Two point charges +8 q and -2 q are located at $\mathrm{X}=0$ (origin) and $\mathrm{X}=\mathrm{L}$…
Two point charges +8 q and -2 q are located at $\mathrm{X}=0$ (origin) and $\mathrm{X}=\mathrm{L}$ respectively. The net electric field due to these two charges is zero at point P on X -axis. The location of point P from the origin is
$\frac{\mathrm{L}}{4}$
2 L
4 L
8 L
Solution
$\begin{array}{ll}
& \frac{1}{4 \pi \varepsilon_0} \frac{8 q}{(L+d)^2}-\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{d^2}=0 \\
& (L+d)^2=4 d^2 \\
\therefore \quad & L=d
\end{array}$
Hence, the distance from the origin is 2 L .