Potential gradient in the first case $=\frac{E_{0}}{l}$

$E=\left(\frac{l}{3}\right) \cdot\left(\frac{E_{0}}{l}\right)=\frac{E_{0}}{3}….(\mathrm{i})$

Potential gradient in the second case.

$=\frac{E_{0}}{3 l / 2}=\frac{2 E_{0}}{3 l}$

$\therefore E=(x) \frac{2 E_{0}}{3 l}$$...(II)$

From equations $(i)$ and $(ii),$ we get

$\frac{E_{0}}{3}=\left(\frac{2 E_{0}}{3 l}\right) x$

$\Rightarrow x=\frac{1}{2}$