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

1. modphysnoob

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))

2. modphysnoob

@Jemurray3

3. modphysnoob

so do I just take derivative

4. Jemurray3

erm

5. Jemurray3

$V(r) = \frac{-ae^2}{4\pi \epsilon_0 r^2} + \lambda e^{-r/p}$?

6. modphysnoob

yes sir

7. Jemurray3

Yes, $\vec{F} = -\vec{\nabla} V$

8. Jemurray3

In your case, since the potential is a function only of r, $\vec{F} = -\frac{dV}{dr} \hat{r}$

9. modphysnoob

so , I get $\frac{a e^2}{2 \pi r^3 \epsilon }-\frac{e^{-\frac{r}{p}} \lambda }{p}$

10. modphysnoob

which must be equal to give F

11. Jemurray3

I don't know... your notation is rather confusing to me so I'm not quite sure what it says.