11:11 PM session.masteringphysics.com 0000 VIVA LTE Item 4 MasteringPhysics Item
ID: 1772875 • Letter: 1
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11:11 PM session.masteringphysics.com 0000 VIVA LTE Item 4 MasteringPhysics Item 4 An electron is projected with an initial speed (texttiplv_10)){v-ON = 1.70x 106((Nrm {rm m/s))) into the uniform field between the parallel plates in the figure K-2.00 cm Uo 0 1.00 cm Q Tap image to zoom .Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates Part A If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.Explanation / Answer
Initial vertical velocity of the electron is zero.
Initial horizontal velocity of the electron is 1.7 x 10 m/s
Time taken to cross the plates = 0.02 / (1.7x10) = 11.7647 ns
Given that the electron enters midway between the plates and it just misses the upper plate as it emerges. Hence the vertical displacement of the electron = 0.5cm = 0.005m
The acceleration of the electron in the vertical direction is calculated as follows.
s = ut + 1/2 at²
0.005 = 0(10) + 1/2 a (11.7647*109)²
a = 0.005 x 2 / (138.408166*1018) = 7.225*10^13 m/s²
Thus, the force acting on the electron due to the electric field is F = eE = ma
E = ma/e = 9.1 x 10³¹ x 7.225*10^13 / (1.6x10¹) = 410.922 V/m
b) When a proton enters the field, the same magnitude of force acts on it but in the vertical direction.
Acceleration of the proton in the vertically downward direction
Hence proton will not hit the plates
d) proton displacement is downward
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