Physics exercises_solution: Chapter 22

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Physics exercises_solution: Chapter 22

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 22

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  1.   22.1: a)   E  A  (14 N/C) (0.250 m 2 ) cos 60  1.75 Nm 2 C. b) As long as the sheet is flat, its shape does not matter. ci) The maximum flux occurs at an angle   0 between the normal and field. cii) The minimum flux occurs at an angle   90 between the normal and field. In part i), the paper is oriented to “capture” the most field lines whereas in ii) the area is oriented so that it “captures” no field lines.    22.2: a)   E  A  EA cos θ where A  An ˆ nS1   ˆ (left) S1  (4  10 N C) (0.1 m) 2 cos (90  36.9)  24 N  m 2 C ˆ j 3 ˆ nS2   k ( top) S2  (4  103 N C) (0.1 m) 2 cos 90  0 ˆ nS3   ˆ (right) S3  (4  103 N C) (0.1 m) 2 cos (90  36.9)  24 N  m 2 C ˆ j ˆ nS4  k (bottom) S4  (4  103 N C) (0.1 m) 2 cos 90  0 ˆ ˆ nS5   i (front) S5   (4  103 N C) (0.1 m) 2 cos 36.9  32 N  m 2 C ˆ ˆ nS6  i (back) S6  (4  103 N C) (0.1 m) 2 cos 36.9  32 N  m 2 C ˆ b) The total flux through the cube must be zero; any flux entering the cube must also leave it.    22.3: a) Given that E  Bi  Cˆ  Dk ,   E  A, edge length L, and ˆ j ˆ  nS1   ˆ  1  E  AnS1  CL2 . ˆ j ˆ  ˆ ˆ nS2   k   2  E  AnS2  DL2 . ˆ  nS3   ˆ   3  E  AnS3   CL2 . ˆ j ˆ  ˆ ˆ nS4  k   4  E  AnS4   DL2 . ˆ  ˆ ˆ nS5   i   5  E  AnS5   BL2 . ˆ  ˆ ˆ nS6  i   6  E  AnS6   BL2 . ˆ b) Total flux  i 1  i  0 6   22.4:   E  A  (75.0 N C) (0.240 m 2 ) cos 70  6.16 Nm 2 C.
  2.   ( 6.00106 C/m) (0.400 m) 22.5: a)   E  A   2 πε0 r (2πrl )  l ε0  ε0  2.71  105 Nm 2 C. b) We would get the same flux as in (a) if the cylinder’s radius was made larger—the field lines must still pass through the surface. c) If the length was increased to l  0.800 m, the flux would increase by a factor of two:   5.42  105 Nm 2 C. 22.6: a)  S1  q1 ε0  (4.00  10 9 C) ε0  452 Nm 2 C. b)  S2  q2 ε0  (7.80  10 9 C) ε0  881 Nm 2 C. c)  S3  (q1  q2 ) ε0  ((4.00  7.80)  10 9 C) ε0  429 Nm 2 C. d)  S4  (q1  q 2 ) ε0  ((4.00  2.40)  10 9 C) ε0  723 Nm 2 C. e)  S5  (q1  q2  q3 ) ε0  ((4.00  7.80  2.40)  10 9 C) ε0  158 Nm 2 C. f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.7: a)   q ε0  (3.60  10 6 C) ε0  4.07  10 5 Nm 2 C. b)   q ε0  q  ε0   ε0 (780 Nm 2 C)  6.90  10 9 C. c) No. All that matters is the total charge enclosed by the cube, not the details of where the charge is located. 22.8: a) No charge enclosed so   0 q  6.00  10 9 C b)  2  12   678 Nm 2 C. ε0 8.85  10 C Nm 2 2 q1  q 2 (4.00  6.00)  10 9 C c)   12  226 Nm 2 C. ε0 8.85  10 C Nm2 2  22.9: a) Since E is uniform, the flux through a closed surface must be zero. That is:      E  dA  εq0  ε10  ρdV  0   ρdV  0. But because we can choose any volume we want, ρ must be zero if the integral equals zero. b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that region.
