The maximum electric field strength E due to a uniformly charged ring of radius r, happens at a distance x, where value of x is (x is measured from the centre of the ring)

Step 1: Given that:

A uniformly charged ring.

Strength of maximum electric field = E

The radius of the ring = r

Distance of the point from the centre of the ring at which the electric field is maximum = x

Step 2: Formula used:

The electric field due to the ring at its axis at a distance of x is given as

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

Where R is the radius of the ring, q is the charge on the ring and x is the distance of the point from the centre of the ring at which the electric field is calculated.

$\frac{d}{dx}\frac{f\left(x\right)}{g\left(x\right)}=\frac{g\left(x\right)\frac{d}{dx}f\left(x\right)-f\left(x\right)\frac{d}{dx}g\left(x\right)}{{\left\{g\left(x\right)\right\}}^2}$

Where, f(x) and g(x) are two different functions in x.

Step 3: Finding the value of x at which the electric field will be maximum:

To find the maximum electric field we will use the concept of maxima and minima

So,

For the maxima;

$\frac{dE}{dx}=0$

$\frac{d}{dx}\frac{kqx}{{(x^2+R^2)}^{\frac{3}{2}}}=0$

$kq\frac{d}{dx}\frac{x}{{(x^2+R^2)}^{\frac{3}{2}}}=0$

Now, using the formula for differentiation;

$kq\frac{{(x^2+R^2)}^{\frac{3}{2}}-\frac{3}{2}{(x^2+R^2)}^{\frac{1}{2}}2x^2}{(x^2+R^2)}=0$

$\frac{{(x^2+R^2)}^{\frac{3}{2}}-\frac{3}{2}{(x^2+R^2)}^{\frac{1}{2}}2x^2}{(x^2+R^2)}=0$

${(x^2+R^2)}^{\frac{3}{2}}-\frac{3}{2}{(x^2+R^2)}^{\frac{1}{2}}2x^2=0$

${(x^2+R^2)}^{\frac{3}{2}}=\frac{3}{2}{(x^2+R^2)}^{\frac{1}{2}}2x^2$

${(x^2+R^2)}^{\frac{3}{2}}=3{(x^2+R^2)}^{\frac{1}{2}}{x}^2$

$\frac{{(x^2+R^2)}^{\frac{3}{2}}}{{(x^2+R^2)}^{\frac{1}{2}}}=3x^2$

${(x^2+R^2)}^{\frac{3}{2}-\frac{1}{2}}=3x^2$

$(x^2+R^2)=3x^2$

Or,

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

$Or,x=\pm \frac{R}{\sqrt{2}}$

Hence,

For the electric field strength to be maximum x must be $\pm \frac{R}{\sqrt{2}}$ .

Thus,

Option C) $x=\frac{R}{\sqrt{2}}$ is the correct option.