Two point charges +10 q and -4 q are located at \(\mathrm{x}=0\) and \(\mathrm{x}=\mathrm{L}\) respectively. What is the location of a point on the x -axis from the origin, which the net electric field due to these two point charges is zero? \((\mathrm{r}=\) required distance \()\)

Let ' \(Q\) ' be the point at a distance \(r\) from \(-4 q\) and \((\mathrm{L}+\mathrm{r})\) from 10 q . wherein net electric field is zero.
\(\begin{aligned}
& E_1=E_2 \\
& \frac{K(10 q)}{(L+r)^2}=\frac{K(4 q)}{r^2}
\end{aligned}\)
\(\begin{aligned}
& \frac{\sqrt{10}}{\mathrm{~L}+\mathrm{r}}=\frac{2}{\mathrm{r}} \\
& \frac{\sqrt{5} \times \sqrt{2}}{\mathrm{~L}+\mathrm{r}}=\frac{\sqrt{2} \times \sqrt{2}}{\mathrm{r}}=\frac{2}{\mathrm{r}} \\
& \sqrt{5} \times \sqrt{2} \times \mathrm{r}=2(\mathrm{~L}+\mathrm{r}) \\
& \sqrt{5} \times \sqrt{2} \times \mathrm{r}=\sqrt{2} \times \sqrt{2}(\mathrm{~L}+\mathrm{r}) \\
& \sqrt{5} \mathrm{r}=\sqrt{2} \mathrm{~L}+\sqrt{2} \mathrm{r} \\
& \sqrt{5}-\sqrt{2} \mathrm{r}=\sqrt{2} \mathrm{~L} \\
& \mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}
\end{aligned}\)
If \(B\) is position of charge \(-4 q\) then point \(Q\) is \(r=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}\) to the right of \(B\)