  3. 22.10: a) If ρ  0 and uniform, then q inside any closed surface is greater than zero.      0   E  dA  0 and so the electric field cannot be uniform, i.e., since an arbitrary surface of our choice encloses a non-zero amount of charge, E must depend on position. b) However, inside a small bubble of zero density within the material with density ρ , the field CAN be uniform. All that is important is that there be zero flux through the surface of the bubble (since it encloses no charge). (See Exercise 22.61.) 22.11:  6sides  q ε0  (9.60  10 6 C) ε0  1.08  10 6 Nm 2 C. But the box is symmetrical, so for one side, the flux is: 1 side  1.80  105 Nm 2 C . b) No change. Charge enclosed is the same. 22.12: Since the cube is empty, there is no net charge enclosed in it. The net flux, according to Gauss’s law, must be zero. 22.13:  E  Qencl ε0 The flux through the sphere depends only on the charge within the sphere. Qencl  ε0  E  ε0 (360 N  m 2 C)  3.19 nC 1 q 1 (2.50  10 10 C) 22.14: a) E (r  0.450 m  0.1 m)    7.44 N C . 4πε0 r 2 4πε0 (0.550 m) 2  b) E  0 inside of a conductor or else free charges would move under the influence of forces, violating our electrostatic assumptions (i.e., that charges aren’t moving). 1 |q| 1 q 1 (0.180  106 C) 22.15: a) | E | r   1.62 m. 4πε0 r 2 4πε0 E 4πε0 614 N C b) As long as we are outside the sphere, the charge enclosed is constant and the sphere acts like a point charge. 22.16: a)   EA  q / ε0  q  ε0 EA  ε0 (1.40  10 5 N C) (0.0610 m 2 )  7.56  10 8 C. b) Double the surface area: q  ε0 (1.40  105 N C) (0.122 m 2 )  1.51  107 C.
  4. 22.17: E  1 q 4 πε 0 r 2  q  4πε0 Er 2  4πε0 (1150 N C) (0.160 m)2  3.27  109 C. So the 9 number of electrons is: ne  13..60 10 19 C  2.04  1010. 27 10 C 22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis. The cylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C at every point on the cylindrical surface and directed perpendicular to the surface. Thus    E  d s  ( E )( Acylinder )  ( E )(2πrL)  (840 N/C) (2π ) (0.400 m) (0.0200 m)  42.2 N  m 2 /C   The field is parallel to the end caps of the cylinder, so for them  E  d s  0 . From Gauss’s law: C2 N  m2 q  ε0  E  (8.854  10 12 ) (42.2 ) N  m2 C  3.74  10 10 C 22.19: 1 λ E 2 0 r
  5. 22.20: a) For points outside a uniform spherical charge distribution, all the charge can be considered to be concentrated at the center of the sphere. The field outside the sphere is thus inversely proportional to the square of the distance from the center. In this case: 2  0.200 cm  E  (480 N C)  0.600 cm   53 N C    b) For points outside a long cylindrically symmetrical charge distribution, the field is identical to that of a long line of charge: λ E , 2πε0 r that is, inversely proportional to the distance from the axis of the cylinder. In this case  0.200 cm  E  (480 N C)  0.600 cm   160 N C    c) The field of an infinite sheet of charge is E  σ / 2ε0 ; i.e., it is independent of the distance from the sheet. Thus in this case E  480 N C . 22.21: Outside each sphere the electric field is the same as all the charge of the sphere if were at its center, and the point where we are to calculate E is outside both spheres.   E1 and E 2 are both toward the sphere with negative charge. | q1 | 1.80  106 C E1  k 2 k 2  2.591  105 N C r1 (0.250 m) | q2 | 3.80  10 6 C E2  k 2  k 2  5.471  105 N C r2 (0.250 m) E  E1  E2  8.06  105 N C , toward the negatively charged sphere.
