An electric field is acting vertically upwards. A small body of mass 1 gm and charge  -$1\mu C$  is projected with a velocity 10 m/s at an angle ${45}^0$ with horizontal. Its horizontal range is 2m then the intensity of electric field is : (g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

As charge is negative and the electric field is vertically upwards the total force acting on the charge is vertically downwards. 
 Hence the net acceleration downwards is $\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}$
$\mathrm{R}=\frac{{\mathrm{V}}^2\sin 2\mathrm{\theta }}{\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}}=\frac{{\mathrm{V}}^2}{\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}}\left(\mathrm{\theta }={45}^{\mathrm{o}}\right)$

$\mathrm{R}=\frac{100}{10+\frac{{10}^{-6}\mathrm{E}}{{10}^{-3}}}=10+{10}^{-3}\mathrm{E}=50$

$\mathrm{E}=40\times {10}^3\mathrm{N}/\mathrm{C}$