Two parallel charged plates are held at a potential difference ?V. Point A is on

Question

Two parallel charged plates are held at a potential difference ?V. Point A is on the plate with the lower potential, and point B is directly across from it on the other plate. An electron (of mass 9.11×10–31 kg) and a proton (1.67×10–27 kg) are released simultaneously from points A and B, respectively. The electron accelerates toward B and the proton accelerates toward A. Assume they do not collide or interact with each other. (a) Which particle reaches the opposite plate first? (b) How many times longer does it take the slower particle to complete this "drag race"?


(a) A) The electron B) The proton C) They take the same amount of time
(b) ________ times longer

Explanation / Answer

As the charges on both particles are same, force effecting on particles are equal.

F = E*q =(V/d)*q (E electric field, d is distance between plates)

According to second law of Newton F= ma

which tells the acceleration is different. Acceleration is inverse proportional to mass.

(a)

As mass of electron is smaller than mass of proton, acceleration of electron is bigger.

So ELECTRON comes first. Answer A)

(b)

distance travelled d = 1/2*a*t^2

so time t is inverse proportional with square root of acceleration. Which means d is proportional with square root of mass(from newtons equation). You can imagine it like a man carrying mass from one place to another. The heavier the mass , the more time it takes to get there.

So it takes (mp/me)^(1/2) = 42.8 times longer.

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