In the figure a uniform, upward electric field of magnitude 1.80 times 10^3 N/C

Question

In the figure a uniform, upward electric field of magnitude 1.80 times 10^3 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 = 1.60 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 theta = 53.0degree with the lower plate and has a magnitude of 6.90 times 10^6 m/s. (a) will the electron strike one of the plates? (b) If so, which plate, top or bottom? (c) How far horizontally from the left edge will the electron strike?

Explanation / Answer

Apply Newton second law along vertical direction

Fy = m.ay

q.E m.g = m.ay

ay = (q.E - m.g)/m

= ( 1.6 * 10^-19 ( 1.80 * 10^3) - 9.1 * 10^-31 ( 9.8)/ 9.1 * 10^-31

= 3.16x1014 m/s2

Apply kinematic equation the maximum vertical distance traveled by the electron is

ymax = (vo.sin)2/2.ay

= ( 6.9 * 10^6 * sin 53)^2/2 (  3.16x1014 m/s2)

= 0.048 m or 4.8 cm

because ymax is greater than d so electron will hit the upper plate.

Apply kinematic equation

d = vo.sin.t - 0.5.ay.t2

1.6x10-2 = (6.9x106.sin53o.t)- 0.5(3.16x1014).t2

1.6x10-2 = 5.51x106t - 1.58x1014t2

solving quadratic equation

t = 3.19 * 10^-9 s

Apply range equation

s = vo.cos.t

= 6.9x106.cos 53.(3.19x10-9)

= 0.013 m

= 1.3cm

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