Question

The ratio of the magnitude of electric field due to +q and – 2q at a point where net potential is zero on positive x-axis is z : 1, then z will be

Answer 1

We know that -
$V=\frac{kq}{r}$
$\frac{kq}{x}-\frac{2kq}{6-x}=0$
$\Rightarrow 6-x=2x$
$\Rightarrow x=2$
$E=\frac{kq}{r^2}$
So electric field due to +q
$E_q=\frac{kq}{(2{)}^2}=\frac{kq}{4}$
Electric field due to –2q
$E_{2q}=\frac{k2q}{(4{)}^2}=\frac{2kq}{16}=\frac{kq}{8}$
ratio is 2 : 1
So z = 2