ π ⎝ ⎠ π ⎣ ⎝ ⎠ ⎦ To evaluate, we first find A = 2π(-0.1x10-9)(6x109)=3.77, and then I = 2.40-1.53=0.87A v= I = 0.87A 2-40 P2.51: Two spherical conductive shells of radius a and b (b > a) are separated by a material with conductivity σ Find an expression for the resistance between the two spheres First find E for a < r < b, assuming +Q at r = a and –Q at r = b From Gauss’s law: Q E= ar 4πε o r Now find Vab: a a Q a r idra r Vab = − ∫ EidL = − ∫ 4πε o r b b −Q dr Q Q ⎛1 1⎞ = = = ⎜ − ⎟ ∫ 4πε o b r 4πε o r b 4πε o ⎝ a b ⎠ Now can find I: Q I = ∫ J idS = σ ∫ EidS = σ ∫ a r ir sin θ dθ dφa r 4πε o r a a = 2π σQ π σQ sin θ d θ dφ = ∫ ∫ 4πε o ε o Finally, R = Vab = 4πσ I ⎛1 1⎞ ⎜ − ⎟ ⎝a b⎠ P2.52: The typical length of each piece of jumper wire on a student’s protoboard is 5.0 cm Assuming AWG-20 (wire diameter 0.812 mm) copper wire, (a) determine the resistance for this length of wire (b) Determine the power dissipated in the wire for 10 mA of current (a) R = L 0.05m = = 1.67 mΩ 5.8 x10 ( S m ) π ( 0.406 x10−3 m )2 σ πa so R = 1.7 mΩ (b) P = I R = (10 x10−3 A ) (1.7 x10−3 Ω ) = 170nW P2.53: A densely wrapped coil of AWG-22 (0.644 mm diameter) copper magnet wire is 150 m long The wire has a very thin insulative sheath Determine the resistance for this length of wire R= L 150m = = 7.94Ω σ πa 5.8 x10 S m π ( 0.322 x10−3 m )2 so R = 7.9Ω 2-41 P2.54: Determine an expression for the power dissipated per unit length in coaxial cable of inner radius a, outer radius b, and conductivity between the conductors σ if a potential difference Vab is applied ⎛b⎞ ln ⎜ ⎟ 2πσ L ⎝ a ⎠ Now for a given potential difference Vab we have V 2πσ LVab2 P 2πσ Vab2 , so = P = ab = R L ln b ln b a a From Eqn(2.84) we have R = ( ) ( ) P2.55: Find the resistance per unit length of a stainless steel pipe of inner radius 2.5 cm and outer radius 3.0 cm R= L , σ π (b − a2 ) ⎞ ⎛ ⎞⎛ R 1 1 mΩ ⎜ ⎟ = 1.05 = = ⎜ ⎟⎜ 2 2 ⎟ L σ π ( b − a ) ⎝ 1.1x10 S m ⎠ π (.030 − 025 ) m m ⎝ ⎠ so R/L = 1.0 mΩ/m so we have P2.56: A nickel wire of diameter 5.0 mm is surrounded by a 0.50 mm thick layer of silver What is the resistance per unit length for this wire? Assuming 1.0 m of this wire carries 1.0 A of current, determine the power dissipated in the nickel portion and in the silver portion of the wire We can treat this wire as two resistors in parallel We have RNi 1 Ω = = 3.4 x10−3 L 1.5 x10 π ( 2.5 x10−3 ) m 1 Ω = 1.87 x10−3 2 − − 3 L m 6.2 x10 π ⎡( x10 ) − ( 2.5 x10 ) ⎤ ⎢⎣ ⎥⎦ Rtotal RNi RAg mΩ = = 1.2 L L L m To find the power dissipated, we first find the potential difference: V = IRtotal = 1.2mV then V2 V2 = 0.42mW , PAg = = 0.77 mW