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### Transcript of Ch24 fundamental physics solution problem

1. 1. 1. (a) An Ampere is a Coulomb per second, so 84 84 3600 30 105 A h C h s s h C = F HG I KJF HG I KJ = . . (b) The change in potential energy is U = qV = (3.0 105 C)(12 V) = 3.6 106 J.
2. 2. 2. The magnitude is U = eV = 1.2 109 eV = 1.2 GeV.
3. 3. 3. The electric field produced by an infinite sheet of charge has magnitude E = /20, where is the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is V V E dx V Exs x s= = z0 , where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by x then their potentials differ in magnitude by V = Ex = (/20)x. Thus, x V = = = 2 2 885 10 50 010 10 88 100 12 2 6 3 . . . . C N m V C m m 2 2 c hb g
4. 4. 4. (a) VB VA = U/q = W/(e) = (3.94 1019 J)/(1.60 1019 C) = 2.46 V. (b) VC VA = VB VA = 2.46 V. (c) VC VB = 0 (Since C and B are on the same equipotential line).
5. 5. 5. (a) E F e= = = 39 10 160 10 2 4 1015 19 4 . . . .N C N Cc h c h (b) V E s= = = 2 4 10 012 2 9 104 3 . . . .N C m Vc hb g
6. 6. 6. (a) By Eq. 24-18, the change in potential is the negative of the area under the curve. Thus, using the area-of-a-triangle formula, we have V E ds x = = = z10 1 2 2 20 0 2 b gb g which yields V = 30 V. (b) For any region within 0 3< < zx E dsm, is positive, but for any region for which x > 3 m it is negative. Therefore, V = Vmax occurs at x = 3 m. V E ds x = = = z10 1 2 3 20 0 3 b gb g which yields Vmax = 40 V. (c) In view of our result in part (b), we see that now (to find V = 0) we are looking for some X > 3 m such that the area from x = 3 m to x = X is 40 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X), we require 1 2 1 20 4 20 40b gb g b gb g+ =X . Therefore, X = 5.5 m.
7. 7. 7. (a) The work done by the electric field is (in SI units) 19 12 210 0 0 120 0 0 (1.60 10 )(5.80 10 )(0.0356) 1.87 10 J. 2 2 2(8.85 10 ) f d i q q d W q E ds dz = = = = = (b) Since V V0 = W/q0 = z/20, with V0 set to be zero on the sheet, the electric potential at P is (in SI units) 12 2 12 0 (5.80 10 )(0.0356) 1.17 10 V. 2 2(8.85 10 ) z V = = =
8. 8. 8. We connect A to the origin with a line along the y axis, along which there is no change of potential (Eq. 24-18: E ds =z 0). Then, we connect the origin to B with a line along the x axis, along which the change in potential is V E ds x dx x = = = F HG I KJzz= 4 00 4 00 4 2 2 0 4 0 4 . . which yields VB VA = 32.0 V.
9. 9. 9. (a) The potential as a function of r is (in SI units) ( ) ( ) ( ) 2 3 30 0 0 0 9 15 2 4 3 0 0 4 8 (8.99 10 )(3.50 10 )(0.0145) 2.68 10 V. 2(0.0231) r r qr qr V r V E r dr dr R R = = = = = (b) Since V = V(0) V(R) = q/80R, we have (in SI units) ( ) 9 15 4 0 (8.99 10 )(3.50 10 ) 6.81 10 V. 8 2(0.0231) q V R R = = =
10. 10. 10. The charge is 9 0 9 2 2 m /C (10m)( 1.0V) 4 1.1 10 C. 8.99 10 N q RV = = =
11. 11. 11. (a) The charge on the sphere is q VR= = = 4 200 8 99 10 33 100 9 9 ( . . V)(0.15 m) N m C C.2 2 (b) The (uniform) surface charge density (charge divided by the area of the sphere) is = = = q R4 33 10 4 015 12 102 9 2 8 . . . . C m C / m2 b g
12. 12. 12. (a) The potential difference is ( ) 2 6 9 2 0 0 3 N m 1 1 1.0 10 C 8.99 10 4 4 C 2.0 m 1.0 m 4.5 10 V. A B A B q q V V r r = = = (b) Since V(r) depends only on the magnitude of r , the result is unchanged.
13. 13. 13. First, we observe that V (x) cannot be equal to zero for x > d. In fact V (x) is always negative for x > d. Now we consider the two remaining regions on the x axis: x < 0 and 0 < x < d. (a) For 0 < x < d we have d1 = x and d2 = d x. Let V x k q d q d q x d x ( ) = + F HG I KJ = + F HG I KJ =1 1 2 2 04 1 3 0 and solve: x = d/4. With d = 24.0 cm, we have x = 6.00 cm. (b) Similarly, for x < 0 the separation between q1 and a point on the x axis whose coordinate is x is given by d1 = x; while the corresponding separation for q2 is d2 = d x. We set V x k q d q d q x d x ( ) = + F HG I KJ = + F HG I KJ =1 1 2 2 04 1 3 0 to obtain x = d/2. With d = 24.0 cm, we have x = 12.0 cm.
14. 14. 14. Since according to the problem statement there is a point in between the two charges on the x axis where the net electric field is zero, the fields at that point due to q1 and q2 must be directed opposite to each other. This means that q1 and q2 must have the same sign (i.e., either both are positive or both negative). Thus, the potentials due to either of them must be of the same sign. Therefore, the net electric potential cannot possibly be zero anywhere except at infinity.
15. 15. 15. A charge 5q is a distance 2d from P, a charge 5q is a distance d from P, and two charges +5q are each a distance d from P, so the electric potential at P is (in SI units) 9 15 4 2 0 0 1 1 1 1 (8.99 10 )(5.00 10 ) 5.62 10 V. 4 2 8 2(4.00 10 ) q q V d d d d d = + + = = = The zero of the electric potential was taken to be at infinity.
