f5ch22 - 620 CHAPTER 22 ELECTRIC CHARGE CHAPTER 22 Answer...

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Unformatted text preview: 620 CHAPTER 22 ELECTRIC CHARGE CHAPTER 22 Answer to Checkpoint Questions 1. 2. 3. 4. C and D attract; B and D attract (a) leftward; (b) leftward; (c) leftward (a) a, c, b; (b) less than 15e (net charge of 30e is equally shared) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. No, only for charged particles, charged particle-like objects all tie a and b (a) between; (b) positively charged; (c) unstable two points: one to the left of the particles and one between the protons 2q2 =40 r2 , upward 6q2 =40 d2 , leftward a and d tie, then b and c tie (a) same; (b) less than; (c) cancel; (d) add; (e) the adding components; (f) positive direction of y; (g) negative direction of y; (h) positive direction of x; (i) negative direction of x (a) positive; (b) negative (a) A, B , and D; (b) all four; (c) connect A and D, disconnect them, then connect one of them to B (there are two other solutions) (a) neutral; (b) negatively (a) possibly; (b) de nitely CHAPTER 22 ELECTRIC CHARGE 621 14. 15. 16. (a) Ground A with the wire, bring the positively charged rod near A but not touching it, remove the wire (A is now negatively charged), remove the rod, connect the wire between A and B (both are now negatively charged by the same amount), remove the wire. (b) Method 1: Connect the two spheres with the wire; bring the rod near sphere A (A now has excess negative charge and B has just as much excess positive charge); remove the wire; remove the rod. Method 2: Ground A with the wire, bring the positively charged rod near A but not touching it, remove the wire (A is now negatively charged), remove the rod, ground B with the wire, bring A near B (A is negatively charged and B is now positively charged by the same amount), remove the wire. same Once enough of the electrons have moved to the opposite end, any other conduction electron is repelled by the electrons at the opposite end as much as it is repelled by the negatively charged rod at the near end. 17. 18. D no (the person and the conductor share the charge) Solutions to Exercises & Problems The charge transferred is 1E q = it = (2:5 104 A)(20 10 6 s) = 0:50 C : 2E (a) (b) 2 :99 109 2 2 q1 F = 4q2 2 = (1:00 C) (8(1:00 m)2 N m =C ) = 8:99 109 N : r 0 2 (8:99 9N 2 2 q1 F = 4q2 2 = (1:00 C)(1:00 10 m)2 m =C ) = 8:99 103 N : r 103 0 3E The magnitude of the force is given by 1 1 F = 4 qrq2 ; 2 0 622 CHAPTER 22 ELECTRIC CHARGE where q1 and q2 are the magnitudes of the charges and r is the distance between them. Thus 2 9 2 6 6 F = (8:99 10 N m /C :)(3:00 210 2 C)(1:50 10 C) = 2:81 N : (12 0 10 m) 4E The magnitude of the force that either charge exerts on the other is given by 1 1 F = 4 qrq2 ; 2 0 where r is the distance between them. Thus r qq 1 r = 4 2F s 0 2 9 2 6 6 :0 = (8:99 10 N m /C )(2670 10 C)(47:0 10 C) = 1:38 m : 5: N (a) From Newtons's third law m1 a1 = m2 a2 , so 7 kg)(7 2 1 m2 = ma a1 = (6:3 109:0 m/s2 :0 m/s ) = 4:9 10 7 kg : 2 5E (b) Since F = q2 =40 r2 = m1 a1 , q = r 40 m1 a1 p 7 kg)(7 0 2 3 = (3:2 10 3 m) (6:8: 10 109 N m:2 =m/s ) = 7:1 10 99 C2 s 11 C: Use jFAB j = jqA qB j=(40 d2 ), etc. (a) Since qA = 2Q and qC = +8Q, 6E j( 2Q)(+8Q)j = 4Q2 : jFAC j = 4 