
$\overrightarrow {\rm{E}} = \frac{{\overrightarrow {\rho {\rm{r}}} }}{{3{{\rm{e}}_0}}} - \frac{{{{\overrightarrow {\rho {\rm{r}}} }^\prime }}}{{3{{\rm{e}}_0}}}$
$\therefore \quad \overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{r}}^{\prime}=\overrightarrow{\mathrm{a}}$
$\therefore \overrightarrow {\rm{E}} = \frac{{\rho \overrightarrow {\rm{a}} }}{{3{ \in _0}}} = $ uniform



Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.