616_PartUniversity Physics Solution

616_PartUniversity Physics Solution - Electric Charge and...

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Electric Charge and Electric Field 21-9 21.28. IDENTIFY: Use constant acceleration equations to calculate the upward acceleration a and then apply q = F E G G to calculate the electric field. SET UP: Let + y be upward. An electron has charge qe = − . EXECUTE: (a) 0 0 y v = and y aa = , so 2 1 00 2 y y yy v t a t −= + gives 2 1 0 2 yy a t −= . Then 2 12 0 26 2 2( ) 2(4.50 m) 1.00 10 m s (3.00 10 s) yy a t == = × × . 2 31 12 19 (9.11 10 kg) (1.00 10 m s ) 5.69 N C 1.60 10 C Fm a E qq ×× == = = × The force is up, so the electric field must be downward since the electron has negative charge. (b) The electron’s acceleration is ~ 11 10 g , so gravity must be negligibly small compared to the electrical force. EVALUATE: Since the electric field is uniform, the force it exerts is constant and the electron moves with constant acceleration. 21.29. (a) IDENTIFY: Eq. (21.4) relates the electric field, charge of the particle, and the force on the particle. If the particle is to remain stationary the net force on it must be zero. SET UP: The free-body diagram for the particle is sketched in Figure 21.29. The weight is mg, downward. For the net force to be zero the force exerted by the electric field must be upward. The electric field is downward. Since the electric field and the electric force are in opposite directions the charge of the particle is negative. mg q E = Figure 21.29 EXECUTE: 32 5 (1.45 10 kg)(9.80 m/s ) 2.19 10 C and 21.9 C 650 N/C mg E μ × = × = (b) SET UP: The electrical force has magnitude . E Fq E e E The weight of a proton is . wm g = E Fw = so eE mg = EXECUTE: 27 2 7 19 (1.673 10 kg)(9.80 m/s ) 1.02 10 N/C. 1.602 10 C mg E e × × This is a very small electric field. EVALUATE: In both cases and ( / ) . qE mg E m q g In part (b) the / mq ratio is much smaller 8 (1 0) than in part (a) 2 so E is much smaller in (b). For subatomic particles gravity can usually be ignored compared to electric forces. 21.30. IDENTIFY: Apply 2 0 1 4 q E r π = P . SET UP: The iron nucleus has charge 26 . e + A proton has charge e + . EXECUTE: (a) 19 11 10 2 0 1 (26)(1.60 10 C) 1.04 10 N/C. 4 (6.00 10 m) E × × × P (b) 19 11 proton 11 2 0 1 (1.60 10 C) 5.15 10 N/C. 4 (5.29 10 m) E × × × P EVALUATE: These electric fields are very large. In each case the charge is positive and the electric fields are directed away from the nucleus or proton. 21.31. IDENTIFY: For a point charge, 2 . q Ek r = The net field is the vector sum of the fields produced by each charge. A charge q in an electric field E G experiences a force . q = F E G G SET UP: The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q 2 and 0.150 m from q 1 . Point B is 0.100 m from q 1 and 0.350 m from q 2 . EXECUTE: (a) The electric fields due to the charges at point A are shown in Figure 21.31a. 9 1 92 2 3 1 22 1 6.25 10 C (8.99 10 N m /C ) 2.50 10 N/C (0.150 m) A q r × ==× = × 9 2 2 4 2 2 12.5 10 C (8.99 10 N m /C ) 1.124 10 N/C (0.100 m) A q r × ==×⋅
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21-10 Chapter 21 Since the two fields are in opposite directions, we subtract their magnitudes to find the net field.
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616_PartUniversity Physics Solution - Electric Charge and...

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