Question

The distance of point P on the axis from the Centre of a uniformly charged circular ring where the electric field is maximum, is $2^{-n}R$(given radius of the ring is R). The value of n in decimal form is___

Answer 1

Once you derive the electric field due to a circular ring at point on the axis you get $E=\frac{1}{4\pi {ϵ}_0}\frac{Qx}{(R^2+x^2{)}^{\frac{3}{2}}}$ .To be sure of this I recommend you to re-watch the derivation video.

So the value of x for which electric field becomes an extreme can be formed out by differentiating E w.r.t x and then putting it equal to zero (points where ‘slope’ becomes zero)

$\frac{dE}{dx}=0\Rightarrow \frac{d}{dx}\left(\frac{KQx}{(x^2+R^2{)}^{\frac{3}{2}}}\right)=0$

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

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

One must find the double derivative and only if it comes out to be negative, this will confirm that that the value obtained is a maxima.

I can tell you another analytical approach. You very well know that the field at the center of the uniformly ring is zero. It is also zero at infinity. The ring is positively charged, so negative field is not possible anywhere in between which suggests that it should attain a maximum value somewhere between the centre of the ring and infinity.

So the distance of point on the axis form the Centre where electric field is maximum is R.$2^{-0.5}$ on comparing with R.$2^{-n}$. So, n = 0.5