Exercise 21.29 University Physics with Modern Physics 14ed After answering Part
ID: 2003245 • Letter: E
Question
Exercise 21.29 University Physics with Modern Physics 14ed After answering Part A, Part B, I am stuck on Part C If the proton would not hit one of the plates, what would be the magnitude of its vertical displacement as it exits the region between the plates? The initial question is An electron is projected with an initial speed v sub 0 = 1.70 x 10^6 m/s into the uniform field between the parallel plates in the figure. 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 which I have asks if the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. E = 410.7 N/C correct. Part B asks suppose that in the figure that in the figure the electron is replaced by a proton with the same initial speed v sub 0. Would the proton hit one of the plates? Answer No. correct. Please answer Part C above. Thank you.
Explanation / Answer
Find acceleration of proton:
a = E q / m
Substitute given values
a = ( 410.7 N/C ) ( 1.602 x 10-19 C ) / 1.67 x 10-27 kg
a = 3.939 x 1010 m/s2
Now, use kinematic equation to find vertical displacement
y = Vo t + 1/2 ( a ) ( t) 2
y = 0 + 1/2 ( 3.939 x 1010 m/s2 ) ( S / vox ) 2
Here, vox = 1.70 x 10^6 m/s, but the length of plates , S is not given,
Substitute S value from the diagram and solve for y
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In this case, the displacements of proton and electron is in opposite to one another since they are oppositely charged to one another, The magnitude of force on both proton and electron is same. As proton mass is larger by a factor 1836 times to that of electron, its acceleration and vertical displacement are smaller by this factor.
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