d2 0 d2 0 After making contact with each other, both A and B now have a charge of [ 2Q + ( 4Q)]=2 = 3Q. When B is grounded its charge is zero. After making contact with CHAPTER 22 ELECTRIC CHARGE 623 C , which has a charge of +8Q, B has a charge of [0 + ( 8Q)]=2 = 4Q, so does charge C . Thus nally QA = 3Q and QB = QC = 4Q. Therefore (b) 3 2 jFAC j = j( 34Q)( d24Q)j = Qd2 ; (c) 0 0 4 2 jFBC j = j( 44Q)( d24Q)j = Qd2 : 0 0 Let the initial charge on either spheres 1 and 2 be q. After being touched by sphere 3, sphere 1 retains only half of q. After being touched by sphere 3, sphere 2 has (q + q=2)=2 = 3q=4 left. So F 0 = c(q=2)(3q=4) = (3=8)cq2 = (3=8)F; where c is a constant. Since q3 is in equilibrium F31 + F32 = 0, i.e., 1 q1 q3 + 1 q2 q3 = 0 ; 40 (2d)2 40 d2 which gives q1 = 4q2 : (a) The magnitude of the force is 6 2 1 1 0 F12 = 4 qdq2 = (8:99 109 N m2 =C2 ) (20:(1 10 2 C) = 1:60 N : 2 :50 m) 0 9E 8P 7E (b) The magnitude of the force on q1 now becomes q 2 2 F1 = F12 + F13 + 2F12 F13 cos p = F12 2 + 2 cos p = (1:60 N) 2 + 2 cos 60 = 2:77 N ; F1 F13 F 12 where we have used F12 = F13 = kq1 q2 =d2 = kq1 q3 =d2 : 10P q1 Put the origin of a coordinate system at the lower left corner of the square. Take the y 624 CHAPTER 22 ELECTRIC CHARGE axis to be vertically upward and the x axis to be horizontal. The force exerted by the charge +q on the charge 2q is F1 y F2 -2q 1 = 4 q(22q) ( j) ; 0 a the force by the charge q to 2q is 1 (2p)( q) q F2 = 40 ( 2a)2 F3 F1 x i + j 1 q2 i j p = 4 a2 p + p ; 2 2 2 0 1 1 2 = 4 (2q)(2 2q) ( i) = 4 4aq2 i : a 0 0 and the force by the charge +2q to 2q is F3 Thus the horizontal (x) component of the resultant force on the charge 2q is 1 q2 1 Fx = F1x + F2x + F3x = 4 a2 p + 4 2 0 7 C)2 1 9 2 2 (1:0 10 = (8:99 10 N m =C ) (5:0 10 2 m)2 p + 4 = 0:17 N; 2 while the vertical (y) component is 1 q2 Fy = F1y + F2y + F3y = 4 a2 0 1 2 + p = 0:046 N : 2 (a) Let the force F on Q be 11P Q 1 F = 4 ( a q1 Q 2)2 + (a q2a=2)2 = 0 a= 0 and solve for q1 : q1 = 9q2 . (b) Now qQ q2 Q 1 1 F = 4 ( a 3a=2)2 + (a 3a=2)2 = 0 ; 0 which gives q1 = 25q2 : CHAPTER 22 ELECTRIC CHARGE 625 Solve for q1 and q2 : q1 , q2 = 1:2 10 5 C and 3:8 10 5 C. 13P Let the two charges be q1 and q2 . Then q1 + q2 = 5:0 10 5 C. Also F12 = q1 q2 =40 r2 , i.e., 109 2 2 1:0 N = q1 q2 (8:99(2:0 m)N m =C ) : 2 12P The negative sign indicates that the spheres attract each other. After the wire is connected the spheres, being identical, have the same charge. Since charge is conserved the total charge is the same as it was originally. This means the charge on each sphere is (q1 + q2 )=2. The force is now one of repulsion and is given by + 2 1 Fb = 4 (q1 4r2q2 ) : 0 Solve the two force equations simultaneously for q1 and q2 . The rst gives 2 q1 q2 = 40 r2 Fa = (0:500 m)9 (0:108 N)2 = 3:00 10 8:99 10 N m2 /C 12 Assume the charge distributions are spherically symmetric so Coulomb's law can be used. Let q1 and q2 be the original charges and choose the coordinate system so the force on q2 is positive if it is repelled by q1 . Take the distance between the charges to be r. Then the force on q2 is 1 1 Fa = 4 qrq2 : 2 0 C2 and the second gives q1 + q2 = r 4(40 )Fb = (0:500 m) Thus and p s 4(0:0360 N) = 2:00 10 6 C : 8:99 109 N m2 /C2 12 q2 = (3:00 q10 1 C2 ) Multiply by q1 to