Two point charge  $q_1\left(\sqrt{10}\mu C\right)$ and $q_2(-25\mu C)$ are placed on the x-axis at $x=1m$ and $x=4,$ respectively. The electric field (in $V/m$) at a point y=3m on y–axis is $(Take,\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{}N{m}^2{C}^{-2})$

Let,  $q_1=\sqrt{10}\mu C=\sqrt{10}\times {10}^{-6}C$

$q_2=-25\mu C=-25\times {10}^{-6}C$

Let $E_{1}and{E}_2$ are the values of electric field due to $q_{1}and{q}_2$ respectively .

Here ,

$E_1=\frac{1}{4\pi {\in }_0}\frac{q_1}{A{C}^2}=\frac{1}{4\pi {\in }_0}\times \frac{\sqrt{10}\times {10}^{-6}}{[1^2+3^2]}$

$=9\times {10}^9\times \sqrt{10}\times {10}^{-7}=9\sqrt{10}\times {10}^2$

$E_1=9\times \sqrt{10}\times {10}^2\left[\cos {\theta }_1(-\widehat{i})+\sin {\theta }_1\widehat{j}\right]$

$\sin {\theta }_1=\frac{3}{\sqrt{10}},\cos {\theta }_1=\frac{1}{\sqrt{10}}$

$\therefore E_1=9\sqrt{10}\times {10}^2\left[\frac{1}{\sqrt{10}}(-\widehat{i})+\frac{3}{\sqrt{10}}\widehat{j}\right]$

$=9\times {10}^2[-\widehat{i}+3\widehat{j}]=(-9\widehat{i}+27\widehat{j})\times {10}^2$

$E_2=\frac{1}{4\pi {\in }_0}\frac{-25\times {10}^{-6}}{(4^2+3^2)}=9\times {10}^{3}v/m$

$E_2=9\times {10}^3[\cos {\theta }_2\widehat{i}-\sin {\theta }_2\widehat{j}]$

$=9\times {10}^3[\frac{4}{5}\widehat{i}-\frac{3}{5}\widehat{j}]=(72\widehat{i}-54\widehat{j})\times {10}^2$

$E=E_1+E_2=(63\tilde{i}-27\widehat{j})\times {10}^{2}v/m$