Electric field due to $$\sqrt {10} \,\mu C$$ charge :

$$\overrightarrow {{E_1}} = - $$ E₁ sin$$\theta $$₁ $$\widehat i$$ + E₁ cos$$\theta $$₁ $$\widehat j$$

Where,

E₁ $$ = {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{q_1}} \right|} \over {{r_1}^2}}$$

$$ = 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}$$

$$ = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m$$

sin $$\theta $$₁ $$=$$ $${1 \over {\sqrt {10} }}$$

and cos$$\theta $$₁ = $${3 \over {\sqrt {10} }}$$

$$ \therefore $$   $$\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)$$

$$ = 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)$$

Electric field due to $$-$$ 25 $$\mu $$C charge,



$$\overrightarrow {{E_2}} = $$ E₂ sin$$\theta $$₂$$\widehat i$$ $$-$$ E₂ cos$$\theta $$₂ $$\widehat j$$

where

E₂ $$ = {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{9_2}} \right|} \over {r_2^2}}$$

$$ = 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}$$

$$ = 9 \times {10^3}$$ V/m

sin$$\theta $$₂ = $${4 \over 5}$$

and cos$$\theta $$₂ = $${3 \over 5}$$

$$ \therefore $$   $$\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)$$

$$ = 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)$$

$$ \therefore $$   Net electric field,

$$\overrightarrow E $$ = $${\overrightarrow E _1}$$ + $${\overrightarrow E _2}$$

$$ = \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m$$