Electric KE and PE question

[matth87]

This is in reference to two plates of a parallel capacitor.

A particle with charge q starts from rest at Plate A and accelerates half the distance to Plate B before colliding with a second particle. What is the kinetic energy of the first particle upon collision? (Ignore the effects of gravity.)

A. qV/2 <---- Correct Answer
B. qV <---- What I chose
C. qV/(2d)
D. 2qV/d

Given Answer
Letting V denote the voltage between the plates, the work done by the electric field on the charged particle as it moves from Plate A to Plate B is qV. However, if the particle moves only half the distance, the work done on the particle will be only half as much: qV/2. (The work done is indeed proportional to the displacement here because the electric force is constant, due to the uniform electric field between the capacitor plates.) By the work&#8211;energy theorem, qV/2 also gives the particle's kinetic energy at the point in question.

My Answer
The answer says let V denote the voltage between the two plates. This is never stated anywhere in the passage. What if I wanted V to be equal to the Voltage between point A and half the distance to plate B. Then I would be right correct? Answer else think this question is bogus?

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[crazybob]

it seems like they want you to assume that V is for the maximum electric potential, so you would have to go with that. in your case, the maximum would be 2qV, and halfway would be qV. so with your reasoning, you could look for two answers that give you a ratio of 2:1 and pick the smaller choice.

 

[G1SG2]

This is in reference to two plates of a parallel capacitor.

A particle with charge q starts from rest at Plate A and accelerates half the distance to Plate B before colliding with a second particle. What is the kinetic energy of the first particle upon collision? (Ignore the effects of gravity.)

A. qV/2 <---- Correct Answer
B. qV <---- What I chose
C. qV/(2d)
D. 2qV/d

Given Answer
Letting V denote the voltage between the plates, the work done by the electric field on the charged particle as it moves from Plate A to Plate B is qV. However, if the particle moves only half the distance, the work done on the particle will be only half as much: qV/2. (The work done is indeed proportional to the displacement here because the electric force is constant, due to the uniform electric field between the capacitor plates.) By the work–energy theorem, qV/2 also gives the particle’s kinetic energy at the point in question.

My Answer
The answer says let V denote the voltage between the two plates. This is never stated anywhere in the passage. What if I wanted V to be equal to the Voltage between point A and half the distance to plate B. Then I would be right correct? Answer else think this question is bogus?

I remember this question. From V=ED, we can see that V depends on D. I thought it was a tricky question.