Derive Electrostatic Force from Coulomb's Law.

[jhosamelly]

Homework Statement

I should derive

[itex]\vec{F_{q}} = \frac{-q'^{2}}{4∏ \epsilon_{0}} \frac{r'/a}{(r'^{2} - a^{2})} \hat{r}[/itex]

from

[itex]\vec{F_{q''q'}} = k\frac{q' q''}{\left|\vec{x'} - \vec{x''}\right|^{2}}[/itex]

Homework Equations

The Attempt at a Solution

I know that [itex] q'' = -q' \frac{a}{r'} [/itex]

and [itex] r'' = \frac{a^{2}}{r'} [/itex]


so, I substituted these. but I can't seem to arrive at the correct formula.

 

[Nate810]

\vec{F_{q''q'}} = k\frac{q' q''}{\left|\vec{x'} - \vec{x''}\right|^{2}}= k\frac{q' (-q'\frac{a}{r'})}{\left|\vec{x'} - \frac{a^{2}}{r'}\hat{r'}\right|^{2}}= k\frac{-q'^{2}\frac{a^{2}}{r'^{2}}}{\left|\vec{r'} - \frac{a^{2}}{r'}\hat{r'}\right|^{2}}= k\frac{-q'^{2}\frac{a^{2}}{r'^{2}}}{\left|\vec{a} - a\hat{r'}\right|^{2}}= k\frac{-q'^{2}\frac{a^{2}}{r'^{2}}}{\left|\vec{a} - a^{2}\frac{1}{r'}\hat{r'}\right|^{2}}= k\frac{-q'^{2}\frac{a^{2}}{r'^{2}}}{\left|\vec{a} - r'\hat{r'}\right|^{2}}= k\frac{-q'^{2}\frac{a^{2}}{r'^{2}}}{(r'^{2} - a^{2})^{2}}I'm stuck here.