Consider a circular loop that is uniformly charged and has a radius a \sqrt2 . Find the position along the positive z -axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in x y plane at the origin :

Electrostatics PYQ — JEE Main 2025 · 2 Apr · Shift 2 · Q66 with Solution

View written solutionFree

Correct answer: A

For a ring of radius $R$ and total charge $Q$ , the electric field at a point on its axis at distance $z$ from the center is

E(z)=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(z^2+R^2)^{3/2}}.

Here the given radius is

R=a\sqrt{2}.

So,

E(z)=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{\left(z^2+2a^2\right)^{3/2}}.

We need to maximize

f(z)=\frac{z}{(z^2+R^2)^{3/2}}.

Differentiate:

\frac{df}{dz}=(z^2+R^2)^{-3/2}+z\left(-\frac{3}{2}\right)(z^2+R^2)^{-5/2}(2z).

Simplifying,

\frac{df}{dz}=(z^2+R^2)^{-5/2}\left[(z^2+R^2)-3z^2\right].

\frac{df}{dz}=(z^2+R^2)^{-5/2}(R^2-2z^2).

For maximum field,

R^2-2z^2=0.

Thus,

2z^2=R^2

z=\frac{R}{\sqrt{2}}.

Since we need the point on the positive $z$ -axis,

z=\frac{R}{\sqrt{2}}.

z=\frac{a\sqrt{2}}{\sqrt{2}}=a.

So the correct option is A.