Question finding Vf of a charge moving through potential difference

AI Thread Summary
A charge of +3C moving through a potential difference of 6V was analyzed to determine its final velocity, starting from rest. The calculations involved using the relationship between electric potential energy and kinetic energy, leading to the conclusion that the change in kinetic energy equals the negative change in potential energy. The final velocity was calculated to be approximately 4.24 m/s, confirming that the approach was valid despite initial uncertainties about the problem's completeness. The discussion highlighted the importance of understanding the relationship between potential energy and kinetic energy in electric fields. Overall, the calculations and reasoning were validated, although the implications of negative kinetic energy were also explored.
pcleary
Messages
9
Reaction score
0

Homework Statement



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

Homework 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

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 canceled 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.

Thanks in advance.
 
Physics news on Phys.org
You should recall what "potential" means.
 
voko said:
You should recall what "potential" means.

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.
 
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?
 
voko said:
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?
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?
 
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?
 
voko said:
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?


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?
 
pcleary said:
Now that I think about it doesn't the KE = PE.

More correctly, ΔKE = -ΔPE.

Or in this case the KE = potential electric energy (U).

Not exactly. ΔPE = Uq.
 
voko said:
More correctly, ΔKE = -ΔPE.



Not exactly. ΔPE = Uq.

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
voko said:
More correctly, ΔKE = -ΔPE.



Not exactly. ΔPE = Uq.

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
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.
 
Back
Top