# Relation F=-dv/dx

1. Jun 27, 2012

### sciboudy

how can we prove this relation F= -dv/dx
could some one explain what we mean by the force equal to the change it potential per distance and

2. Jun 27, 2012

### ZapperZ

Staff Emeritus
Your question is ambiguously presented, because it appears as if "v" is velocity, rather than "V" as in electrical potential difference.

You should know that the electric field E is $E = -d\Phi /dx$ in 1-dimension. Since F=qE, then $F=qE= -q d\Phi /dx$, where $\Phi$ is the electrostatic potential. But $q \Phi$ is V, the potential difference. Thus, F=- dV/dx.

In 3D, the derivative in 1D is replaced by the grad operator.

Zz.

3. Jun 27, 2012

### truesearch

If you accept that energy (work done) is basically Force x distance then:
Force x distance = change in energy
F x dx = dE so F = dE/dx

4. Jun 28, 2012

### sciboudy

OK thank you sir ZApperz and thank you truesearch
now i have another question based in the meaning of potential in the quantum mechanics
what is the potential for example whe i say a particle have V=0 ? is that mean the particle will never stop any time ?? or what