PNi = RNi RAg RAg = 2-42 11 Dielectrics P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m Determine the electric flux density ( D = ε E = 194.5 x10−12 F m a )(12V m ) FVC = 2.3 nC m x P2.58: MATLAB: A 20 nC point charge at the origin is embedded in Teflon (εr = 2.1) Find and plot the magnitudes of the polarization vector, the electric field intensity and the electric flux density at a radial distance from 0.1 cm out to 10 cm We use the following equations: Q E= , P = χ eε o E , D = ε r ε o E 4πε r ε o r % % % % % % % % % % % % % M-File: MLP0258 Plot E, P and D vs distance r from a point charge Q at the origin with a dielectric Variables Q charge (C) eo free space permittivity (F/m) r radial distance (m) Chi electric susceptibility E electric field intensity(V/m) D electric flux density (C/m^2) P polarization vector (C/m^2) % initialize variables Q=20e-9; er=2.1; eo=8.854e-12; Chi=er-1; % perform calculations r=0.001:.001:0.100; rcm=r.*100; E=Q./(4*pi*r.^2); P=Chi*eo*E; D=er*eo*E; % plot data 2-43 subplot(2,1,1) loglog(rcm,P,' k',rcm,D,'-k') legend('P','D') ylabel('C/m^2') grid on subplot(2,1,2) loglog(rcm,E) ylabel('V/m') xlabel('radial distance (cm)') grid on Fig P2.58 P2.59: Suppose the force is very carefully measured between a pair of point charges separated by a dielectric material and is found to be 20 nN The dielectric material is removed without changing the position of the point charges, and the force has increased to 100 nN What is the relative permittivity of the dielectric? F1 = Q1Q2 QQ , F2 = 2 , 4πε r ε o R 4πε o R F2 100 = εr = =5 20 F1 P2.60: The potential field in a material with εr = 10.2 is V = 12 xy2 (V) Find E, P and D E = −∇V = − ∂ (12 xy ) ∂x ax − ∂ (12 xy ) ∂y nC D = ε r ε o E = -1.1 y 2a x − 2.2 xya y m χ e = ε r − = 9.2 a y = −12 y 2a x − 24 xya y P = χ eε o E = ( 9.2 ) ( 8.854 x10−12 ) E = -9.8 y 2a x − 2.00 xya y V m nC m2 P2.61: In a mineral oil dielectric, with breakdown voltage of 15 MV/m, the potential function is V = x3 – 6x2 –3.1x (MV) Is the dielectric likely to breakdown, and if so, where? 2-44 E = −∇V = ( −3 x + 12 x + 3.1) a x MV m dE d 2E = −6 x + 12, = −6, so from 6x – 12 = we find the maximum electric field dx dx occurs at x = 2m At x = 2m, we have E = -12+24+3.1 = 15.1 MV/m, exceeding the breakdown voltage 12 Boundary Conditions P2.62: For y < 0, εr1 = 4.0 and E1 = 3ax + 6πay + 4az V/m At y = 0, ρs = 0.25 nC/m2 If εr2 = 5.0 for y > 0, find E2 (g) E2 = 3ax + 20.7ay + 4az V/m E1 = 3ax + 6πay + 4az V/m (a) EN1 = 6πay (f) EN2 = DN2/5εo = 20.7ay (b) ET1 = 3ax + 4az (c) ET2 = ET1 = 3ax + 4az (e) DN2 = 0.92 ay (d) DN1 = εr1εoEN1 = 