Charge Q is uniformly distributed over a ring of radius r .Electric field at a point on the axis is maximum at a distance x from the centre. x is equal to

The correct option is A $\frac{r}{\sqrt{2}}$

The electric field due to the ring on its axis at a distance x is given by:-

$E=\frac{kqx}{(x^2+R^2{)}^{3/2}}$

To find maximum electric field, we will use the concept of maximum and minimum :-

$\frac{dE}{dx}=kq\frac{(x^2+R^2{)}^{3/2}-3/2(x^2+R^2{)}^{1/2}.2x^2}{(x^2+R^2)}$

Now, $\frac{dE}{dx}=0\Rightarrow (x^2+R^2{)}^{3/2}=\frac{3}{2}\times 2x^2(x^2+R^2{)}^{1/2}$

$\Rightarrow x^2+R^2=3x^2$

$\Rightarrow 2x^2=R^2$

$\Rightarrow x^2=\frac{R^2}{2}$

$\Rightarrow x=\pm \frac{R}{\sqrt{2}}$