  6. 22.22: For points outside the sphere, the field is identical to that of a point charge of the same total magnitude located at the center of the sphere. The total charge is given by charge density  volume: 4 q  (7.50 n C m 3 )( π )(0.150 m)3  1.60  1010 C 3 a) The field just outside the sphere is q (9  109 N  m 2 /C 2 ) (1.06  10 10 C) E   42.4 N C 4πε0 r 2 (0.150 m) 2 b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9  109 N  m 2 / C 2 ) (1 / 8) (1.06  10 10 C) E  21.2 N C (0.075 m) 2 22.23: The point is inside the sphere, so E  kQr / R 3 (Example 22.9) ER 3 (950 N C) (0.220 m)3 Q   10.2 nC kr k (0.100 m) 22.24: a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its magnitude is the same as the cavity charge: qinner   6.00 nC, since E  0 inside a conductor. b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “left behind” when the  6.00 nC moved to the inner surface: qtot  qinner  qouter  qouter  qtot  qinner  5.00 nC  6.00 nC   1.00 nC. 22.25: S 2 and S 3 enclose no charge, so the flux is zero, and electric field outside the plates is zero. For between the plates, S1 shows that: EA  q ε0  σ A ε0  E  σ ε0 .
  7. 22.26: a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so:   q q 7.50  10 9 C  E  dA  E 2 A   E  ε0  2ε0 A 2ε0 (0.800 m) 2  662 N C . b) At a distance of 100 m from the center, the sheet looks like a point, so: 1 q 1 (7.50  10 9 C) E 2  2  6.75  10 3 N C . 4πε0 r 4πε0 (100 m) c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly over both faces, giving it half the charge density on any as the insulator (σ :). Ec  εσ0  2σε:0 near one face. Unlike a conductor, the insulator is the charge density in some sense. Thus one shouldn’t think of the charge as “spreading over each face” for an insulator. Far away, they both look like points with the same charge. Q Q Q 22.27: a) σ    σ 2πR  λ. A 2πRL L Q σ 2πRL σR b)  E  d A  E (2πrL)  ε0  ε0  E  rε0 . c) But from (a), λ   2 R, so E  2 πε0r , same as an infinite line of charge. λ
  8. 22.28: All the σ ' s are absolute values. σ σ σ σ (a) at A : E A  2 ε20  2 ε30  2 ε40  2 ε10 1 EA  (σ 2  σ 3  σ 4  σ1 ) 2ε0 1  (5 μ C m 2  2 μ C m 2  4 μ C m 2  6 μ C m 2 ) 2ε0  2.82  105 N C to the left. (b) σ1 σ σ σ 1 EB   3  4  2  (σ 1  σ 3  σ 4  σ 2 ) 2 ε 0 2ε 0 2ε 0 2ε 0 2 ε 0 1  (6 μ C m 2  2 μ C m 2  4 μ C m 2  5 μ C m 2 ) 2ε 0  3.95  105 N C to the left. (c) σ2 σ σ σ 1 EC   3  4  1  (σ 2  σ 3  σ 4  σ1 ) 2ε0 2ε0 2ε0 2ε0 2ε0 1  (5 μ C m 2  2 μ C m 2  4 μ C m 2  6 μ C m 2 ) 2ε0  1.69  105 N C to the left 22.29: a) Gauss’s law says +Q on inner surface, so E  0 inside metal. b) The outside surface of the sphere is grounded, so no excess charge. c) Consider a Gaussian sphere with the –Q charge at its center and radius less than the inner radius of the metal. This sphere encloses net charge –Q so there is an electric field flux through it; there is electric field in the cavity. d) In an electrostatic situation E  0 inside a conductor. A Gaussian sphere with the  Q charge at its center and radius greater than the outer radius of the metal encloses zero net charge (the  Q charge and the  Q on the inner surface of the metal) so there is no flux through it and E  0 outside the metal. e) No, E  0 there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive and negative mass (the gravity force is always attractive), so this cannot be done for gravity.
