The E-x graph due to the field of two equal and opposite charges given below is 
 

                          
 

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

$\overline{\mathrm{E}}=\overline{{\mathrm{E}}_1}+\overline{{\mathrm{E}}_2}$

At a point x < -a, field is 

$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{Q}}{{(\mathrm{x}-\mathrm{a})}^2}-\frac{\mathrm{Q}}{{(\mathrm{x}+\mathrm{a})}^2}\right)$

At a point -a <x < a, field is 

$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{Q}}{{(\mathrm{x}-\mathrm{a})}^2}+\frac{\mathrm{Q}}{{(\mathrm{x}+\mathrm{a})}^2}\right)$

At a point x > a, field is 

$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{Q}}{{(\mathrm{x}+\mathrm{a})}^2}-\frac{\mathrm{Q}}{{(\mathrm{x}-\mathrm{a})}^2}\right)$

From the obtained relations, the graph will be