Find how far a proton travels between capacitor plates

AI Thread Summary
To determine how far a proton travels between capacitor plates, it's essential to recognize that the electric field is uniform in this region, allowing for simpler calculations. The force on the proton can be calculated using the electric field and its charge, rather than relying on non-uniform field equations. Conservation of energy can be effectively applied, using the relationship between electric potential and distance, specifically V = Ex, where V is the potential and x is the distance from the negatively charged plate. The initial potential energy can be calculated using the known voltage and the uniform electric field. Understanding these principles clarifies the approach to solving the problem accurately.
r-swald
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Homework Statement
A proton is fired with an initial speed (V_0) of 100,000 m/s from the midpoint of a capacitor toward the positively charged plate. The plates are .4m apart with the negatively charged plate having 0v, and the positively charged plate having 500v. Find the location (in cm) where the proton stops and turns around (using the negative plate as x=0).
Relevant Equations
(sum of) F = ma
(sum of) E = W_nc
F = kQq / r^2
F = qE
E = sigma / epsilon_0
Vf = Vi + at
Vf^2 = Vi^2 +2ax
X = Vit + at^2 / 2
PE = qV
KE = mv / 2
I first attempted to use kinematics, but to find acceleration I would need to find the force either by finding the e-field first and multiplying it with q, or just kQq / r^2. However, I was not given second charge or surface charge density. Is there another way I could go about making the kinematics first or would I have to go to conservation of energy?

I did try conservation of energy but the initial potential energy is where I get stuck. I would try to use PE = qV, but I'm not sure what the potential would be halfway between the capacitor plates. 500 / 2? but I thought the the potential lines were closer together at some points and further at other so I didn't think that just dividing by 2 would be accurate? (image won't attach, hopefully I explained it well enough)
 
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The units of E field are V/m. You are given the capacitor voltage and separation, so you have the E-field in the volume between the plates. Does that help?
 
There are a few points about the electric field and potential between capacitor plates that you have missed.
1. The field is uniform, meaning it has the same value and direction everywhere between the plates.
2. This means that you cannot use F = kQq / r^2 which is a non-uniform field due to a point charge.
3. The electric potential is given by ##V=Ex## where ##x## is the distance from the negatively charged plate where we are told that the potential is zero.

I suggest that you put these ideas together in an energy conservation argument as you started to do. It's the quickest path to the answer.
 
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Thanks for the replies, pretty sure I got it. I was just missing the V = -Ed (had it in an odd location on formula sheet) and the constant E-field. 👍
 
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