Jackson equation 2.9; How to derive it? I wanted to understand the derivation of Jackson's equation (2.9) that you can find here on page 31 (50/661) from the potential given earlier. This equation is written as:
$$
\textbf{F}=\frac{q}{y^2}\left[Q-\frac{qa^3(2y^2-a^2)}{y(y^2-a^2)^2}\right]\frac{\textbf{y}}{y}
$$
Does anybody know how to do this as there are apparently many steps left out in between?

The problem is to find the force a charge $q$ feels from the image charge it produces on a conducting isolated sphere containing a total charge $Q$. We are given in 2.8 that the potential is
\begin{equation}
\Phi(\mathbf{x})=\frac{q}{|\mathbf{x} - \mathbf{y}|} - \frac{rq}{|\mathbf{x} - r^2 \mathbf{y}|} + \frac{Q + rq}{|\mathbf{x}|},
\end{equation}
where $r$ the radius of the sphere in units of the distance of the charge $q$ from the sphere.
To find the force on the charge $q$, we should differentiate this expression to get the electric field. However, we need not consider the first term since it is the term generated by the charge $q$ and so it represents a self-force and must be zero.
Keep the other to terms we find the electric field from the sphere at a point $\mathbf{x}$ is 
\begin{equation}
\mathbf{E}(\mathbf{x})=\left( - \frac{rq}{(\mathbf{x} - r^2 \mathbf{y})^2} + \frac{Q+rq}{x^2}\right)\hat{y}
\end{equation}
Since we are interested in the force on the charge $q$, we should evaulate this expression at the point $\mathbf{y}$. We find
\begin{equation}
\begin{aligned}
\mathbf{E}(\mathbf{y})&=\left( - \frac{rq}{(\mathbf{y} - r^2 \mathbf{y})^2} + \frac{Q+rq}{y^2}\right)\hat{y} \\
&=\left( - \frac{rq}{(1 - r^2)^2} + Q+rq\right)\frac{\hat{y}}{y^2} \\
&= \left(  \frac{-rq + rq(1-r^2)^2}{(1 - r^2)^2} + Q\right)\frac{\hat{y}}{y^2} \\
&= \left(Q + rq\frac{(1-r^2)^2-1}{(1 - r^2)^2} \right)\frac{\hat{y}}{y^2} \\
&= \left(Q + rq\frac{r^4-2r^2}{(1 - r^2)^2} \right)\frac{\hat{y}}{y^2} \\
&= \left(Q + r^3q\frac{r^2-2}{(1 - r^2)^2} \right)\frac{\hat{y}}{y^2} \\
&= \left(Q - r^3q\frac{2-r^2}{(1 - r^2)^2} \right)\frac{\hat{y}}{y^2}
\end{aligned}
\end{equation}
Plugging in $\mathbf{F} = q\mathbf{E}$, we get 2.9.
