# Electric Potential can be added together

1. Mar 3, 2004

### Moose352

My textbook says electric potential (a scalar quantity) can be added together instead of vectorially adding the electric field to find the voltage when there are two fields. I am a little confused by this. In one example, two opposite charges (+/- 50 uC) of equal magnitude are seperated by a small distance (52cm). What is the voltage at a point equally distant from the two charges. According to the book, you would simply at the two potentials two get the a final answer of 0 V. This makes no sense to me at all, since the field line diagram shows that there is a force exerted at the point equidistant from both charges. So what is the problem here?

2. Mar 3, 2004

### chroot

Staff Emeritus
Truth be told, the question you asked has no answer. You are free to define the "zero voltage" anywhere you like. You can call the point equidistant between the two charges zero volts if you'd like, but you could just as well call it 40 * 10^9 V. What matters is not the actual value of the potential at a point, but the gradient or slope of the potential at that point.

To give an analogy, if you rest a ball on the top of a symmetrical hill, it won't move. It certainly is at a higher (gravitational) potential than the valley below. It does not move, though, because in its immediate vicinity, the hill is approximately flat. All that matters to the ball is the slope in its immediate vicinity, not how high or low it is relative to something else. No gradient -> no force.

In the case of your two charges, a test particle will definitely feel a force -- a negative test particle will be attracted to the positive side and repelled by the negative side. Therefore, there exists a potential gradient at that point. The actual value of the potential, however, is arbitrary.

- Warren

3. Mar 3, 2004

### Moose352

I understand what you are saying, and I am still thinking about it, but I'm not sure if you answered my real question: Can you add the eletric potentials without concern for direction instead of using the electric field vectors?

4. Mar 3, 2004

### chroot

Staff Emeritus
Yes, the potential is a vector field, and can be superposed. You simply add together the potentials due to several individual sources.

- Warren

5. Mar 4, 2004

### HallsofIvy

Staff Emeritus
Uhh, Warren, did you mean to say "a scalar field"?

6. Mar 4, 2004

### Moose352

I'm confused...

7. Mar 4, 2004

### Sammywu

In a high school or university grade of Physics, E potential is treated as a scalar field; the spatial differential of the scalar field is a vector called E field.

E field times by E charge is the force felt by the object.

The potential is zero at the point, but the infinitesmal close point has a different potentail. This difference creates the force field.

8. Mar 4, 2004

### lethe

it is the derivative of the potential that tells you the force, not the value itself.

so even though the potential is 0, its derivative is not, so there is a force.

9. Mar 4, 2004

### lethe

of course this question has an answer...

yes, there is a degree of freedom in the zero of potential, but there is also a convention for fixing this degree of freedom, so that people can make meaningful statements about the value of the electric potential.

that convention is to set V=0 at infinite radius from all charges in your system.

with this convention, the question makes perfect sense, and has a well-defined answer.

10. Mar 4, 2004

### lethe

perhaps chroot was thinking that the electric potential is the scalar component of a Lorentz vector.

while true, this seems rather extraneous for an electrostatics problem, however.

11. Mar 4, 2004

### Moose352

I don't understand how the derivative can be nonzero but the potential can be zero.

I am looking at it from a (perhaps overly) simplistic view. According to the book's answer, the eletric potential (and therefore the potential energy of a charge) at that point is zero. This, however, seems contradictory to the fact that a charge placed in the field will in fact gain a kinetic energy (a force is exerted on it), and therefore must have had potential energy. What am I missing here??

I am still not convinced with the scalar field. I can not beleive that the direction of the potential can altogether be disregarded when calculating potential.

12. Mar 4, 2004

### lethe

yes, that is right

think of potential energy (and also electric potential) like a hill, that the charged particles roll up and down.

if the hill that we are rolling down looks like y=x2, then when the ball is at the bottom and not moving then it will not feel any force. the derivative of this function is zero at the bottom.

but what if the function is like

$$y=\frac{x}{x^2-1}$$

this function is pretty much what the potential looks like between two charges. notice that when x=0, y is also =0, so the potential right between the two charges is 0

on the other hand, the derivative of this function is nonzero at x=0. this corresponds to the fact that the force is nonzero, even though the potential is zero.

or, take a much simpler example: if the electric field is constant, like inside a capacitor, then the electric potential rises linearly, it looks like V=ax. this is zero at x=0, but still the force is not zero there.

think of the picture of a hill. the graph of y=ax just keeps falling off, there is lots of room to keep falling. it is the same with the potential between the two charges.

scalars don't have direction. i am not sure what you mean here.

13. Mar 4, 2004

### turin

Unlike kinetic energy, which must always be positive semidefinite in order to have a real-valued speed, potential energy may be as negative as it wants to be. This is sometimes called guage invariance. What ever you call it, the point is that you can set zero potential whereever you want. For instance, it is customary to set zero altitude/elevation at sea level. But Death Valley in California, the Dead Sea Valley in Isreal, and probably a few other famous places on Earth have a negative elevation. This is not physically impossible, nor does it mean that these places are literally under water. It is an artifact of the convention to set the zero point at oceanic sea level. And even this can change drastically in the presence of drastic tides (or the great flood of Noah). If you put a ball on the edge of Massada, it will have no problem rolling down to the valley floor below.

The bottom line is that the electric charges don't know what the value of the potential is at all; they have a rather bad case of myopia. All they know is that there is a direct they can go to lower their potential, so that's the direction they want to go.

14. Mar 4, 2004

### Staff: Mentor

I think lethe and others have explained things well, but let me add a few comments.
That's like saying you don't understand how a function (Y = f(x)) could have an non-zero slope at the point where y = 0.

Perhaps you are thinking that potential = 0 is the rock bottom. But no, it's just 0 relative to a standard reference point; the potential can certainly be negative.
I does have potential energy! The potential is zero in the middle, but closer to the negative charge it's even less. (It's negative.) So the particle just rolls down the potential hill. No mystery there.
The potential, being a scalar field, has no "direction". But the direction of the electric field is certainly contained in the potential, so you aren't disregarding it.

Think this way. You find the field by taking the derivative of the potential function. If you have two functions (two potentials), does it matter if you add them first, then take the derivative? Of course not. (The derivative of the sum equals the sum of the derivatives.)