$V_A=\frac{k(100)}{a} \quad V_B=\frac{k(100)}{b}$
$a < b$
$V_a > V_B$
Charge flows from $A$ to $B$.

$V\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, < \, - d}\\
{ - {V_0}{{\left( {1 + \frac{x}{d}} \right)}^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\, - \,d\, \le x < 0}\\
{ - {V_0}\left( {1 + 2\frac{x}{d}} \right)\,\,\,\,\,\,\,\,\,\,\,for\,0\, \le x < d}\\
{ - 3{V_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, > \,d}
\end{array}} \right.$
where $-V_0$ is the potential at the origin and $d$ is a distance. Graph of electric field as a function of position is given as
$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$
$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$