24πεo ay (e) a 21 i( D1 − D2 ) = ρ s , -a y i( DN − DN ) a y = ρ s , DN − DN = ρ s DN = ρ s + DN = 0.25 ⎛ 10−9 F ⎞ nC nC nC 24 + ⎜ ⎟ π = 0.92 2 m m ⎝ 36π m ⎠ m P2.63: For z ≤ 0, εr1 = 9.0 and for z > 0, εr2 = 4.0 If E1 makes a 30° angle with a normal to the surface, what angle does E2 make with a normal to the surface? Refer to Figure P2.63 ET = E1 sin θ1 , ET = E2 sin θ , and ET = ET also DN = ε r1ε o E1 cos θ1 , DN = ε r 2ε o E2 cos θ , and DN = DN ( since ρ s = ) Therefore ⎛ε ⎞ ET ET = , and after routine math we find θ = tan −1 ⎜ r tan θ1 ⎟ DN DN ⎝ ε r1 ⎠ Using this formula we obtain for this problem θ2 = 14° Fig P2.63 2-45 P2.64: A plane defined by 3x + 2y + z = separates two dielectrics The first dielectric, on the side of the plane containing the origin, has εr1 = 3.0 and E1 = 4.0az V/m The other dielectric has εr2 = 6.0 Find E2 We first use gradient to find a normal to the planar surface Let F = 3x + 2y + z – = ∇F = 3a x + 2a y + a z , and ∇F = 14, ∇F = 0.802a x + 0.534a y + 0.267a z ∇F Now we can work the boundary condition problem E1 = 4a z , E N1 = ( E1 ia N ) a N = 0.857a x + 0.570a y + 0.285a z so a N = ET = E1 − E N = −0.857a x − 0.570a y + 3.715a z , ET = ET D N = ε r1ε o E N = ε o ⎡⎣ 2.571a x + 1.710a y + 0.855a z ⎤⎦ , and D N = D N EN = DN ε r 2ε o = DN = 0.429a x + 0.285a y + 0.143a z 6ε o Finally we have E2 = ET + E N = −0.43a x − 0.29a y + 3.8a z V m P2.65: MATLAB: Consider a dielectric-dielectric charge free boundary at the plane z = Construct a program that will allow the user to enter εr1 (for z < 0), εr2, and E1, and will then calculate E2 (Just for fun, you may want to have the program calculate the angles that E1 and E2 make with a normal to the surface) % M-File: MLP0265 % % Given E1 at boundary between a pair of % dielectrics with no charge at boundary, % calculate E2 Also calculates angles % clc clear % enter variables disp('enter vector quantities in brackets,') disp('for example: [1 3]') er1=input('relative permittivity in material 1: '); er2=input('relative permittivity in material 2: '); a12=input('unit vector from mtrl to mtrl 2: '); E1=input('electric field intensity vector in mtrl 1: '); % perform calculations 2-46 En1=dot(E1,a12)*a12; Et1=E1-En1; Et2=Et1; Dn1=er1*En1; %ignores eo since it will factor out Dn2=Dn1; En2=Dn2/er2; E2=Et2+En2 % calculate the angles th1=atan(magvector(Et1)/magvector(En1)); th2=atan(magvector(Et2)/magvector(En2)); th1r=th1*180/pi th2r=th2*180/pi Now run the program: enter vector quantities in brackets, for example: [1 3] relative permittivity in material 1: relative permittivity in material 2: unit vector from