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Chapter 21 1. Use Coulomb’s law to calculate the magnitude of the force. F k 2. Q1Q2 r 2 8.988 109 N m 2 C 2 19 C 26 1.602 10 19 C 12 1.5 10 m 2 2.7 10 3 N Use the charge per electron to find the number of electrons. 1 electron 38.0 10 C 1.602 10 6 19 3. 1.602 10 2.37 1014 electrons C Use Coulomb’s law to calculate the magnitude of the force. F k Q1Q2 r2 8.988 10 N m C 9 2 2 25 10 C 2.5 10 C 7200 N 6 3 0.28 m 2 14. (a)If the force is repulsive, both charges must be positive since the total charge is positive. Call the total charge Q. Q1 Q2 Q F Q Q2 4 Q1 2 kQ1Q2 kQ1 Q Q1 d2 Q12 QQ1 Fd 2 k 0 Fd 2 Fd 2 Q Q2 4 k k 2 12 90.0 106 C d2 90.0 10 C 6 2 4 8.988 109 N m2 C2 12.0N 1.16 m 2 60.1 106 C , 29.9 106 C (b)If the force is attractive, then the charges are of opposite sign. The value used for F must then be negative. Other than that, the solution method is the same as for part (a). Q1 Q2 Q F d2 Fd 2 Q Q 4 2 Q1 kQ1Q2 k kQ1 Q Q1 d2 Q Q 4 2 2 90.0 106 C 1 2 Q QQ1 2 1 Fd 2 k 0 Fd 2 k 2 90.0 10 C 6 2 4 12.0N 1.16 m 2 8.988 10 9 N m 2 C2 106.8 10 6 C , 16.8 106 C 15. Determine the force on the upper right charge, and then use the symmetry of the configuration to determine the force on the other three charges. The force at the upper right corner of the square is the vector sum of the forces due to the other three charges. Let the variable d Q1 Q4 Q3 Q2 F41 k Q d 2 F41 x k Q2 F42 k 2d 2 Q2 F43 k Q 2 , F41 y 0 d2 F42 x k Q2 2d 2 cos45o k F43 x 0 , F43 y k d2 2Q 2 4d 2 , F42 y k 2Q 2 4d 2 Q2 d2 Add the x and y components together to find the total force, noting that F4 x F41x F42 x F43 x k F4 F42x F42y k Q2 2 1 8.988 10 N m C tan 1 9 F4 y F4 x 2 2 d 2 2 Q2 4 2 2k d Q2 2Q 2 4d 2 0 k Q2 2 1 F4 y d 4 2 1 2 d 2 4.15 10 C 3 0.100 m k 2 2 45o above the x-direction. 2 1 2.96 107 N 2 F41 d represent the 0.100 m length of a side of the square, and let the variable Q represent the 4.15 mC charge at each corner. 2 F42 F43 F4 x F4 y . For each charge, the net force will be the magnitude determined above, and will lie along the line from the center of the square out towards the charge. 17. The spheres can be treated as point charges since they are spherical, and so Coulomb’s law may be used to relate the amount of charge to the force of attraction. Each sphere will have a magnitude Q of charge, since that amount was removed from one sphere and added to the other, being initially uncharged. F k Q1Q2 r2 k Q2 r2 F Qr 0.12 m k 1.7 10 2 N 8.988 109 N m 2 C 2 1 electron 12 1.0 10 electrons 19 1.602 10 C 1.650 10 7 C 18. The negative charges will repel each other, and so the third charge Q Q0 4Q0 must put an opposite force on each of the original charges. Consideration x l–x of the various possible configurations leads to the conclusion that the third charge must be positive and must be between the other two charges. See l the diagram for the definition of variables. For each negative charge, equate the magnitudes of the two forces on the charge. Also note that 0 x l . left: k k k Q0Q x 2 Q0Q x 2 Q0Q x2 k k k 4Q02 l2 4Q0Q l x 4Q02 l 2 2 right: k l x 2 4Q02 l2 Q 4Q0 x2 l 2 Q0 4 9 Q0 4 3 2 94 Q0 , and a distance 1 3 l from Q0 towards 4Q0 Use Eq. 21–3 to calculate the force. Take east to be the positive x direction. E 22. k x 13 l Thus the charge should be of magnitude 21. 4Q0Q F q F qE 1.602 1019 C 1920 N C ˆi 3.08 10 16 N ˆi 3.08 10 16 N west Use Eq. 21–3 to calculate the electric field. Take north to be the positive y direction. . E F q 2.18 1014 N ˆj 1.36 105 N C ˆj 1.36 105 N C south 19 1.602 10 C Q1 0 28. The electric field due to the negative charge will point toward the negative charge, and the electric field due to the positive charge will point away from the positive charge. Thus both fields point in the same direction, towards the negative charge, and so can be added. Q1 E E1 E2 k k 2 1 r Q2 r 4 8.988 109 N m 2 C2 0.080 m 2 k 2 2 Q1 l / 2 8.0 10 k 2 6 l / 2 2 E2 l 2 4k l2 Q 1 Q2 C 5.8 106 C 7.8 107 N C towards the negative charge The direction is Q2 Q2 0 E1 . 