Two charges $+5\mathrm{\mu C}$ and $+10\mathrm{\mu C}$  are placed 20 cm apart. The net electric field at the mid-Point between the two charges is 

From following figure, 

${\mathrm{E}}_{\mathrm{A}}=\text{ Electric field at mid point }\mathrm{M}\text{ due to }+5\mathrm{\mu C}\text{ charge }$

$=9\times {10}^9\times \frac{5\times {10}^{-6}}{(0.1{)}^2}=45\times {10}^5\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_{\mathrm{B}}= \mathrm{Electric} \mathrm{field} \mathrm{at} \mathrm{M} \mathrm{due} \mathrm{to} +10\mathrm{\mu C} \mathrm{charge}$

$=9\times {10}^9\times \frac{10\times {10}^{-6}}{(0.1{)}^2}=90\times {10}^5\mathrm{N}/\mathrm{C}$

Net electric field at 

$\mathrm{M}=\left|{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}\right|-\left|{\overrightarrow{\mathrm{E}}}_{\mathrm{A}}\right|=45\times {10}^5\mathrm{N}/\mathrm{C}=4.5\times {10}^6\mathrm{N}/\mathrm{C},$

in the direction  of  ${\mathrm{E}}_{\mathrm{B}} \mathrm{i}.\mathrm{e}.\mathrm{towards} +5\mathrm{\mu C} \mathrm{charge}$