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



This Question is Closed

modphysnoob
Best Response
You'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))

modphysnoob
Best Response
You've already chosen the best response.
0
@Jemurray3

modphysnoob
Best Response
You've already chosen the best response.
0
so do I just take derivative

Jemurray3
Best Response
You've already chosen the best response.
0
erm

Jemurray3
Best Response
You've already chosen the best response.
0
\[ V(r) = \frac{ae^2}{4\pi \epsilon_0 r^2} + \lambda e^{r/p} \]?

modphysnoob
Best Response
You've already chosen the best response.
0
yes sir

Jemurray3
Best Response
You've already chosen the best response.
0
Yes,
\[ \vec{F} = \vec{\nabla} V \]

Jemurray3
Best Response
You've already chosen the best response.
0
In your case, since the potential is a function only of r,
\[ \vec{F} = \frac{dV}{dr} \hat{r} \]

modphysnoob
Best Response
You've already chosen the best response.
0
so , I get
\[\frac{a e^2}{2 \pi r^3 \epsilon }\frac{e^{\frac{r}{p}} \lambda }{p}\]

modphysnoob
Best Response
You've already chosen the best response.
0
which must be equal to give F

Jemurray3
Best Response
You'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.