Difference between Electric Potential energy and Potential energy

[SiriusAboutAstronomy]

Homework Statement

A proton is released from rest in a uniform electric field of magnitude 8x10^4 V/m. After the proton has moved 0.5 meters
a) What is the change in electric potential?
b) What is the change in potential energy?
c) What is the speed of the proton?

Homework Equations

U(elec)=U(naught)+qEs
s=change in distance
q=charge of particle
E= electric field

The Attempt at a Solution

I'm more or less looking for clarification of the problem... on part a), do you think he means what is the change in electric potential energy?
If not, how does the electric potential change? Maybe I don't know the difference well enough.
In part b) if part a) is talking about electric potential energy of the particle, does the particle gain or lose mechanical potential energy?

 

[Simon Bridge]

I'm more or less looking for clarification of the problem... on part a), do you think he means what is the change in electric potential energy?

No. He has clearly asked for the change in electric potential.
You do need to go through your course notes for these definitions.
http://en.wikipedia.org/wiki/Electric_potential
http://en.wikipedia.org/wiki/Electric_potential_energy

if part a) is talking about electric potential energy of the particle, does the particle gain or lose mechanical potential energy?

What would "mechanical potential energy" be? If the particle loses potential energy from anywhere, it gains kinetic energy.

 

[SiriusAboutAstronomy]

Okay, but if the electric potential is independent of the charge of the particle, and the electric field is uniform, wouldn't there be no change in the electric potential?
U(elec)=qV ==> V=U/q
I guess I am confused why he asked a) before b), because to solve it, U=qEs, and V=U/q.
So in this case V=Es? I just don't understand why there would be a change in electric potential in an infinite uniform electric field...

 

[Simon Bridge]

Okay, but if the electric potential is independent of the charge of the particle, and the electric field is uniform, wouldn't there be no change in the electric potential?

The field is the gradient of the potential.

You should have something like $\vec{E}=\vec{\nabla}V$ in your notes.
If the gradient is a constant - what is the function?

 

[Nickie]

I would like to provide a clear explanation of the difference between electric potential energy and potential energy. Electric potential energy refers to the energy that a charged particle possesses due to its position in an electric field. This energy is dependent on the magnitude of the electric field and the charge of the particle. On the other hand, potential energy is a more general term that refers to the energy that a particle possesses due to its position in a force field. This can include gravitational potential energy, elastic potential energy, etc.

Now, moving on to the given problem. When a proton is released from rest in a uniform electric field, it will experience a force due to the electric field and will accelerate. This means that the proton will gain kinetic energy and its speed will increase. Since the proton is moving in the direction of the electric field, its electric potential energy will decrease as it moves closer to the source of the field.

a) The change in electric potential refers to the difference in electric potential energy before and after the proton has moved. In this case, the proton has moved 0.5 meters in the direction of the electric field, which means it has moved closer to the source of the field. Therefore, the change in electric potential will be negative.

b) The change in potential energy refers to the difference in potential energy before and after the proton has moved. In this case, the proton has gained kinetic energy as it accelerates in the electric field. Therefore, the change in potential energy will be positive.

c) The speed of the proton can be calculated using the equation for kinetic energy, where the change in kinetic energy is equal to the change in potential energy. We can also use the equation for the work done by the electric field, which is equal to the change in kinetic energy. Solving for the speed, we get:

KE = qΔV = (1/2)mv^2
v = √(2qΔV/m)

Using the given values, we get:

v = √(2(1.6x10^-19 C)(-8x10^4 V/m)(0.5 m)/(1.67x10^-27 kg))
v = 1.6x10^7 m/s

In conclusion, the change in electric potential is negative, the change in potential energy is positive, and the speed of the proton is 1.6x10^7 m/s.