mtrl to mtrl 2: [0 1] electric field intensity vector in mtrl 1: [3 5] E2 = th1r = 45 th2r = 68.1986 P2.66: A 1.0 cm diameter conductor is sheathed with a 0.50 cm thickness of Teflon and then a 2.0 cm (inner) diameter outer conductor (a) Use Laplace’s equations to find an expression for the potential as a function of ρ in the dielectric (b) Find E as a function of ρ (c) What is the maximum potential difference that can be applied across this coaxial cable without breaking down the dielectric? (a) Since V is only a function of ρ, 2-47 ∇ 2Vcyl = ∂ ⎛ ∂V ρ ρ ∂ρ ⎜⎝ ∂ρ ⎞ ⎟=0 ⎠ ∂V =A ∂ρ or V = A ln ρ + B where A and B are constants Now we apply boundary conditions BC1: = A ln b + B, B = − A ln b, so ρ ⎛ρ⎞ ∴V = A ln ⎜ ⎟ ⎝b⎠ Va ⎛a⎞ BC2: Va = A ln ⎜ ⎟ , A = ⎝b⎠ ln a ( b) , V = Va Fig P2.66 ( b) ln ρ ( b) ln a or V = −1.443Va ln (100 ρ ) (b) E = −∇V = − 1.443Va ∂V aρ = aρ ρ ∂ρ (c) 1.443Va = 288.5Va = Ebr = 60 x106 , 005 60 x106 so Va = = 208kV , ∴ (Vab )max = 210kV 288.5 Emax = P2.67: A 1.0 m long carbon pipe of inner diameter 3.0 cm and outer diameter 5.0 cm is cut in half lengthwise Determine the resistance between the inner surface and the outer surface of one of the half sections of pipe One approach is to consider the resistance for the half-section of pipe is twice the resistance for a complete cylindrical section, given by Eqn (2.84) But we’ll used the LaPlace equation approach instead Laplace: ∇ 2Vcyl = ∂ ⎛ ∂V ρ ρ ∂ρ ⎜⎝ ∂ρ ⎞ ⎟ = ; here we see V only depends on ρ ⎠ ∂V = A; V = A ln ρ + B ; ∂ρ where A and B are constants Now apply boundary conditions BC1: So: ρ 2-48 Vb = = A ln b + B; B = − A ln b; ⎛ρ⎞ V = A ln ⎜ ⎟ ⎝b⎠ BC2: Va ⎛a⎞ Va = A ln ⎜ ⎟ ; A = ⎝b⎠ ln a V = Va ( b) ( b) ; ln ρ ( b) ln a E = −∇V = − Va ∂V aρ = − ∂ρ ln a ( b) ρ aρ Fig P2.67 J=σE I = ∫ J i dS = − R= Va I σ Va π π Lσ Va ∫ ρ ρ dφ ∫ dz = ln ( b) ln ( b ) a = 5.4 µΩ = ln a L 0 (b a) πσ L P2.68: For a coaxial cable of inner conductor radius a and outer conductor radius b and a dielectric εr in-between, assume a charge density ρ v = ρ o ρ is added in the dielectric region Use Poisson’s equation to derive an expression for V and E Calculate ρs on each plate ∇ 2V = − ρv ∂ ⎛ ∂V ⎞ − ρ o =− ρ = ε ρ ∂ρ ⎜⎝ ∂ρ ⎟⎠ ερ so ⎛ ∂V ⎞ ρo ∂ ⎛ ∂V ⎞ ρo ∂V ρ o = ρ + A , where A is a constant ⎜ρ ⎟ = ; ∫d⎜ρ ⎟ = ∫ d ρ; ρ ε ε ∂ρ ⎝ ∂ρ ⎠ ε ∂ρ ⎝ ∂ρ ⎠ ρ ρ ∂V ρ o A A = + ; dV = ∫ o d ρ + ∫ d ρ ; V = o ρ + A ln ρ + B , where B is a constant ∂ρ ε ρ ε ρ ε Now apply boundary conditions: V = Va at ρ = a and V = at ρ = b Applying the second one gives us: V= ( ) ρo ( ρ − b ) + A ln ρ b ε Applying the first one: 2-49 Va = ρo ( a − b ) + A ln a b ; A = ε ( ) Va + ρo (b − a ) ε ( b) ln a Therefore, V= ρo (b − a ) ⎛ ρ ⎞ ρ ε ln ⎜ ⎟ + o ( ρ − b ) a ⎝b⎠ ε ln Va + ( b) E = −∇V = − ρ ⎞ ∂V ∂ ⎛ ⎛ρ⎞ ρ aρ = − K ln ⎜ ⎟ + o ρ − o b ⎟ a ρ ⎜ ∂ρ ∂ρ ⎝ ε ⎠ ⎝b⎠ ε ⎛ K ρ ⎞ E = ⎜ − − o ⎟ aρ ⎝ ρ ε ⎠ where K= Va + ρo (b − a ) ε , ( b) ln a ⎡ ⎛ ρo ⎤ ⎞ ⎢ − ⎜ Va + ε ( b − a ) ⎟ ρ ⎥ ⎠ − o ⎥a so E = ⎢ ⎝ a ε ⎥ ρ ⎛ ⎞ ⎢ ρ ln ⎜ ⎟ ⎢⎣ ⎥⎦ ⎝b⎠ ⎡ ⎛ ρo ⎤ ⎞ ⎢ − ⎜ Va + ε ( b − a ) ⎟ ρ ⎥ ⎠− o⎥=ρ DN = ρ s ; DNa = ε E ρ = a = ⎢ ⎝ sa a ε ⎥ ⎛ ⎞ ⎢ a ln ⎜ ⎟ ⎢⎣ ⎥⎦ ⎝b⎠ ⎡ ⎛ ρo ⎤ ⎞ ⎢ − ⎜ Va + ε ( b − a ) ⎟ ρ ⎥ ⎠− o⎥=ρ DNb = ε E ρ =b = ⎢ ⎝ sb a ε ⎥ ⎛ ⎞ ⎢ b ln ⎜ ⎟ ⎢⎣ ⎥⎦ ⎝b⎠ P2.69: For the parallel plate capacitor given in Figure 2.51, suppose a charge density ⎛πz ⎞ ρ v = ρ o sin ⎜ ⎟ ⎝ 2d ⎠ is added between the plates Use Poisson’s equation to derive a new expression for V and E Calculate ρs on each plate 2-50 ( ) πz ∂ 2V ( z ) − ρ v − ρo sin 2d = = ε ε ∂z 2ρ d ∂V ( z ) − ρ o = sin π z dz = o cos π z +A ∫ 2d 2d ε πε ∂z 2ρ d 2ρ d V ( z ) = o ∫ cos π z dz + A∫ dz = o2 sin π z + Az + B 2d 2d πε π ε Now apply the boundary conditions: 2ρ d Vd − o2 2ρ d π ε + Ad ; A = Va = = B; Vd = o2 sin π d 2d π ε d 2ρ d ⎛ V 2ρ d ⎞ + ⎜ d − 2o ⎟ z V ( z ) = o2 sin π z d π ε π ε ⎠ ⎝d ( ( ) ) ( ( ( ( E = −∇V = − ) ) ) ) ⎡ ∂ ⎛ 2ρ d ⎞ ∂V ∂ ⎛ ⎛ V 2ρ d ⎞ ⎞⎤ a z = ⎢ − ⎜ o2 sin π z + − ⎜ ⎜ d − 2o ⎟ z ⎟ ⎥ a z ⎟ 2d ∂z ∂z ⎝ ⎝ d π ε ⎠ ⎠⎦ ⎠ ⎣ ∂z ⎝ π ε ( ) V 2ρ d ⎞ ⎛ ρd E = ⎜ − o cos π z − d + 2o ⎟ a z d d π ε ⎠ ⎝ πε ( ) at z = 0, DN = ε E , so ρ s z =0 at z = d, DN = ε E , so ρ s z =d ρo d εVd ρo d − + = ρs π π d εV ρ d = − d + o2 = ρ s π d =− 13 Capacitors P2.70: A parallel plate capacitor is constructed such that the dielectric can be easily removed With the dielectric in place, the capacitance is 48 nF With the dielectric removed, the capacitance drops to 12 nF Determine the relative permittivity of the dielectric C1 = ε rε o A d ; C2 = εo A d ; C1 48 = εr = = 4.0 C2 12 P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate separation, and a dielectric with relative permittivity of 1200 has a 12 V potential difference across the plates (a) What is the minimum allowed dielectric strength for this capacitor? Calculate (b) the capacitance, and (c) the magnitude of the charge density on one of the plates (a) E = 12V kV kV ; (a) Ebr = =6 m m 0.002m 2-51 (b) C = (c) C = ε rε o A d = (1200 ) (8.854 x10−12 F / m )(1m2 ) 0.002m = 5.3µ F Q C ; Q = CV = ( 5.3x10−6 F ) (12V ) = 64µ C V FV P2.72: A conical section of material extends from 2.0 cm ≤ r ≤ 9.0 cm for ≤ θ ≤ 30° with εr = 9.0 and σ = 0.020 S/m Conductive plates are placed at each radial end of the section Determine the resistance and capacitance of