Consider a circular loop that is uniformly charged and has a radius $\mathrm{a} \sqrt{2}$. 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 :

<p>To determine the position along the positive $ z $-axis where the electric field due to a uniformly charged circular loop is at its maximum, we follow these steps:</p> <p><p><strong>Expression for Electric Field ($ E $):</strong> </p> <p>The electric field $ E $ at a point along the $ z $-axis for a charged circular loop can be expressed as:</p> <p>$ E = \frac{KQr}{(x^2 + R^2)^{3/2}} $</p> <p>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.</p></p> <p><p><strong>Maximizing the Electric Field:</strong> </p> <p>To find where $ E $ is maximum, we take the derivative of $ E $ with respect to $ x $ and set it to zero:</p> <p>$ \frac{dE}{dx} = 0 $</p></p> <p><p><strong>Solve for $ x $:</strong> </p> <p>Solving the equation from the derivative, we find:</p> <p>$ x = \frac{R}{\sqrt{2}} $</p></p> <p><p><strong>Substitute Given Radius:</strong> </p> <p>Given the radius $ R = a\sqrt{2} $, substituting into the expression for $ x $:</p> <p>$ x = \frac{\sqrt{2}a}{\sqrt{2}} = a $</p></p> <p>Thus, the position along the positive $ z $-axis where the electric field is maximum is at $ x = a $.</p>