JEE Main 2025 — Electrostatics Question with Solution

To determine the position along the positive $z$-axis where the electric field due to a uniformly charged circular loop is at its maximum, we follow these steps:

Expression for Electric Field ($E$):

The electric field $E$ at a point along the $z$-axis for a charged circular loop can be expressed as:

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

Here, $K$ is the Coulomb's constant, $Q$ is the total charge, $r$ is a constant involving charge distribution, $x$ is the distance along the $z$-axis, and $R$ is the radius of the loop.

Maximizing the Electric Field:

To find where $E$ is maximum, we take the derivative of $E$ with respect to $x$ and set it to zero:

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

Solve for $x$:

Solving the equation from the derivative, we find:

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

Substitute Given Radius:

Given the radius $R=a\sqrt{2}$, substituting into the expression for $x$:

$x=\frac{\sqrt{2}a}{\sqrt{2}}=a$

Thus, the position along the positive $z$-axis where the electric field is maximum is at $x=a$.