Confused about Electric potential energy / potential difference / electric potential

  • #1

Homework Statement


My textbook is no help, my teacher is no help, so Ive found myself here. I am having a difficult time understanding the differences in
1.Electric Potential energy
2. Potential Difference
3. Electric potential
Im sure this question has been asked tons of times but I need someone to try to describe this in laymans terms.

Homework Equations




The Attempt at a Solution


Ill tell you what I think I know so far:
Electric potential energy=F⋅ds=qE⋅ds, where ds is the displacement. Just like for gravity U=mgh.

The definition my textbook gives for electric potential is:
V(electric potantial)=U/q, is this U electric potential energy?

I assume potential difference is ΔV

any help would be much appreciated!
thanks
 

Answers and Replies

  • #2
RPinPA
Science Advisor
Homework Helper
571
319
V(electric potantial)=U/q, is this U electric potential energy?

I assume potential difference is ΔV
Yes and yes.

Part of the distinction lies in the distinction between the force on a particle or the energy of that particle, and the things CAUSING that particle to have energy or to experience a force. Let's go back to the Coulomb force, ##F = k q_1 Q / r^2## as the force of charge ##q_1## on charge ##Q##. Of course, it's also the force of ##Q## on ##q_1##, but I'm going to treat ##Q## as my "test particle".

This is of the form F = (stuff) * Q. The force on Q can be expressed as a product of Q and a bunch of other stuff. And that's true when Q is being affected by a whole collection of other charges, ##F = \sum \left( k q_i Q/r_i^2 \right) = \left( \sum kq_i/r_i^2\right) Q##. So we can define something called the "electric field" E which is a property of all the other charges ##q_i##. It's ##F/Q##, the force per unit charge on Q, and it has no dependence on Q. And it can be used to predict the force on any other charge we bring up instead of Q.

Pedantic Edit: Force and electric field are vectors, that should really be a vector sum. ##\vec F = \left( \sum kq_i \hat r_i/r_i^2\right) Q##. But the point is still that it's ##\vec F = \text{(stuff)} * Q##.

The gravitational analog is ##g = GM/R^2##, the gravitational field of a planet ##M## of radius ##R## which is felt by any mass ##m## on that planet and is characterized by the planet ##M##, with no dependence on ##m##.

So that's the motivation for dividing by Q, that it gives a quantity which characterizes some charge distribution and can be used to predict the effects on other charges.

Now, any conservative force (don't worry about what that means if you haven't been taught it yet), which gravitational and electrical forces are (that's all you need to know right now), can be expressed as the negative gradient, the derivative with respect to position, of potential energy. That means that things want to decrease in energy. Where there's a downhill in energy, there's a force pointing down the hill. That gives rise to the integral relation you quoted ##dU = F \cdot ds##. Force is the derivative of potential energy, potential energy is the integral of force.

If you divide the force by Q so you only have the force per unit charge (they usually talk about a "test charge"), then that's related to the energy per unit test charge. F/Q gives you E, a quantity which characterizes the charge distribution and can be multiplied by any test charge to give the force on that test charge. U/Q gives you V, a quantity which also characterizes the charge distribution and which can be multiplied by any test charge to give the energy on that test charge.

You may or may not remember that potential energies are relative. You can put the zero wherever you want. Sometimes for gravitational potential energy the zero is on the floor and can the energy be positive (above the floor) or negative (below the floor). Sometimes it's infinitely far away from planet earth and relative to that all potential energies are negative. All that matters is the difference in potential energy between two points. And that's the PE of a particular particle. That tells you how much work you need to move the particle from point A to point B.

Analogously, U/q = V is a relative quantity. It's called the electric potential, but we can put the zero anywhere we want. All that matters is the difference. We call that potential difference or voltage difference and it's measured in Volts. Usually, the ground is a convenient reference point for zero voltage but it doesn't have to be. A bird can sit on a high-voltage line 10000 V above ground. because both feet are at 10,000 V. The potential difference across the bird is 0.

Does that help? Short answer:

Force on a test charge q vs Electric field = F/q which is all the terms caused by other things on that test charge
Potential energy U of a test charge vs Potential V = U/q which is all the terms caused by other things on that test charge
 
Last edited:
  • Like
Likes cookiemnstr510510 and cookiemnstr510
  • #3
Yes and yes.

Part of the distinction lies in the distinction between the force on a particle or the energy of that particle, and the things CAUSING that particle to have energy or to experience a force. Let's go back to the Coulomb force, ##F = k q_1 Q / r^2## as the force of charge ##q_1## on charge ##Q##. Of course, it's also the force of ##Q## on ##q_1##, but I'm going to treat ##Q## as my "test particle".

This is of the form F = (stuff) * Q. The force on Q can be expressed as a product of Q and a bunch of other stuff. And that's true when Q is being affected by a whole collection of other charges, ##F = \sum \left( k q_i Q/r_i^2 \right) = \left( \sum kq_i/r_i^2\right) Q##. So we can define something called the "electric field" E which is a property of all the other charges ##q_i##. It's ##F/Q##, the force per unit charge on Q, and it has no dependence on Q. And it can be used to predict the force on any other charge we bring up instead of Q.

Pedantic Edit: Force and electric field are vectors, that should really be a vector sum. ##\vec F = \left( \sum kq_i \hat r_i/r_i^2\right) Q##. But the point is still that it's ##\vec F = \text{(stuff)} * Q##.

The gravitational analog is ##g = GM/R^2##, the gravitational field of a planet ##M## of radius ##R## which is felt by any mass ##m## on that planet and is characterized by the planet ##M##, with no dependence on ##m##.

So that's the motivation for dividing by Q, that it gives a quantity which characterizes some charge distribution and can be used to predict the effects on other charges.

Now, any conservative force (don't worry about what that means if you haven't been taught it yet), which gravitational and electrical forces are (that's all you need to know right now), can be expressed as the negative gradient, the derivative with respect to position, of potential energy. That means that things want to decrease in energy. Where there's a downhill in energy, there's a force pointing down the hill. That gives rise to the integral relation you quoted ##dU = F \cdot ds##. Force is the derivative of potential energy, potential energy is the integral of force.

If you divide the force by Q so you only have the force per unit charge (they usually talk about a "test charge"), then that's related to the energy per unit test charge. F/Q gives you E, a quantity which characterizes the charge distribution and can be multiplied by any test charge to give the force on that test charge. U/Q gives you V, a quantity which also characterizes the charge distribution and which can be multiplied by any test charge to give the energy on that test charge.

You may or may not remember that potential energies are relative. You can put the zero wherever you want. Sometimes for gravitational potential energy the zero is on the floor and can the energy be positive (above the floor) or negative (below the floor). Sometimes it's infinitely far away from planet earth and relative to that all potential energies are negative. All that matters is the difference in potential energy between two points. And that's the PE of a particular particle. That tells you how much work you need to move the particle from point A to point B.

Analogously, U/q = V is a relative quantity. It's called the electric potential, but we can put the zero anywhere we want. All that matters is the difference. We call that potential difference or voltage difference and it's measured in Volts. Usually, the ground is a convenient reference point for zero voltage but it doesn't have to be. A bird can sit on a high-voltage line 10000 V above ground. because both feet are at 10,000 V. The potential difference across the bird is 0.

Does that help? Short answer:

Force on a test charge q vs Electric field = F/q which is all the terms caused by other things on that test charge
Potential energy U of a test charge vs Potential V = U/q which is all the terms caused by other things on that test charge
It is starting to make more sense. Let me chew on this for a little while and get back to you!
But thank you!
Appreciate your time and effort on this!
 
Top