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modphysnoob
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V(r)=  a e^2 lambda exp(r/p)
 +
4 pi epilson r
 one year ago
 one year ago
modphysnoob Group Title
V(r)=  a e^2 lambda exp(r/p)  + 4 pi epilson r
 one year ago
 one year ago

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modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
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

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
@Jemurray3
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
so do I just take derivative
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
\[ V(r) = \frac{ae^2}{4\pi \epsilon_0 r^2} + \lambda e^{r/p} \]?
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
yes sir
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
Yes, \[ \vec{F} = \vec{\nabla} V \]
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
In your case, since the potential is a function only of r, \[ \vec{F} = \frac{dV}{dr} \hat{r} \]
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
so , I get \[\frac{a e^2}{2 \pi r^3 \epsilon }\frac{e^{\frac{r}{p}} \lambda }{p}\]
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
which must be equal to give F
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
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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