  9.  ˆ ˆ 22.30: Given E  (5.00 ( N C)  m) x i  (3.00 ( N C)  m) z k , edge length  L  0.300 m, L  0.300 m, and ns1   ˆ  1  E  nS1 A  0. ˆ j ˆ  ˆ ˆ nS1   k   2  E  nS2 A  (3.00 ( N C)  m)(0.300 m)2 z  (0.27 ( N C)m) z  ˆ (0.27 ( N C)m)(0.300 m)  0.081 ( N C) m 2 .  nS3   ˆ   3  E  nS3 A  0. ˆ j ˆ  ˆ ˆ nS4  k   4  E  nS4 A  (0.27 ( N C)  m) z  0 ( z  0). ˆ  ˆ ˆ n   i    E  n A  (5.00 ( N C)  m)(0.300 m) 2 x  (0.45( N/C)  m) x ˆ S5 5 S5   (0.45 ( N/C)  m)(0.300 m)   (0.135 ( N C)  m 2 ).  ˆ ˆ nS6  i   6  E  nS6 A   (0.45 ( N C)  m) x  0 ( x  0). ˆ b) Total flux:    2   5  (0.081  0.135) ( N C)  m 2  0.054 Nm 2 C q  ε0   4.78  10 13 C 22.31: a) b) Imagine a charge q at the center of a cube of edge length 2L. Then:   q / ε0 . Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24 of the flux. That is,   q 24 0 . 22.32: a)   EA  (125 N C)(6.0 m 2 )  750 N  m 2 C . b) Since the field is parallel to the surface,   0. c) Choose the Gaussian surface to equal the volume’s surface. Then: 750 – EA= q ε0  E  6.01m 2 (2.40  108 C 0  750)  577 N C , in the positive x -direction. Since q  0 we must have some net flux flowing in so EA   E A on second face. d) q  0 but we have E pointing away from face I. This is due to an external field that does not affect the flux but affects the value of E.
  10. 22.33: To find the charge enclosed, we need the flux through the parallelepiped: 1  AE1 cos 60  (0.0500 m)(0.0600 m)(2.50  10 4 N C) cos 60  37.5 N  m 2 C  2  AE2 cos 120  (0.0500 m)(0.0600 m)(7.00  10 4 N C) cos 60   105 N  m 2 C So the total flux is   1   2  (37.5  105) N  m 2 C  67.5 N  m 2 C , and q  ε0  (67.5 N  m 2 C)ε0  5.97  10 10 C. b) There must be a net charge (negative) in the parallelepiped since there is a net flux flowing into the surface. Also, there must be an external field or all lines would point toward the slab. 22.34: The  particle feels no force where the net electric field is zero. The fields can cancel only in regions A and B. Eline  Esheet λ σ  2πε0 r 2ε0 50 C/m r  λ    0.16m  16cm  (100 C/m 2 ) The fields cancel 16 cm from the line in regions A and B.
  11. 22.35:  The electric field E1 of the sheet of charge is toward the sheet, so the electric  field E 2 of the sphere must be away from the sheet. This is true above the center of the sphere. Let r be the distance above the center of the sphere for the point where the electric field is zero. σ 1 Q2 r E1  E2 so 1  2ε0 4πε0 R 3 2πσ1R3 2π (8.00  109 C/m 2 )(0.120 m)3 r   0.097 m Q2 0.900  10 9 C
  12. 22.36: a) For r  a, E  0, since no charge is enclosed. For a  r  b, E  1 q 4 πε 0 r 2 , since there is +q inside a radius r. For b  r  c, E  0, since now the –q cancels the inner +q. For r  c, E  1 q 4 πε 0 r 2 , since again the total charge enclosed is +q. b) c) Charge on inner shell surface is –q. d) Charge on outer shell surface is +q. e) 22.37: a) r  R, E  0, since no charge is enclosed. b) R  r  2 R, E  1 Q 4 πε 0 r 2 , since charge enclosed is Q. r  2 R, E  1 2Q 4 πε 0 r 2 , since charge enclosed is 2Q.
