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HW1solutions

# HW1solutions - Physics 213 HW#1 Solutions q3= 5 q1= Spring...

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Physics 213 HW #1 Solutions Spring 2009 0 2 4 q1=? q3=+5 q2=-3 21.13. We solve this problem by applying Coulomb’s law. The two forces on 3 q must have equal magnitudes and opposite directions. The force 2 F that 2 q exerts on 3 q has magnitude 2 3 2 2 2 q q F k r and is in the + x direction. 1 F must be in the x direction, so 1 q must be positive. 1 2 F F gives 1 3 2 3 2 2 1 2 q q q q k k r r . 2 2 1 1 2 2 2.00 cm (3.00 nC) 0.750 nC 4.00 cm r q q r . The result for the magnitude of 1 q doesn’t depend on the magnitude of 3 q . 21.33. [Electron in an Electric Field] Use q F E to relate the force on the particle to its charge and the electric field between the plates. The force is constant and produces constant acceleration. The motion is similar to projectile motion, so use constant acceleration equations for the horizontal and vertical components of the motion. For an electron . q e   and q q F E < 0 F and E are in opposite directions, so F is upward. Free-body diagram for the electron: y y F ma eE ma We need to find the acceleration of the electron: Just misses upper plate 0 2.00 cm x x when 0 0.500 cm. y y   x- component: 6 0 0 0 1.60 10 m/s, 0, 0.0200 m, ? x x v v a x x t 2 1 0 0 2 x x x x v t a t 8 0 6 0 0.0200 m 1.25 10 s 1.60 10 m/s x x x t v In this same time t the electron travels 0.0050 m vertically. y- component: 8 0 0 1.25 10 s, 0, 0.0050 m, ? y y t v y y a   2 1 0 0 2 y y y y v t a t 13 2 0 2 8 2 2( ) 2(0.0050 m) 6.40 10 m/s (1.25 10 s) y y y a t Now find 31 13 2 19 (9.109 10 kg)(6.40 10 m/s ) 364 N/C 1.602 10 C ma E e Note: The acceleration produced by the electric field is >> g , so it is OK to neglect the gravitational force on the electron. (d) (b) 19 10 2 27 (1.602 10 C)(364 N/C) 3.49 10 m/s 1.673 10 kg p eE a m in the same direction as the electric field (downward). Still << g. (d) This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the proton won’t hit the plates. The proton has the same initial speed, so it takes the same time 8 1.25 10 s t to travel horizontally the length of the plates. Displacement: 2 10 2 8 2 6 1 1 0 0 2 2 ( 3.49 10 m/s )(1.25 10 s) 2.73 10 m. y y y y v t a t   = 6 2.73 10 m, downward. (c) The displacements are in opposite directions because the electron has negative charge and the proton has positive charge. The electron and proton have the same magnitude of charge, so the force due the electric field has the same magnitude for both charges.

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HW1solutions - Physics 213 HW#1 Solutions q3= 5 q1= Spring...

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