Two point charges q 1 10   μC and q 2 - 25   μC are placed on the x -axis at x = 1 &#160…

Electric field due to $\sqrt{10} \mathrm{\mu C}$


${\overrightarrow{E}}_1=\frac{1}{4\pi {\in }_0}\frac{\sqrt{10}\times {10}^{-6}}{r_1^3} {\overrightarrow{r}}_1$

$=\frac{1}{4\pi {\in }_0}\frac{\sqrt{10}\times {10}^{-6}}{{\left(\sqrt{10}\right)}^3}\left(-\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}\right)$

Similarly, electric field due to $-25 \mathrm{\mu C}$

${\overrightarrow{E}}_2=\frac{1}{4\pi {\in }_0}\frac{25\times {10}^{-6}}{{\left(5\right)}^3}\left(4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}\right)$

Net electric field, 
$\overrightarrow{\mathrm{E}}=9\times {10}^9\times {10}^{-6}\left(\frac{-\widehat{\mathrm{i}}}{10}+\frac{3\widehat{\mathrm{j}}}{10}+\frac{4}{5}\widehat{\mathrm{i}}-\frac{3}{5}\widehat{\mathrm{j}}\right)$

$=9\times {10}^3\left(\frac{7\widehat{\mathrm{i}}}{10}-\frac{3\widehat{\mathrm{j}}}{10}\right)$

$=\left(63\widehat{\mathrm{i}}-27\widehat{\mathrm{j}}\right)\times {10}^2 \frac{\mathrm{N}}{\mathrm{C}}$