obtain the quadratic equation 2 q1 (2:00 10 6 C)q1 3:00 10 12 10 q1 3:00 q 1 12 C2 = 2:00 10 6 C : C2 = 0 : 626 CHAPTER 22 ELECTRIC CHARGE The solutions are 6 2:00 10 C (2:00 10 6 C)2 + 4(3:00 10 12 C2 ) q1 = : 2 If the plus sign is used q1 = 1:00 10 6 C and if the minus sign is used q1 = 3:00 10 6 C. Use q2 = ( 3:00 10 12 )=q1 to calculate q2 . If q1 = 1:00 10 6 C then q2 = 3:00 10 6 C and if q1 = 3:00 10 6 C then q2 = 1:00 10 6 C. Since the spheres are identical the solutions are essentially the same: one sphere originally had charge 1:00 10 6 C and the other had charge 3:00 10 6 C. Another solution exists. If the signs of the charges are reversed the forces remain the same, so a charge of 1:00 10 6 C on one sphere and a charge of 3:00 10 6 C on the other also satis es the conditions of the problem. Let the third charge q3 be located on the line joining the two charges, a distance x from q1 = 1:0C. Then the net force exerted on q3 is 14P q q3 q1 + q2 F3 = 4 x2 (r x)2 = 0 : 0 Substitute q1 = +1:0C, q2 = 3:0C and r = 10 cm into the above equation and solve for x to obtain x = 14 cm. (a) 15P 1 1 qq F21 = 4 jqr12q2 j = 4 (x x )j2 1+2(jy y )2 0 0 2 1 2 1 12 9 N m2 =C2 )j(3:0 10 6 C)( 4:0 10 6 C)j 10 = (8:99 [( 2:0 3:5)2 + (1:5 0:5)2 ](10 4 m2 ) = 36 N : The direction of F21 is such that it makes an angle with the positive x axis, where y1 = tan 1 1:5 cm 0:5 cm = 10 : 1 y2 = tan x x 2:0 cm 3:5 cm 2 1 (b) Let the third charge be located at (x3 ; y3 ), a distance r from q2 . Since q1 , q2 and q3 must be on the same line, we have x3 = x2 r cos and y3 = y2 r sin . The force on q2 exerted by q3 is F23 = q2 q3 =40 r2 , which must cancel with the force exerted by q1 on q2 : q2 F23 = 4jq3 j2 = F21 ; 0r CHAPTER 22 ELECTRIC CHARGE 627 which gives So x3 = x2 r cos = 2:0 cm (6:3 cm) cos( 10 ) = 8:3 cm and y3 = y2 r sin = 1:5 cm (6:3 cm) sin( 10 ) = 2:6 cm: (a) If the system of three charges is to be in equilibrium the force on each charge must be zero. Let the third charge be q0 . It must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q0 could not be in equilibrium. Suppose q0 is a distance x from q, as shown on the diagram to the right. The force acting on q0 is then given by 16P 6 6 C)(8 9 2 2 jq 3 r = 42 qFj = (4:0 10 C)(4:0 1036 N :99 10 N m =C ) = 6:3 cm : 0 12 s r x q q0 L-x 4q 1 0 F0 = 4 qq20 (L4qqx)2 = 0 ; x 0 where the positive direction was taken to be toward the right. Solve this equation for x. Canceling common factors yields 1=x2 = 4=(L x)2 and taking the square root yields 1=x = 2=(L x). The solution is x = L=3. The force on q is 1 qq0 + 4q2 = 0 : Fq = 4 x2 L2 Solve for q0 : q0 = 4qx2 =L2 = (4=9)q, where x = L=3 was used. The force on 4q is 1 4q2 + 4qq0 = 1 4q2 + 4( 4=9)q2 F4q = 4 L2 (L x)2 4 L2 (4=9)L2 0 0 1 4q2 4q2 = 0 : = 4 L2 L2 0 With q0 = (4=9)q and x = L=3 all three charges are in equilibrium. (b) If q0 moves toward q the force of attraction exerted by q is greater in magnitude than the force of attraction exerted by 4q and q0 continues to move toward q and away from its initial position. The equilibrium is unstable. (a) The magnitudes of the gravitational and electrical forces must be the same: 1 q2 = G Mm Me ; 40 d2 d2 em em 17P 0 628 CHAPTER 22 ELECTRIC CHARGE where q is the charge on either body, dem is the center-to-center