Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=\mathrm{L}$ respectively. The location of a…
Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=\mathrm{L}$ respectively. The location of a point on the $x$-axis from the origin, at which the net electric field due to these two point charges is zero is
$\frac{\mathrm{L}}{4}$
4 L
8 L
2 L
Solution
Let ' $A$ ' be the point at a distance $r$ from $-2 q$ and at $(L+r)$ from $+8 q$ where net electric field is zero.
$\mathrm{E}_1=\mathrm{E}_2$
$\begin{array}{ll}
\therefore & \frac{\mathrm{K}(8 q)}{(L+r)^2}=\frac{K(2 q)}{r^2} \\
\therefore \quad & \frac{4}{(L+r)^2}=\frac{1}{r^2} \\
& 2 r=L+r \\
& r=L \\
& A=L+L=2 L
\end{array}$