Question

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

Answer 1

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))

Answer 2

@Jemurray3

Answer 3

so do I just take derivative

Answer 4

erm

Answer 5

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

Answer 6

yes sir

Answer 7

Yes,
\[ \vec{F} = -\vec{\nabla} V \]

Answer 8

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

Answer 9

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

Answer 10

which must be equal to give F

Answer 11

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