F = ma vs. F = qE ("Projectile Motion")

[ilovemcat]

So I was looking at this unusual problem in EK regarding the projectile motion of a negative charge shot a velocity "v" across a single-charged positive plate. One of the questions asked to derive an equation for the vertical velocity of the object in terms of the variables they provided. The way they reasoned it went something like this:

F = mg and so for a projectile in a gravitational field v = sqrt (2gh)

"Then for a constant E-field, the following must be true":

F = qE ; Therefore since F = mg, g must also = qE/m
So then they just sub'd that in for "g" and viola!

Maybe this was obvious for everyone else, ...but this was a what-the-heck moment for me. I'm having a hard time trying to make sense of that. Since when is Electrical Force necessarily equal to Gravitational Force? How where they able to make this assumption? Do you guys understand what I'm trying to ask, lol.

I approached this from a different perspective (the longer way).

I figured the Voltage due to the E-field due to some outside charge (-q) in this case was the height it traveled (ie. "d") and then I used the handy dandy V = Ed formula to find the Potential Difference. Multiplying this times q, I was able to find PE. Then I equated this to KE and solved for v. I'm not sure if this is right, but I would think that this "v" is strictly the vertical velocity since a charge moving perpendicular to an E-field experiences no Potential Difference. Maybe my understanding is flawed. I would appreciate it if someone could correct me if I'm wrong.

---

Members don't see this ad.

 

[Rabolisk]

It's not that mg = qE leading to g = qE/m. It's that ma = qE leading to a = qE/m. You assume (correctly) that the only force that matters is the coulombic force between the plate and the charged particle. They are only using mg to give you an analogy. This works because, in gravitational projectile motion, the acceleration is always downwards and constant (g). Analogously, in this sort of projectile motion, the acceleration is always downwards and constant (qE/m) because the plate is below the negatively charged particle. Since v = sqrt(2gh) comes from v = sqrt(2ay) (y = displacement) which is true for any constant acceleration (from kinematics equations), you can use v = sqrt (2 * qE/m * h).

Going deeper into this, why is acceleration constant? We know that v = sqrt(2ay) only works when acceleration is constant, and we know that to be the case for gravitational projectile motion. In general, coulombic forces increase dramatically as charges are brought together (F = kq1q2/r^2). In order for acceleration to be constant, the force has to be constant, which means that the electric field has to be constant. This can be proven using Gauss's Law.

 

[ilovemcat]

Ah, that makes perfect sense! Great explanation as always. Thanks so much.

By the way, were you able to understand the way I solved it originally. The reason I ask is because in projectile motion problems: at the peak there is PE (due to height) and KE (due to horizontal velocity of the projectile). Here I just took all the change in PE (found by multiplying qV ==> qEh) and basically equated that to "KE = 1/2mv^2" (where v in this case is vertical velocity - what they're asking for). Realistically, I know that you have to consider the KE contributed by the horizontal velocity as the peak. But I kinda just disregarded that and took this KE to equal the max KE due to the vertical velocity. Does that make sense? If so, is this a safe assumption to make?

Edit: Now that I think more about it KE for Vx just cancels out of the equation.

PE + KE (due to Vx at Peak) = KE (due to Vy just before it hits) + KE (due to Vx at Peak)
PE = KE (due to Vy just before it hits)