  13. 22.38: a) r  a, E  1 Q 4 πε 0 r 2 , since the charge enclosed is Q. a  r  b, E  0, since the –Q on the inner surface of the shell cancels the +Q at the center of the sphere. r  b, E   4πε 0 2Q , since the total enclosed charge is –2Q. 1 r2 b) The surface charge density on inner surface: σ   4πa 2 . Q c) The surface charge density on the outer surface: σ   42Q2 . πb d) e)
  14. 22.39: a)(i) r  a, E  0, since Q  0 (ii) a  r  b, E  0, since Q  0. (iii) b  r  c, E  1 2q 4 πε 0 r 2 , since Q   2q. (iv) c  r  d , E  0, since Q  0. (v) r  d , E  1 6q 4 πε 0 r 2 , since Q   6q. b)(i) small shell inner: Q0 (ii) small shell outer: Q   2q (iii) large shell inner: Q   2q (iv) large shell outer: Q   6q 22.40: a)(i) r  a, E  0, since the charge enclosed is zero. (ii) a  r  b, E  0, since the charge enclosed is zero. (iii) b  r  c, E  1 2q 4 πε 0 r 2 , since charge enclosed is  2q. (iv) c  r  d , E  0, since the net charge enclosed is zero. (v) r  d , E  0, since the net charge enclosed is zero. b)(i) small shell inner: Q0 (ii) small shell outer: Q   2q (iii) large shell inner: Q   2q (iv) large shell outer: Q0
  15. 22.41: a)(i) r  a, E  0, since charge enclosed is zero. (ii) a  r  b, E  0, since charge enclosed is zero. (iii) b  r  c, E  1 2q 4 πε 0 r 2 , since charge enclosed is  2q. (iv) c  r  d , E  0, since charge enclosed is zero. (v) r  d , E   4πε 0 1 2q r2 , since charge enclosed is  2q. b)(i) small shell inner: Q0 (ii) small shell outer: Q   2q (iii) large shell inner: Q   2q (iv) large shell outer: Q   2q 22.42: a) We need: 4 ρ  28π ρR 3 3Q Q  ((2 R)3  R 3 )  Q   ρ . 3 3 28πR  b) r  R, E  0 and r  2 R, E  0, since the net charges are zero. Q 4π ρ 3 Q ρ R  r  2 R,   E (4πr 2 )   (r  R 3 )  E  2  2 (r 3  R 3 ). ε0 3ε0 4πε0 r 3ε0 r Substituting ρ from (a) E  2 Q 7 πε 0 r 2  28πε R 3 . Qr 0 c) We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere—but we see a smooth transition from the uniform insulator to the outside.
  16. 22.43: a) The sphere acts as a point charge on an external charge, so: F  qE  4πε 0 qQ , radially inward. 1 r2 (b) If the point charge was inside the sphere (where there is no electric field) it would feel zero force. q q 3q  1  22.44: a) ρinner   4 3 4 3   Vb  Va 3 πb  3 πa 4π  b3  a 3  q q  3q  1  outer   4 3 4 3   Vd  Vc 3 πd  3 πc 4π  d 3  c3    b) (i) r  a  E  dA  0  E  0.   1 4 (ii) a  r  b  E  dA    inner dV  E 4 r 2  π (r 3  a 3 )  inner ε0 3ε0 1 (r 3  q 3 ) q (r 3  a 3 ) E  inner  3ε0 r2 4ε0 (b 3  a 3 )   q q q (iii) b  r  c  E  dA   E 4πr 2   E  ε0 ε0 4πε0 r 2   q 1 (iv) c  r  d  E  dA    ρouter dV  ε0 ε0 q 4π 3 q q(r 3  c 3 ) E 4 r  2 (r  c ) ρouter, so E  3  ε0 3ε0 4πε0 r 2 4πε0 r 2 (d 3  c 3 )   q q (v) r  d  E  d A    0  E  0 ε0 ε0
  17. 1 2λ 22.45: a) a  r  b, E  , radially outward, as in 22.48 (b). 4 0 r b) r  c, E  1 4 πε0 r , radially outward, since again the charge enclosed is the 2λ same as in part (a). c) d) The inner and outer surfaces of the outer cylinder must have the same amount of charge on them: λl  λ inner l  λ inner   λ, and λ outer  λ. q αl α 22.46: a) (i) r  a, E (2πrl )   E . ε0 ε0 2πε0 r (ii) a  r  b, there is no net charge enclosed, so the electric field is zero. q 2αl α (iii) r  b, E (2πrl )   E . ε0 ε0 πε0 r b) (i) Inner charge per unit length is  α. (ii) Outer charge per length is  2 .
