Electric Potentials

[MedPR]

This is from NOVA Physics. I don't understand what the text outside of the box means.

Is it saying that as a charge q approaches a different charge Q, that graph represents the energy needed as distance, d, decreases? Is this graph showing the energy only for a positive charge q in an area with positive potential Q?

Also, V is electric potential right? And as you decrease the distance from charge Q, the potential increases? How does figure 14-23 show the value of V? For instance, looking only at the line on the left, the distance from Q is about the same at both endpoints of the line, so does that mean the potential is the same at those two points? Does the end point nearer the top of the figure have a greater potential than the end point nearer to the left side of the figure?

http://imgur.com/iTT6n

---

Members don't see this ad.

 

[milski]

Yes, it's showing the energy needed to move the charge from infinity (where the potential is considered 0, by convention) to distance d from the other charge. The graph shown is for charges with the same sign - the closer you get, the more energy you need to move there.

The graph for charges with opposite signs will be the mirror image of this one below the x-axis - getting closer in that case decreases your total energy. That's pretty much the same case as potential energy from gravity - same shape of the curve, same reasoning.

 

[MedPR]

.

 

[MedPR]

Yes, it's showing the energy needed to move the charge from infinity (where the potential is considered 0, by convention) to distance d from the other charge. The graph shown is for charges with the same sign - the closer you get, the more energy you need to move there.

The graph for charges with opposite signs will be the mirror image of this one below the x-axis - getting closer in that case decreases your total energy. That's pretty much the same case as potential energy from gravity - same shape of the curve, same reasoning.

So is this graph showing two different, positive charges approaching charge Q? Each line is the energy of one charge, q?

 

[TXKnight]

So is this graph showing two different, positive charges approaching charge Q? Each line is the energy of one charge, q?

Imagine you were to put that figure on a Cartesian plane (V vs distance) and you have distance (d) on your x axis, and that figure (14-23) is shown above your axis; then point Q can be your origin. Having said that, the figure is showing the energy as a charge approaches from the negative x to the point where d=0 (that is Q), then as the charge moves away from point Q to more positive values for x. So, it only shows the potential difference at point p with respect to Q as a positive charge q approaches and then moves away from Q.

 

[milski]

So is this graph showing two different, positive charges approaching charge Q? Each line is the energy of one charge, q?

Or one charge approaching from plus infinity, going through Q and disappearing in -infinity. It's the same thing anyway - look at what TXKnight said, he explained it in a slightly different way.

 

[MedPR]

Imagine you were to put that figure on a Cartesian plane (V vs distance) and you have distance (d) on your x axis, and that figure (14-23) is shown above your axis; then point Q can be your origin. Having said that, the figure is showing the energy as a charge approaches from the negative x to the point where d=0 (that is Q), then as the charge moves away from point Q to more positive values for x. So, it only shows the potential difference at point p with respect to Q as a positive charge q approaches and then moves away from Q.

Or one charge approaching from plus infinity, going through Q and disappearing in -infinity. It's the same thing anyway - look at what TXKnight said, he explained it in a slightly different way.

I understand this much. I'm confused about where the energy is indicated on those lines. It appears that the lines are showing the path a positive charge would take as it is approaches Q from infinity. So it starts on infinity in the x plane, then moves towards infinity in the y plane.

 

[milski]

I understand this much. I'm confused about where the energy is indicated on those lines. It appears that the lines are showing the path a positive charge would take as it is approaches Q from infinity. So it starts on infinity in the x plane, then moves towards infinity in the y plane.

There is no path shown on the graph. Not only that but the path should be irrelevant to energy, since we are talking about a conservative force.

The x axis is distance, or probably better to say, displacement from the Q charge. (Displacement since it has a sign). The y axis is the potential at such displacement and is the same things as the work that has to be done to bring the charge from infinity to that point.

If I was doing the graph, I would have done only the right half and called d a distance. It's not any more useful, the way they have it shown.

 

[MedPR]

There is no path shown on the graph. Not only that but the path should be irrelevant to energy, since we are talking about a conservative force.

The x axis is distance, or probably better to say, displacement from the Q charge. (Displacement since it has a sign). The y axis is the potential at such displacement and is the same things as the work that has to be done to bring the charge from infinity to that point.

If I was doing the graph, I would have done only the right half and called d a distance. It's not any more useful, the way they have it shown.

That makes sense, thank you. I was interpreting the graph to show the charge approaching Q in the x direction, then moving away from Q in the y direction.

So in this case, since the line is asymptotic, can we say that d is never 0?

 

[milski]

That makes sense, thank you. I was interpreting the graph to show the charge approaching Q in the x direction, then moving away from Q in the y direction.

So in this case, since the line is asymptotic, can we say that d is never 0?

The graph tells you what E is for any given d. For d being 0, E is infinity.

 

[TXKnight]

There is no path shown on the graph. Not only that but the path should be irrelevant to energy, since we are talking about a conservative force.

The x axis is distance, or probably better to say, displacement from the Q charge. (Displacement since it has a sign). The y axis is the potential at such displacement and is the same things as the work that has to be done to bring the charge from infinity to that point.

If I was doing the graph, I would have done only the right half and called d a distance. It's not any more useful, the way they have it shown.

This. Energy is a conservative force, and yes the x is displacement, so not really a diagram of the path taken (or something like that). Good luck.

 

[TXKnight]

That makes sense, thank you. I was interpreting the graph to show the charge approaching Q in the x direction, then moving away from Q in the y direction.

So in this case, since the line is asymptotic, can we say that d is never 0?

I believe that's correct.