Question

Two point charges q${}_1$ and q${}_2$ are placed at a distance 'd' apart as shown in the figure. The electric field intensity is zero at a point 'P' on the line joining them as shown. If the ratio $\frac{q_1}{q_2}$ is $-{\left(\frac{xr+d}{r}\right)}^2$. Find the value of $x$.

Answer 1

Here, the electric field at P is $E_P=0$

1) Since $E=-\frac{dV}{dx},$ so $\frac{d{V}_P}{dx}=0$ . Thus the potential at P is maximum or minimum.

2) $E_P=k\frac{q_2}{r^2}+k\frac{q_1}{(r+d{)}^2}=0$

$\Rightarrow \frac{q_1}{q_2}=-{\left(\frac{r+d}{r}\right)}^2$