Answer
(a) The figure is shown below.
(b) $V = \dfrac{q}{4\pi \epsilon_{\circ}} \left[\dfrac{1}{|x|} - \dfrac{2}{|x-a|} \right]$
(c) ($x$ = -$a$) and $(x = \dfrac{a}{3})$
Work Step by Step
(a) The positions of the charges are shown in the figure below.
(b) At $x>a $, \\
For charge $q$, the distance is $x$, and for $-2q$ the distance is $(x-a)$. Hence, the potential is
\begin{gather*}
V = \dfrac{1}{4\pi \epsilon_{\circ}} \sum_{i} \dfrac{q_{i}}{r_{i}}\\
\boxed{V = \dfrac{1}{4\pi \epsilon_{\circ}} \left[\dfrac{q}{x} + \dfrac{-2q}{x-a} \right]} \\
\end{gather*}
