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

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

Physics
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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))
so do I just take derivative

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erm
\[ V(r) = \frac{-ae^2}{4\pi \epsilon_0 r^2} + \lambda e^{-r/p} \]?
yes sir
Yes, \[ \vec{F} = -\vec{\nabla} V \]
In your case, since the potential is a function only of r, \[ \vec{F} = -\frac{dV}{dr} \hat{r} \]
so , I get \[\frac{a e^2}{2 \pi r^3 \epsilon }-\frac{e^{-\frac{r}{p}} \lambda }{p}\]
which must be equal to give F
I don't know... your notation is rather confusing to me so I'm not quite sure what it says.

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