In Fig. 22-63, a uniform, upward electric field of magnitude 1.60 x 10 3 N/C has
ID: 2065765 • Letter: I
Question
In Fig. 22-63, a uniform, upward electric field of magnitude 1.60 x 103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 12.0 cm and separation d = 1.90 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity of the electron makes an angle = 44.0° with the lower plate and has a magnitude of 6.50 x 106 m/s. (a) Will the electron strike one of the plates? Type the number of the correct answer: 1 = yes, 2 = no. (b) If so, which plate? Type the number of the correct answer: 1 = top, 2 = bottom. (c) How far horizontally from the left edge will the electron strike?
Explanation / Answer
vx = 6.5 *10E6 * cos 44 = 4.68 * 10E6 m/s horizontal velocity vy = 6.5 * 10E6 * sin 44 = 4.52 * 10E6 m/s vertical velocity a = F / m = E q / m = 1600 * 1.6 * 10E-19 / 9.11 * 10E-31 = 2.81 * 10E14 m/s^2 t = .12 / vx = .12 / 4.68 * 10E6 = 2.57 * 10E-8 time to cross plates h = vy t - 1/2 a t^2 height to which electron rises vy t = 1/2 a t^2 for h = 0 (strikes bottom plate) t = 2 vy / a = 3.22 * 10E-8 sec This is greater than time needed to cross plates so it does not strike the bottom plate. How high will it rise, will it reach .019 m h = vy t - 1/2 a t^2 t^2 - 2 vy / a + 2 h /a =0 time to reach upper plate 2 vy / a = 3.22 * 10E-8 2 h / a = 1.35 * 10E-16 sec Solve quadratic for t (1.1 * 10E-7 sec??) Since this is greater than 3.22 * 10E-7 (time to cross plates) it will not reach upper plate (but check the math)