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Relation F=-dv/dx

  1. Jun 27, 2012 #1
    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. jcsd
  3. Jun 27, 2012 #2

    ZapperZ

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    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 [itex]E = -d\Phi /dx[/itex] in 1-dimension. Since F=qE, then [itex]F=qE= -q d\Phi /dx[/itex], where [itex]\Phi[/itex] is the electrostatic potential. But [itex]q \Phi [/itex] is V, the potential difference. Thus, F=- dV/dx.

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

    Zz.
     
  4. Jun 27, 2012 #3
    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
     
  5. Jun 28, 2012 #4
    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
     
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