In the figure a uniform, upward electric field of magnitude 2.10 times 10^3 N/C

Question

In the figure a uniform, upward electric field of magnitude 2.10 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 = 17.0 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 = 45.0degree with the lower plate and has a magnitude of 6.10 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

The electric field is upward in the diagram and the charge is negative, so the force of the field on it is downward. The magnitude of the acceleration a = eE/m

E = magnitude of electric field = 2.1 x 10^3

m = mass of electron

e = charge on electron

a = 3.69 x 10^14 m/s^2

x = u*t*costheta

y = u*t*sintheta-at^2/2

vy = u*sintheta - a*t

First, we find the greatest y coordinate attained by the electron.If it is less than d, the electron does not hit the upper plate. If it is greater than d, it will hit the upper plate if the corresponding x coordinate is less than L.

The greatest y coordinate occurs when vy = 0

ymax = u^2*sin^2theta/2a

ymax = 0.0252 m

this is greater than d = 1.6 cm

part b )

x coordinate of the position of the electron when y = d. Since

usintheta = 6.1 x 10^6 * sin45 = 4.31 x 10^6 m/s

2ad = 1.1808 x 10^13 m^2/s^2

d = utsintheta - at^2/2

t = [usintehat - sqrt((usintheta)^2-2ad)]a

t = 4.62 x 10^-9 s

x = ut*cos45

x = 0.02 m

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