16. 16. 16. In applying Eq. 24-27, we are assuming V 0 as r . All corner particles are equidistant from the center, and since their total charge is 2q1 3q1+ 2 q1 q1 = 0, then their contribution to Eq. 24-27 vanishes. The net potential is due, then, to the two +4q2 particles, each of which is a distance of a/2 from the center. In SI units, it is 9 12 2 2 2 0 0 0 4 4 161 1 16(8.99 10 )(6.00 10 ) 2.21 V. 4 / 2 4 / 2 4 0.39 q q q V a a a = + = = =
17. 17. 17. (a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius R of the drop are related by V = q/40R. Thus R q V = = = 4 8 99 10 30 10 500 54 10 0 9 12 4 . / . N m C C V m. 2 2 c hc h (b) After the drops combine the total volume is twice the volume of an original drop, so the radius R' of the combined drop is given by (R')3 = 2R3 and R' = 21/3 R. The charge is twice the charge of original drop: q' = 2q. Thus, = = = = V q R q R V 1 4 1 4 2 2 2 2 500 790 0 0 1 3 2 3 2 3 / / / ( V) V.
18. 18. 18. When the charge q2 is infinitely far away, the potential at the origin is due only to the charge q1 : V1 = 1 04 q d = 5.76 107 V. Thus, q1/d = 6.41 1017 C/m. Next, we note that when q2 is located at x = 0.080 m, the net potential vanishes (V1 + V2 = 0). Therefore, 2 1 0 0.08 m kq kq d = + Thus, we find q2 = 1( / )(0.08 m)q d = 5.13 1018 C = 32 e.
19. 19. 19. We use Eq. 24-20: V p r = = = 1 4 8 99 10 147 334 10 52 0 10 163 10 0 2 9 30 9 2 5 . . . . . N m2 C2 C m m V. e jc h c h
20. 20. 20. From Eq. 24-30 and Eq. 24-14, we have (for i = 0) Wa = qV = e p cos 4o r2 p cos i 4o r2 = e p 4o r2 (cos 1) . where r = 20 109 m. For = 180 the graph indicates Wa = 4.0 1030 J, from which we can determine p. The magnitude of the dipole moment is therefore 5.6 1037 C. m.
21. 21. 21. (a) From Eq. 24-35, in SI units, 2 2 0 2 2 9 12 2 / 2 ( / 4) 2 ln 4 (0.06/ 2) (0.06) / 4 (0.08) 2(8.99 10 )(3.68 10 )ln 0.08 2.43 10 V. L L d V d + + = + + = = (b) The potential at P is V = 0 due to superposition.
22. 22. 22. The potential is (in SI units) 9 12 2rod rod 0 0 0 1 1 (8.99 10 )(25.6 10 ) 6.20 V. 4 4 4 3.71 10 P dq Q V dq R R R = = = = = We note that the result is exactly what one would expect for a point-charge Q at a distance R. This coincidence is due, in part, to the fact that V is a scalar quantity.
23. 23. 23. (a) All the charge is the same distance R from C, so the electric potential at C is (in SI units) 9 12 1 1 1 2 0 0 6 51 5(8.99 10 )(4.20 10 ) 2.30 V, 4 4 8.20 10 Q Q Q V R R R = = = = where the zero was taken to be at infinity. (b) All the charge is the same distance from P. That distance is 2 2 ,R D+ so the electric potential at P is (in SI units) 2 1 1 1 2 2 2 2 2 0 0 9 12 2 2 2 2 6 51 4 4 5(8.99 10 )(4.20 10 ) 1.78 V. (8.20 10 ) (6.71 10 ) Q Q Q V R D R D R D = = + + + = = +
24. 24. 24. Since the charge distribution on the arc is equidistant from the point where V is evaluated, its contribution is identical to that of a point charge at that distance. We assume V 0 as r and apply Eq. 24-27: 1 1 1 1 0 0 9 12 2 4 21 1 1 1 4 4 2 4 4 (8.99 10 )(7.21 10 ) 3.24 10 V. 2.00 Q Q Q Q V R R R R 0 + + = + + = = =
25. 25. 25. The disk is uniformly charged. This means that when the full disk is present each quadrant contributes equally to the electric potential at P, so the potential at P due to a single quadrant is one-fourth the potential due to the entire disk. First find an expression for the potential at P due to the entire disk. We consider a ring of charge with radius r and (infinitesimal) width dr. Its area is 2r dr and it contains charge dq = 2r dr. All the charge in it is a distance 2 2 r D+ from P, so the potential it produces at P is 2 2 2 2 0 0 1 2 . 4 2 rdr rdr dV r D r D = = + + The total potential at P is 2 2 2 2 02 20 0 0 0 . 2 2 2 R Rrdr V r D R D D r D = = + = + + The potential Vsq at P due to a single quadrant is (in SI units) 15 2 2 2 2 12 0 5 (7.73 10 ) (0.640) (0.259) 0.259 4 8 8(8.85 10 ) 4.71 10 V. sq V V R D D = = + = + =
26. 26. 26. The dipole potential is given by Eq. 24-30 (with = 90 in this case) V = p cos 4o r2 = 0 since cos(90) = 0 . The potential due to the short arc is 1 0 1/ 4q r and that caused by the long arc is 2 0 2/ 4q r . Since q1 = +2 C, r1 = 4.0 cm, q2 = 3 C, and r2 = 6.0 cm, the potentials of the arcs cancel. The result is zero.
27. 27. 27. Letting d denote 0.010 m, we have (in SI units) 9 9 4