  18. 22.47: a) (i) r  a, E (2πrl )  q ε0  αl ε0 E α 2 πε 0 r , radially outward. (ii) a  r  b, there is not net charge enclosed, so the electric field is zero. (iii) r  b, there is no net charge enclosed, so the electric field is zero. b) (i) Inner charge per unit length is   . (ii) Outer charge per length is ZERO. q ρπr 2 l ρr 22.48: a) r  R, E (2πrl )  ε0  ε0 E 2ε 0 , radially outward. ρπR 2l ρR 2 b) r  R, and λ  ρπR 2 , E (2πrl )  q ε0  ε0 E 2 ε0 r  λ 2 πε0 r  2 kλ r . c) r  R. the electric field for BOTH regions is E  ρR 2ε 0 , so they are consistent. d) 22.49: a) The conductor has the surface charge density on BOTH sides, so it has twice the enclosed charge and twice the electric field. b) We have a conductor with surface charge density σ on both sides. Thus the electric field outside the plate is   E (2 A)  (2σA) ε0  E  σ ε0 . To find the field inside the conductor use a Gaussian surface that has one face inside the conductor, and one outside. Then:   Eout A  Ein A  (σA) ε0 but Eout  σ ε0  Ein A  0  Ein  0.
  19. 22.50: a) If the nucleus is a uniform positively charged sphere, it is only at its very center where forces on a charge would balance or cancel   q e  r3  er b)    E  dA   E 4πr 2   3 E  R  ε0 ε0   4πε0 R 3 1 e2r  F  qE   . 4πε0 R 3 So from the simple harmonic motion equation: 1 e2r 1 e2 1 1 e2 F  mω2 r   3 ω 3  f  . 4πε0 R 4πε0 mR 2π 4πε0 mR3 1 1 e2 c) If f  4.57  1014 Hz  2 4πε0 mR 3 1 (1.60  10 19 C) 2 R3 31  3.13  10 10 m. 4πε0 4π (9.11  10 kg)(4.57  10 Hz) 2 14 2 ractual r Thompson  1 d) If r  R then the electron would still oscillate but not undergo simple harmonic motion, because for r  R, F  1 r 2 , and is not linear. 22.51: The electrons are separated by a distance 2d , and the amount of the positive nucleus’s charge that is within radius d is all that exerts a force on the electron. So: ke 2 Fe  2  Fnucleus  2ke 2 Rd3  d 3  R 3 / 8  d  R / 2. (2d )
  20. Q  2 r / a0 4Q r 2 2 x / a0 22.52: a) Q(r )  Q   ρdV  Q  e a0 0 3 r 2 dr sin  dθ d  Q  3 x e dx πa0 4Qe  r  Q( r )  Q  (2e r  α 2 r 2  2 r  2)  Qe 2 r / a0 [2(r / a0 ) 2  2(r / a0 )  1]. a0  3 3 Note if r  , Q(r )  0. b) The electric field is radially outward, and has magnitude: kQe2 r / a0 E 2 (2(r a0 ) 2  2(r a0 )  1). r 19 2 1 (82 )(1.6  10 C) 22.53: a) At r  2 R, F  qe E  1 qe qFe 4 πε0 4 R 2  4 πε0 4 ( 7.1  1015 m)2  94 N. So: a  F m  94 N 9 .11  1031 kg  1.0  1032 m/s 2 . b) At r  R, a  4a(a)  4.1  1032 m/s 2 . c) At r  R 2 , Q  1 (82e) ( 1 because the charge enclosed goes like r 3 ) so with the 8 8 radius decreasing by 2, the acceleration from the change in radius goes up by (2) 2  4, but the charge decreased by 8, so a  8 a( b )  2.1  1032 m/s2 . 4 d) At r  0, Q  0, so F  0.
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