An electron is projected horizontally with a speed of 5.0 $\times$ 10⁶ m exactly in the middle of two parallel plates at potentials of -40 V and 104 V,respectively.The plate separation distance is 3.0 cm.The plates are long enough that the proton,rather than passing completely between the plates,eventually strikes the plate sitting at 104 V potential.What is the speed of the proton when it strikes the plate? The mass of the electron is 9.11 $\times$ 10^-31 kg.

Question

Quizplus · Accepted Answer

The electric potential difference between the plates is 104 V - (-40 V) = 144 V or 144 J/C. The electric field between the plates is E = V/d = 144/0.03 = 4800 V/m or 4800 N/C.

The electron experiences a downward force due to the electric field, given by F = qE, where q is the charge of the electron (-1.6 x 10^-19 C). This force causes the electron to accelerate downward, and its horizontal speed remains constant.

We can use the vertical motion of the electron to determine the time it takes for it to reach the lower plate. Using the equation y = v0t + 0.5at^2, where y is the distance the electron falls (half the plate separation distance), v0 is the initial vertical velocity (0 m/s), and a is the acceleration due to gravity (-9.8 m/s^2), we get:

0.015 m = 0.5(-9.8)t^2
t = sqrt(0.00306) = 0.055 s

In this time, the electron has traveled a horizontal distance of d = vt = (5.0 x 10^16 m/s)(0.055 s) = 2.75 x 10^15 m.

The proton starts with a horizontal velocity of 5.0 x 10^16 m/s and experiences no electric field, so it continues with constant horizontal velocity until it strikes the plate. We can use the same time, t = 0.055 s, to find its horizontal distance:

d = vt = (5.0 x 10^16 m/s)(0.055 s) = 2.75 x 10^15 m

The proton has traveled the same horizontal distance as the electron and therefore strikes the plate with the same speed. The correct answer is B) 7.1 x 10^6 m/s (rounded to two significant figures).

An Electron Is Projected Horizontally with a Speed of 5 ×\times×

An Electron Is Projected Horizontally with a Speed of 5 $\times$