HW1solutions - Physics 213 HW #1 Solutions Spring 2008...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 213 HW #1 – Solutions Spring 2008 21.33. [Electron in an Electric Field] IDENTIFY: Use ± F = q ± 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. (a) SET UP: For an electron q = ± e . Figure 21.33a ± F = q ± E and q < 0 ± ± F and ± E are in opposite directions, so ± F is upward. Free-body diagram for the electron: EXECUTE: F y = ma y ± eE = ma Figure 21.33b We need to find the acceleration of the electron: Just misses upper plate ± x ± x 0 = 2.00 cm when y ± y 0 =+ 0.500 cm. x- component: v 0 x = v 0 = 1.60 ± 10 6 m/s, a x = 0, x ² x 0 = 0.0200 m, t = ? x ± x 0 = v 0 x t + 1 2 a x t 2 t = x ± x 0 v 0 x = 0.0200 m 1.60 ² 10 6 m/s = 1.25 ² 10 ± 8 s In this same time t the electron travels 0.0050 m vertically. y- component: t = 1.25 ± 10 ² 8 s, v 0 y = y ² y 0 0.0050 m, a y = y ± y 0 = v 0 y t + 1 2 a y t 2 a y = 2( y ± y 0 ) t 2 = 2(0.0050 m) (1.25 ² 10 ± 8 s) 2 = 6.40 ² 10 13 m/s 2 Now find E = ma e = (9.109 ± 10 ² 31 kg)(6.40 ± 10 13 m/s 2 ) 1.602 ± 10 ² 19 C = 364 N/C Note: The acceleration produced by the electric field is >> g , so it is OK to neglect the gravitational force on the electron. (d) (b) a = eE m p = (1.602 ± 10 ² 19 C)(364 N/C) 1.673 ± 10 ² 27 kg = 3.49 ± 10 10 m/s 2 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 t = 1.25 ± 10 ² 8 s to travel horizontally the length of the plates. Displacement: y ± y 0 = v 0 y t + 1 2 a y t 2 = 1 2 ( ± 3.49 ² 10 10 m/s 2 )(1.25 ² 10 ± 8 s) 2 = ± 2.73 ² 10 ± 6 m. = 2.73 ± 10 ² 6 m, downward. (c) EVALUATE: 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/31/2008 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell University (Engineering School).

Page1 / 4

HW1solutions - Physics 213 HW #1 Solutions Spring 2008...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online