# Question finding Vf of a charge moving through potential difference

1. Oct 6, 2012

### pcleary

1. The problem statement, all variables and given/known data

A 2kg charge of +3C moves through a potential difference of 6V and begins at rest. What is its final velocity?

2. Relevant equations

m = 2kg
q = +3C
dV = 6V
Vi = 0 m/s
Vf = ?

F = m*a

(Vf)2 = (Vi)2 + 2(a)(d)

dV = (ke)(q)/(r)

1 N/C is equal to 1 V/m

3. The attempt at a solution

Okay I am not sure if this was some mistake given by our professor. This is one of the review questions at the end of his presentation. These are to be done as extra practice. We didn't go over anything like this in class so I finagled my way to what I believe is the answer.

I used the propert 1 N/C = 1 V/m to move over and get 1 N = (V)(C)/(m)

so F (newtons) = (V)(C)/(m)

now from here I said **** I don't have a distance. What equation could I use to find it (r = distance is assumed here)

Using the dV = ke(q)/r I solved for r = d = 4.5x10^9 m

Plugging into my finagled force formula

F = (6N*m/C)(3C) / (4.5x10^9m) --> F = 4.0x10^-9 N

a = F / m --> a = (4.0x10^-9kg*m/s^2) / (2kg) --> a = 2.0x10^-9 m/s^2

putting this back into my kinematic equation (from christ..3 years ago) I got

Vf = sqrt [2(2.0x10^9m/s^2)(4.5x10^9m)] --> sqrt [18m^2 / s^2]

Vf = 4.24 m/s

Now I know this is an outrageous stretch here. And chances are that I am a fool and wasted my time trying to solve an incomplete / misplaced problem. However being that I calculated all of this using proper units and they all cancelled out accordingly I think I am right. Can anyone please chime in on whether or not this is right and if not how it could (if at all) be calculated differently. This is more for my own personal sanity and pride because like I said we didn't go over this type of problem and I don't expect it on the test.

2. Oct 6, 2012

### voko

You should recall what "potential" means.

3. Oct 6, 2012

### pcleary

Potential difference is voltage is it not?

Electrical potential energy is (U) in J no?

Can you please explain how to do the problem as it seems inferred that I did it wrong.

4. Oct 6, 2012

### voko

So the "electric potential" is the potential energy due to the electric field. When a particle moves under the influence of a potential force from A to B, what can be said about the change of its kinetic energy from A to B?

5. Oct 6, 2012

### pcleary

So the "voltage" is the U due to the E(lectric field). Sorry I am trying to get it in terms that I understand. What is the potential force you're talking about?

If I am reading it right I would say the dKE is 0 because once it reaches B it stops moving / has used all its KE?

6. Oct 7, 2012

### voko

The potential force is the force due to the electric field. Which simply means that there is potential energy related to the force.

"Has used all its KE" makes no sense because we do not even know, in general, if there WAS any KE to begin with. In this problem, what is the initial KE?

7. Oct 7, 2012

### pcleary

Now that I think about it doesn't the KE = PE. Or in this case the KE = potential electric energy (U).

So could I have just skipped all that **** and just used the KE = 1/2 mv^2?

Setting U = 1/2 mv^2 and solving for v?

8. Oct 7, 2012

### voko

More correctly, ΔKE = -ΔPE.

Not exactly. ΔPE = Uq.

9. Oct 7, 2012

### pcleary

Oh you're right. So ΔPE = potential energy of the charge (Uq).

Could I say that ΔPE = -ΔKE? So it would be

ΔKE = -1/2 mv^2?

Just curious was my answer to velocity correct? If the sign was wrong how could I have anticipated that when doing it my (long and unnecessary) way?

10. Oct 7, 2012

### pcleary

So working it out that way I found.

ΔV = ΔU / q

ΔU = Uq = (6V)(+3C) = 18J

KE = -PE = -Uq

KE = -18J

KE = 1/2*m*v^2

-18J = 1/2 (2kg) (v^2)

v = i sqrt(18)

V = 4.24 m/s , 4.24i m/s

How do I account for the negative (unreal) component?

11. Oct 7, 2012

### voko

You can only get negative ΔKE if (initial KE) + ΔKE > 0. If initial KE is not big enough, then this motion is impossible. Physically, that means that a positive charge can move through a positive potential difference only of its initial velocity is high enough. However, it can move in the opposite direction with any initial velocity.