Negative sign when finding forces from derivative of potential?

[21joanna12]

Hi! I'm currently reading a book where they give the Coulomb potential, gravitational potential and harmonic potential as

+Q1Q2/4∏εx

-Gm1m2/x

+(1/2)qx²

I think I get the signs as they are used here, but when I am trying to find the force by taking the derivative of these with respect to x, I don't know whether to to f= +dV/dx or f= -dV/dx for any of them and I can't figure out the significance of the sign- which way is defined as +ve/-ve and why?

Sorry if I posted this is the wrong place or if it is not very clear :/ This is my first post of PF...

Thank you for any replies! :)

 

[MisterX]

F = -dV/dx

This way the force points "down" the slope to a lower potential.

 

[Doc Al]

You might find this short discussion helpful: Negative Signs in Potential

 

[21joanna12]

Thank you! :)

 

[HalfLostAndSearching]

As a fellow scientist, I can assure you that you are not alone in your confusion about the negative sign when finding forces from the derivative of potential. In fact, this is a common source of confusion for many students and researchers in the field of physics.

The key to understanding the negative sign is to remember that forces are vectors, meaning they have both magnitude and direction. The negative sign simply represents the direction of the force.

In the case of the Coulomb potential, the negative sign indicates that the force between two charged particles is attractive, pulling them towards each other. Similarly, the negative sign in the gravitational potential indicates an attractive force between two masses.

On the other hand, the positive sign in the harmonic potential indicates a repulsive force, pushing the particle away from the equilibrium position.

Therefore, when taking the derivative of these potentials to find the force, it is important to include the negative sign to correctly represent the direction of the force.

I hope this helps clarify the significance of the negative sign and how it is defined in the context of finding forces from the derivative of potential. Keep up the good work in your studies and don't hesitate to ask for clarification when needed. Good luck!