Two point charges $q_1\left(\sqrt{10}\mu c\right)$  and  $q_2\left(-25\mu c\right)$ are placed on the  $x-axis$ at $x=1m$  and  $x=4m$ respectively the electric field $\left(inV/m\right)$  at point $y=3m$ on $y-axis$  is  $\left(take\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N-m^2/C^2\right)$

$$\begin{gathered} Let E_1 \& E_2 are the values of electric field due to q_1 and q_2 respectively \\ Magnitude of E_2 =\frac{q_2}{4\mathrm{\pi }\in {\mathrm{r}}^2} \\ \Rightarrow E_2= \frac{9\times {10}^9\times 25\times {10}^{-6}}{(4^2+3^2)} \\ \Rightarrow E_2 = 9\times {10}^3 \\ In vector form, E_2 = 9\times {10}^3 (\cos {\theta }_2 \hat{i} - \sin {\theta }_2 \hat{j}) \\ u\sin g trigonometry, \cos {\theta }_2 = 4/5 and \sin {\theta }_2 = 3/5 \\ Hence, E_2 = (72\hat{i}-54\hat{j}) \times {10}^2 \\ Magnitude of E_1 =\frac{q_1}{4\mathrm{\pi }\in {\mathrm{r}}^2} \\ \Rightarrow E_1= \frac{9\times {10}^9\times \sqrt{10}\times {10}^{-6}}{(1^2+3^2)} \\ \Rightarrow E_1 = 9\sqrt{10}\times {10}^2 \\ In vector form, E_1 = 9\sqrt{10}\times {10}^2 (-\cos {\theta }_1 \hat{i} + \sin {\theta }_1 \hat{j}) \\ u\sin g trigonometry, \cos {\theta }_1 = 1/\sqrt{10} and \sin {\theta }_1 = 3/\sqrt{10} \\ Hence, E_1 = (-9\hat{i}+27\hat{j}) \times {10}^2 \\ So, net electric field is the vector sum \\ {E}_{net}=E_1+E_2 \\ = (63\hat{i}-27\hat{j}) \times {10}^2 \end{gathered}$$