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modphysnoob

V(r)= - a e^2 lambda exp(-r/p) ---------------- + 4 pi epilson r

  • one year ago
  • one year ago

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  1. modphysnoob
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    show that force can be expressed as F(r) = a e^2 (-(r_0)^2/r) ------------------- 4 pi epilson r_0^2 + exp(- r -r_0/ rho))

    • one year ago
  2. modphysnoob
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    @Jemurray3

    • one year ago
  3. modphysnoob
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    so do I just take derivative

    • one year ago
  4. Jemurray3
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    erm

    • one year ago
  5. Jemurray3
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    \[ V(r) = \frac{-ae^2}{4\pi \epsilon_0 r^2} + \lambda e^{-r/p} \]?

    • one year ago
  6. modphysnoob
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    yes sir

    • one year ago
  7. Jemurray3
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    Yes, \[ \vec{F} = -\vec{\nabla} V \]

    • one year ago
  8. Jemurray3
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    In your case, since the potential is a function only of r, \[ \vec{F} = -\frac{dV}{dr} \hat{r} \]

    • one year ago
  9. modphysnoob
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    so , I get \[\frac{a e^2}{2 \pi r^3 \epsilon }-\frac{e^{-\frac{r}{p}} \lambda }{p}\]

    • one year ago
  10. modphysnoob
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    which must be equal to give F

    • one year ago
  11. Jemurray3
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    I don't know... your notation is rather confusing to me so I'm not quite sure what it says.

    • one year ago
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