separation of the Earth and Moon, G is the universal gravitational constant, Me is the mass of the Earth, and Mm is the mass of the Moon. Solve for q: p q = 40 GMm Me : According the text, Me = 5:98 1024 kg, and Mm = 7:36 1022 kg, so q= s (6:67 10 11 = 5:7 1013 C : N m2 /kg2 )(7:36 1022 kg)(5:98 1024 kg) 8:99 109 N m2 /C2 Notice that the distance r cancels because both the electric and gravitational forces are proportional to 1=r2 . (b) The charge on a hydrogen ion is e = 1:60 10 19 C so there must be q = 5:7 1013 C = 3:6 1032 ions : e 1:6 10 19 C Each ion has a mass of 1:67 10 27 kg so the total mass needed is 27 (3:6 1032 )(1:67 10 kg) = 6:0 105 kg : 18P The magnitude of the force of either of the charges on the other is given by 1 F = 4 q(Qr2 q) ; 0 where r is the distance between the charges. You want the value of q that maximizes the function f (q) = q(Q q). Set the derivative df=dq equal to zero. This yields 2q Q = 0, or q = Q=2. (a) Let the side length of the square be a. Then the magnitude of the force on each charge 1 FQ = 4 0 19P Q is p2Qq Q2 : a2 + (p2a)2 CHAPTER 22 ELECTRIC CHARGE 629 Let FQ = 0, we get Q = 2 2q. p p (b) For both FQ and Fq to vanish we would require Q = 2 2q and q = 2 2Q. These two equations cannot be simultaneously valid unless Q = q = 0. So this is impossible. (a) A force diagram for one of the balls is shown to the right. The force of gravity mg acts downward, the electrical force Fe of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle to the vertical. The ball is in equilibrium, so its acceleration is zero. The y component of Newtons second law yields T cos mg = 0 and the x component yields T sin Fe = 0. Solve the rst equation for T (T = mg= cos ) and substitute the result into the second to obtain mg tan Fe = 0. Now 20P p y T Fe mg x When these two substitutions are made in the equation mg tan = Fe , it becomes so tan = p 2 x=2 2 : L (x=2) If L is much larger than x we may neglect x=2 in the denominator and write tan x=2L. This is equivalent to approximating tan by sin . The magnitude of the electrical force of one ball on the other is q2 Fe = 4 x2 : 0 mgx = 1 q2 ; 2L 40 x2 1 q2 x3 = 4 2mgL ; 0 and q2 L 1=3 : x = 2 mg 0 (b) Solve x3 = (1=40 )(2q2 L=mg) for q: s r 40 mgx3 = (0:010 kg)(9:8 m/s2 )(0:050 m)3 = 2:4 10 8 C : q= 2L 2(8:99 109 N m2 /C2 )(1:20 m) 21P If one of them is discharged, there would no electrostatic repulsion between the two balls and they would both come to the position = 0, making contact with each other. A 630 CHAPTER 22 ELECTRIC CHARGE redistribution of charges would then occur, with each of the balls getting q=2. They would then again be separated due to electrostatic repulsion, which results in the new separation 1 1=3 1 1=3 2 1=3 0 = (q=2) L = x= (5:0 cm) = 3:1 cm : x 20 mg 4 4 (a) Since the rod is in equilibrium the net force acting on it is zero and the net torque about any point is also zero. Write an expression for the net torque about the bearing, equate it to zero, and solve for x. The charge Q on the left exerts an upward force of magnitude (1=40 )(qQ=h2 ), at a distance L=2 from the bearing. Take the torque to be positive. The attached weight exerts a downward force of magnitude W , at a distance L=2 x from the bearing. This torque is positive. The charge Q on the