Two point charges +10 q and -4 q are located at $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}$ respectively.…
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 $)$
$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ right to pt.B
$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ left to pt.A
$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ right to pt.B
$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ left to pt.A
Solution
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 $B$