# Attractive potential

1. Jun 15, 2010

### Cosmossos

Hello
I'm so confused. Attractive potential is positive or negative?
My intuition is that an Attractive potential makes a particle go from low potential to high potential so it should be negative.
Is that right?
Do you have another intuitive explanation?

thanks

2. Jun 15, 2010

### nickjer

Potential energy? Or electric potential? Those can give you quite different answers. As for potential energy, unless you have a reference potential to compare with then you can't say for sure if it is positive or negative. You can only say that a particle at rest subjected to that potential will prefer to go to a lower potential.

For example a spring has a positive potential energy (1/2 kx^2). But it is still attractive. You can only say a particle at rest will want to go to the lowest point of the curve.

3. Jun 16, 2010

### Cosmossos

But I'm talking about a potential in the meaning of quantum mechanics.
If I'm told that a particle is in an attractive potenial . what's it sign?
thanks

4. Jun 16, 2010

### nickjer

A quantum harmonic oscillator is an attractive potential and it has a positive sign. A constant electric field can be attractive for certain charges and it's potential can also be positive. The only thing that holds true is that the particle at rest will want to go to a lower potential energy than it is currently at.

Most often though, the reference potential is 0 at r=infinity. That means to be attractive, the potential must be negative at 'r' less than infinity.

5. Jun 16, 2010

### Cosmossos

But why?
I got a question which says that a particle is in a repulsive potential -Vo (for r<a)
Why is -Vo a repulsive potential?

6. Jun 16, 2010

### nickjer

I just told you that the sign of the potential is meaningless. The only thing that matters is the shape of the potential. If the potential energy decreases as it approaches the object then it is an attractive potential. If it increases then it is a repulsive potential.

7. Jun 17, 2010

### ross_tang

Just the gradient of the potential matters

Please refer to the below formula:

$$\vec{F} = - \nabla V$$

Therefore we are just interested in the negative gradient of the potential, not it's sign.