Accelerated Electron and its Potential Energy

[Abu]

Homework Statement

An electron acquires 3.16*10^-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential.

Homework Equations

w =Δv * q

The Attempt at a Solution

Since kinetic energy is equal to the negative of potential energy
Using the above formula and conservation of energy, it is simply
Δv = w/q
Δv = -3.16*10^-16 / -1.6*10^-19
Δv = 1975 Volts
So this means that when the electron went from point A to point B, the electric potential increased by 1975 volts. This means that point B is the higher potential.

And this makes sense because if you do w = Δv * q where Δv is 1975 Volts and q is negative 1.6*10^-19, you get negative 3.16*10^-16.

Is this right? And if so, if the work is equal to the change in potential energy, then that means the work done is negative 3.16*10^-16. What does this negative work mean exactly?

Side note: I am using the convention where w =Δv * q, not w = negative ΔV * q

 

[TSny]

w =Δv * q

What does w represent? Is it the potential energy, or is it the change in potential energy, or is it something else?

Since kinetic energy is equal to the negative of potential energy

This statement is not true in general. For a conservative system, the change in KE equals the negative of the the change in potential energy.

Using the above formula and conservation of energy, it is simply
Δv = w/q
Δv = -3.16*10^-16 / -1.6*10^-19
Δv = 1975 Volts
So this means that when the electron went from point A to point B, the electric potential increased by 1975 volts. This means that point B is the higher potential.

This looks correct.

And this makes sense because if you do w = Δv * q where Δv is 1975 Volts and q is negative 1.6*10^-19, you get negative 3.16*10^-16.
Is this right?

If w represents the change in potential energy, then, yes.

And if so, if the work is equal to the change in potential energy, then that means the work done is negative 3.16*10^-16. What does this negative work mean exactly?

The work done by the electric force is the negative of the change in potential energy. Think about the direction of the force on the electron as it accelerates. Is the force in the same direction as the displacement? If so, is the work done by the force positive or negative?

 

[Abu]

What does w represent? Is it the potential energy, or is it the change in potential energy, or is it something else?

This statement is not true in general. For a conservative system, the change in KE equals the negative of the the change in potential energy.

This looks correct.

If w represents the change in potential energy, then, yes.

The work done by the electric force is the negative of the change in potential energy. Think about the direction of the force on the electron as it accelerates. Is the force in the same direction as the displacement? If so, is the work done by the force positive or negative?

Thank you for your response. For the first question, I thought it represents the change in electric potential energy. Pertaining to your last question, when it is said like that I assume that the work is the opposite sign of the change of potential energy. So since in this case the change in potential energy is negative, the work done by the electric force is positive.

Also, I found the concept that the electron gains electric potential (voltage) yet loses electric potential energy confusing to understand. Are you able to explain that to me? Thank you very much

 

[TSny]

Thank you for your response. For the first question, I thought it represents the change in electric potential energy.

OK, that would be right. That is, w = q Δv is correct if w is the change in potential energy. It might be appropriate to use Δw to denote change in potential energy instead of w. It's also a little confusing to use w for potential energy, since a lot of people would use w for work. Many textbooks use U for potential energy. Thus, ΔU = q Δv.

Pertaining to your last question, when it is said like that I assume that the work is the opposite sign of the change of potential energy. So since in this case the change in potential energy is negative, the work done by the electric force is positive.

Yes. That should make sense. The only way the electron can gain kinetic energy is if the net work done on it is positive.

Also, I found the concept that the electron gains electric potential (voltage) yet loses electric potential energy confusing to understand. Are you able to explain that to me?

It's important not to confuse electric potential and electric potential energy. The best thing to do is to study carefully the definition of electric potential v. It is defined such that Δv = ΔU/q, where ΔU is the change in electric potential energy of the charge q, and Δv is the corresponding potential difference. So, you can see that if q is negative, ΔU and Δv have opposite signs.

 

[Abu]

OK, that would be right. That is, w = q Δv is correct if w is the change in potential energy. It might be appropriate to use Δw to denote change in potential energy instead of w. It's also a little confusing to use w for potential energy, since a lot of people would use w for work. Many textbooks use U for potential energy.

Yes. That should make sense. The only way the electron can gain kinetic energy is if the net work done on it is positive.

It's important not to confuse electric potential and electric potential energy. The best thing to do is to study carefully the definition of electric potential. It is defined such that ΔU = Δv/q, where ΔU is the change in potential energy of the charge q. So, you can see that if q is negative, ΔU and Δv have opposite signs.

I thought that ΔU = Δv*q not Δv/q. I realize that through the formula it is apparent that they will have opposite signs but it's hard for me to imagine a scenario where that is applicable and thus I cannot explain the idea

 

[TSny]

I thought that ΔU = Δv*q not Δv/q.

Yes, I initially wrote it wrong. Thanks.

I realize that through the formula it is apparent that they will have opposite signs but it's hard for me to imagine a scenario where that is applicable and thus I cannot explain the idea

It takes a while to get these concepts sorted out. Potential V is associated with points in space and is independent of q. But potential energy U depends on both V and q through the equation U = qV.

ΔV between two points is determined solely by the electric field between the two points. You've seen pictures of electric field lines. Points of higher electric potential V are located "upstream" in the electric field while points of lower V are located "downstream".

So, point a is at a higher potential than b. For example, point a might be at a potential V_a = 100 V. Point b might be at a potential V_b = 40 V. In going from a to b, the potential drops by 60 V. That is, ΔV = V_b - V_a = -60 V.

If I now take a point charge q and place it at point a, it will have a potential energy U = qV_a. Likewise, if it is placed at b it will have a potential energy U_b = qV_b. So, if the charge is moved from a to b, the change in potential energy is ΔU = qΔV. You can see that in going from a to b, the sign of the change in potential energy ΔU depends on both the sign of the charge q as well as the sign of the potential difference ΔV. But ΔV does not depend on q. It depends only on the electric field between the points a and b.

Hope this helps.