Explanation

(a) From following figure,
$E_{A} =$ Electric field at mid point $M$ due to $+ \ 5\mu C$ charge
$= 9 \times 10^{9} \times \frac{5 \times 10^{- 6}}{(0.1)^{2}} = 45 \times 10^{5}N/C$
$E_{B} =$ Electric field at $M$ due to $+ \ 10\mu C$ charge
$= 9 \times 10^{9} \times \frac{10 \times 10^{- 6}}{(0.1)^{2}} = 90 \times 10^{5}N/C$
[image:1_image362.emf]
Net electric field at
$M = \left| \overrightarrow{E_{B}} \right| - \left| \overrightarrow{E_{A}} \right| = 4.5 \times 10^{5}N/C = 4.5 \times 10^{6}N/C,$ in the direction of $E_{B}\ i.e.$ towards $+ \ 5\mu C$ charge