the section ∂ ⎛ ∂V ⎞ A ∂V = A; V = − + B , where A and B are constants ⎜r ⎟ = 0; r ∂r r ∂r ⎝ ∂r ⎠ r Boundary conditions: r = a, V = and r = b, V = Vb ⎛1 1⎞ ⎜ − ⎟ a r⎠ V = Vb ⎝ ⎛1 1⎞ ⎜ − ⎟ ⎝a b⎠ ∂V E = −∇V = − ar ∂r −Vba r −ε r ε oVba r E= ,D = ⎛1 1⎞ ⎛1 1⎞ ⎜ − ⎟r ⎜ − ⎟r ⎝a b⎠ ⎝a b⎠ ∇ 2V = Fig P2.72 ρ sb = −ε r ε oVb ; ⎛1 1⎞ ⎜ − ⎟b ⎝a b⎠ Q = ∫ ρ s dS ; dS = r sin θ dθ dφ 2-52 ε r ε oVb 30o 2π ⎛ 1 ⎞ ∫0 ⎜ − ⎟b a b ⎝ ⎠ −12 = 1.73x10 Vb Q ε ε C = b = 1.7 pF ; RC = ; R = = 2.3k Ω Vb Cσ σ Qb = b sin θ dθ ∫ dφ P2.73: An inhomogeneous dielectric fills a parallel plate capacitor of surface area 50 cm2 and thickness 1.0 cm You are given εr = 3(1 + z), where z is measured from the bottom plate in cm Determine the capacitance Place +Q at z = d and –Q at z = Q Q Q ρs = , D = − a z , E = − a S S ε rε o S z d d Vdo = − ∫ EidL = − ∫ 0 -Q ε rε o S d a z idza z = Q dz ε o S ∫0 ε r evaluating the integral: d d dz dz 1 ∫0 ε r = ∫0 (1 + z ) = ln (1 + z ) = ln cm −12 2 Q 3ε o S ( 8.854 x10 F m )( 50cm ) ⎛ m ⎞ C= = = ⎜ ⎟ = 19 pF Vdo ln ( ln ( ) cm ) ⎝ 100cm ⎠ P2.74: Given E = 5xyax + 3zaz V/m, find the electrostatic potential energy stored in a volume defined by ≤ x ≤ m, ≤ y ≤ m, and ≤ z ≤ m Assume ε = εo 1 WE = ε o ∫ EiEdv = ε o ⎡⎣ ∫ 25 x y dxdydz + ∫ z dxdydz ⎤⎦ 2 1 1 ⎤ ⎡ WE = ε o ⎢ 25∫ x dx ∫ y dy ∫ dz + 9∫ dx ∫ dy ∫ z dz ⎥ = 125 pJ ⎣ 0 0 0 ⎦ P2.75: Suppose a coaxial capacitor with inner radius 1.0 cm, outer radius 2.0 cm and length 1.0 m is constructed with different dielectrics When oriented along the z-axis, εr for 0° ≤ φ ≤ 180° is 9.0, and for 180° ≤ φ ≤ 360° is 4.0 (a) Calculate the capacitance (b) If 9.0 V is applied across the conductors, determine the electrostatic potential energy stored in each dielectric for this capacitor (a) a coaxial line, 2-53 2π Lε r ε o ln b a But for only half the line, π Lε r ε o C= ln b a So π Lε r1ε o = 361 pF C1 = ln b a and π Lε r 2ε o = 161 pF C2 = ln b a So Fig P2.75 CTOT = C1 + C2 = 522 pF 1 (b) WE1 = C1V = 14.6nJ ; WE = C2V = 6.5nJ 2 C= ( ) ( ) ( ) ( ) ... coordinate location of center of a differential charge element x,y,z cartesian coord location of charge % element R vector from charge element to P Rmag magnitude of R aR unit vector of R dr,dtheta,dphi... origin? (b) Suppose instead of the 8.00 nC charge of part (a) that you locate a charge Q at (0.00, 6.00m, 0.00) What value of Q will result in a total electric field intensity of zero at the origin?... thickness and an inner radius of 4.00 cm is centered on the z-axis and has an evenly distributed 3.00 C of charge per meter length of pipe Plot Dρ as a function of radial distance from the z-axis