30. Assuming the electric force is the only force on the electron, then Newton’s second law may be used to find the electric field strength. ma Fnet ma qE E 33. q 1.673 10 27 9.80 m s 0.18 N C C kg 1.8 106 1.602 10 19 2 The field at the upper right corner of the square is the vector sum of the fields due to the other three charges. Let the variable l represent the 1.0 m length of a side of the square, and let the variable Q represent the charge at each of the three occupied corners. E1 k E2 k E3 k Q l E1 x k 2 Q 2l Q l 2 2 Q l2 E2 x k , E1 y 0 Q 2l 2 cos45o k E3 x 0 , E1 y k E3 Q1 E1 l Q3 Q2 2Q 4l 2 , E2 y k E2 2Q 4l 2 Q l2 Add the x and y components together to find the total electric field, noting that Ex E y . E x E1 x E2 x E3 x k E E x2 E y2 k Ex k 2Q 4l 2 0k Q 2 1 Ey l 4 2 2 Q 1 1 2 k 2 2 2 l 4 l 2 Ey l 2 Q 8.988 109 N m 2 C2 tan 1 Q 2.25 10 C 6 1.22 m 2 1 4 2 2.60 10 N C 2 45.0 from the x-direction. 37. Make use of Example 21-11. From that, we see that the electric field due to the line charge 1 ˆ E1 i. 2 0 x along the y axis is In particular, the field due to that line of charge has no y dependence. 1 ˆ E2 j. 2 0 y In a similar fashion, the electric field due to the line charge along the x axis is Then the total field at x, y is the vector sum of the two fields. E E1 E2 1 ˆ 1 ˆ 1 ˆ 1 ˆ i j i y j 2 0 x 2 0 y 2 0 x 1 E 2 0 1 x 2 1 y 2 Ey x 2 0 y x 2 y 2 ; tan 1 tan 1 tan 1 1 2 0 xy Ex y 2 0 x 45. Because the distance from the wire is much smaller than the length of the wire, we can approximate the electric field by the field of an infinite wire, which is derived in Example 21-11. 4.75 106 C 6 1 1 2 N m2 2.0 m 1.8 10 N C, E 8.988 109 2 0 x 4 0 x C2 2.4 102 m away from the wire 2 49. Select a differential element of the arc which makes an angle of with the x axis. The length of this element is Rd , and the charge on that element is dq Rd . The magnitude of the field produced by that element is Rd R dEbottom dEtop 0 x dE 1 Rd 4 0 R2 . From the diagram, considering pieces of the arc that are symmetric with respect to the x axis, we see that the total field will only have an x component. The vertical components of the field due to symmetric portions of the arc will cancel each other. So we have the following. dEhorizontal 1 Rd 4 0 R2 0 Ehorizontal cos 1 4 0 cos Rd 0 R2 0 2 sin 0 cos d sin 0 sin 0 4 0 R 4 0 R 4 0 R 0 E The field points in the negative x direction, so 2 sin 0 ˆ i 4 0 R Choose a differential element of the rod dx a distance x from the origin, as shown in the Q dq dx . l diagram. The charge on that differential element is The variable x is treated as positive, so 53. dE that the field due to this differential element is the rod to find the total field. dx dx 1 dq 4 0 x x 2 dx Q 4 0 l x x 2 . Integrate along l 1 Q 1 1 E dE 2 2 4 0 l x x 4 0 l 0 x x 4 0 l x x 0 4 0 l x x l 0 l Q Q l Q Q 4 0 x x l 58. (a)The electron will experience a force in the opposite direction to the electric field. Since the electron is to be brought to rest, the electric field must be in the same direction as the initial velocity of the electron, and so is to the right. (b)Since the field is uniform, the electron will experience a constant force, and therefore have a constant acceleration. Use constant acceleration relationships to find the field strength. F qE ma a qE m v 2 v02 2ax v02 2 qE m x E 60. m v 2 v02 2qx mv 9.109 10 kg 7.5 10 m s 2qx 2 1.602 10 C 0.040 m 31 2 0 5 19 2 40 N C 2 sig. fig. Since the field is constant, the force on the electron is constant, and so the acceleration is constant. Thus constant acceleration relationships can be used. The initial conditions are y0 0, vx 0 1.90 m s, and x0 0, v y 0 0. F ma qE a q m E x x0 v x 0t 12 a x t 2 v x 0t e m eE x 2m E ; ax e m Ex , a y e m Ey t2 1.60 10 C 2.00 10 N C 2.0s 3.2 m 1.90 m s 2.0s 2 9.11 10 kg eE 1.60 10 C 1.20 10 N C 2.0s y y v t a t t 2m 2 9.11 10 kg 19 11 2 31 19 0 y0 1 2 2 y y 11 2 31 2 4.2 m