right exerts an upward force of magnitude (1=40 )(2qQ=h2 ), at a distance L=2 from the bearing. This torque is negative. The equation for rotational equilibrium is 1 qQ L + W L x 40 h2 2 2 The solution for x is 22P 1 2qQ L = 0 : 40 h2 2 L 1 + 1 qQ : x= 2 40 h2 W (b) The net force on the rod vanishes. If N is the magnitude of the upward force exerted by the bearing then 1 qQ 1 W 4 qQ 4 2h2 N = 0 : 2 0 h 0 Solve for h so that N = 0. The result is r 1 qQ h = 4 3W : 0 The magnitude of the force is 1 e2 = (8:99 109 N m2 =C2 ) (1:60 10 F = 4 r2 (2:82 10 0 24E 23E C)2 = 2:89 10 9 N : 10 m)2 19 9N 2= 2 q2 e= 2 F = 4 r2 = (4 3)2 = (8:99 10 9(2:m C )(1:60 10 6 10 15 m)2 0 0r 19 C)2 = 3:8 N : CHAPTER 22 ELECTRIC CHARGE 631 The mass of an electron is me = 9:11 10 with total mass M = 75:0 kg is 25E 31 kg so the number of electrons in a collection M 0 N = m = 9:11 75:10kg31 kg = 8:23 1031 electrons : e The total charge of the collection is q = Ne = (8:23 1031 )(1:60 10 1013 C : 26E 19 C) = 1:32 Q = NA q = (6:02 1023 )(2)(1:60 10 19 C) = 1:9 105 C = 0:19 MC : 27E (a) The magnitude of the force between the ions is given by q2 F = 4 r2 ; 0 where q is the charge on either of them and r is the distance between them. Solve for the charge: q = r 40 F = (5:0 10 p 10 m) s 3:7 10 9 N = 3:2 10 8:99 109 N m2 /C2 19 C: (b) Let N be the number of electrons missing from each ion. Then Ne = q, or 19 2 N = q = 13::60 10 19C = 2 : e 10 C (a) the number of electrons to be removed is 7 :0 n = q = 1160 10 19CC = 6:3 1011 : e : 10 28E (b) The fraction is 11 n 6 frac = NZ = (2:95:3 10 )(29) = 7:4 10 1022 13 : 632 CHAPTER 22 ELECTRIC CHARGE 29E (a) The magnitude of the force is given by q2 F = 4 r2 ; 0 where q is the magnitude of the charge on each drop and r is the center-to-center separation of the drops. Thus (8:99 109 N m2 /C2 )(1:00 10 F= (1:00 10 2 m)2 16 C)2 = 8:99 10 19 N: (b) If N is the number of excess electrons (of charge e each) on each drop then N= q= e Let q2 =40 r2 = mg, we get 30E 1:00 10 16 C = 625 : 1:60 10 19 C r = q 4 mg = (1:60 10 0 31E r 1 19 8 9 N m2 2 C) (1:67:99 10 10 kg)(9:=C 2 ) = 0:119 m : 27 8 m/s s The second electron must be placed a distance d under the rst one to produce an upward force that balances the weight of the rst one. So 1 e2 W = me g = 4 d2 ; 0 which gives r 1 d = e 4 m1 g = (1:6 10 0 e 19 8 99 9 N m2 = 2 C) (9:11 : 10 10 kg)(9:80C 2 ) = 5:1 m : 31 m/s s The current intercepted would be q 2 i = (4R ) t = 4(6:37 106 m)2 (1500= s)(1:6 10 32P 19 C) = 0:122 A : CHAPTER 22 ELECTRIC CHARGE 633 If charge of magnitude q passes through the lamp in time t the current is i = q=t. The charge is the total charge on one mole of electrons, or q = NA e, where NA is the Avogadro constant. Thus i = NA e=t, or 33P NA e = (6:02 1023 )(1:60 10 t= i 0:83 A This is equivalent to 1:3 da. 34P 19 C) = 1:2 105 s : The numer of moles of H2 O molecules in a glass of water (of volume V ) is n = V=M , where is the density of water and M is its molar mass. Each H2 O molecule has 10 protons. Thus Q = 10enNA = 10eV NA M 10(1:6 10 19 C)(1:00 103 kg/m3 )(250 10 6 m3 )(6:02 1023 = mol) = 1:8 10 2 kg/mol = 1:3 107 C : 35P (a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine ion at the cube center. Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube. Since the two ions in such a pair exerts forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero. (b) Rather than remove a cesium ion, superpose charge e at the position of one cesium ion. This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion. The forces of the 8 cesium ions at the cube corners sum to zero, so the only force on the chlorine p is the force of the added charge. ion length of a cube edge. The length of a body diagonal of a cube is 3a, where a is the p Thus the distance from the center of the cube to a corner is d = ( 3=2)a. The force has magnitude 2 1 1 e2 F = 4 d2 = 4 (3=e4)a2 0 0 9 N m2 /C2 )(1:60 10 = (8:99 10 =4)(0:40 10 9 m)2 (3 19 C)2 = 1:9 10 9 N : 634 CHAPTER 22 ELECTRIC CHARGE Since both the added charge and the chlorine ion are negative the force is one of repulsion. The chlorine ion is pushed away from the site of the missing cesium ion. In this case the net charge on each penny would be qnet = q, where q = 137000 C and = 0:00010%. The force is then 2 qnet = [(0:00010%)(137000 C)]2 (8:99 109 N m2 =C2 ) = 1:7 108 N : F = 4 r2 (1:0 m)2 0 Therefore the magnitudes of the positive charge on the proton and negative charge on the electron cannot possibly dier by as much as 0:00010%. The net charge carried by John whose mass is m is roughly 37P 36P q = (0:01%) mNA Ze M = 9 105 C ; : 1023 = mol)(18)(1:6 10 = (0:01%) (200 lb)(0:4536 kg/lb)(6002 kg/mol :018 19 C) and the net charge carried by Mary is half of that. So the electrostatic force between them is estimated to be 105 C)2 1 F 4 q(q=2) = (8:99 109 N m2 =C2 ) (92(30 m)2 = 4 1018 N : 2 0 d Apply the principle of charge conservation. The results are: (a) an electron; (b) a proton. None of the reactions given include a beta decay, so the number of protons, the number of neutrons, and the number of electrons are each conserved. Atomic numbers (numbers of protons and numbers of electrons) and molar masses (combined numbers of protons and neutrons) can be found in Appendix F of the text. (a) 1 H has 1 proton, 1 electron, and 0 neutrons and 9 Be has 4 protons, 4 electrons, and 9 4 = 5 neutrons, so X has 1 + 4 = 5 protons, 1 + 4 = 5 electrons, and 0 + 5 1 = 4 neutrons. One of the neutrons is freed in the reaction. X must be boron with a molar mass of 5 + 4 = 9 g= mol: 9 B. 39E 38E CHAPTER 22 ELECTRIC CHARGE 635 (b) 12 C has 6 protons, 6 electrons, and 12 6 = 6 neutrons and 1 H has 1 proton, 1 electron, and 0 neutrons, so X has 6 + 1 = 7 protons, 6 + 1 = 7 electrons, and 6 + 0 = 6 neutrons. It must be nitrogen with a molar mass of 7 + 6 = 13 g=mol: 13 N. (c) 15 N has 7 protons, 7 electrons, and 15 7 = 8 neutrons; 1 H has 1 proton, 1 electron, and 0 neutrons; and 4 He has 2 protons, 2 electrons, and 4 2 = 2 neutrons; so X has 7 + 1 2 = 6 protons, 6 electrons, and 8 + 0 2 = 6 neutrons. It must be carbon with a molar mass of 6 + 6 = 12: 12 C. (a) (b) 40E 9 2= 2 :60 1 2 F = 4 q2 = (8:99 10 N(9m C )[2(1m)2 10 15 :0 10 0r 19 C)]2 = 11 N : F 11 N a = m = 4(1:60 10 27 = 1:7 1027 m/s2 : kg) (a) F = (Q2 =40 d2 )(1 ); (c) 0.5; (d) 0.15 and 0.85 (a) Let J = qQ=40 d2 . For < 0, F = J [1=2 + 1=(1 + jj)2 ]; for 0 < < 1, F = J [1=2 1=(1 )2 ]; for 1 < , F = J [1=2 + 1=( 1)2 ] 42 41 ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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