For a uniformly charged ring of radius $R$, the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is :

Question

Derive the expression for Electric Field Intensity on the axial and equatorial line of a dipole.

Answer 1

To determine the distance $h$ from the center of a uniformly charged ring where the electric field on its axis is maximized, we start by considering the following concept:

Concept: The electric field at a point on the axis of a uniformly charged ring can be derived and has a specific formula.

The electric field $E$ at a point on the axis of a ring a distance $h$ from the center, with a total charge $Q$ and radius $R$, is given by:

E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Qh}{(R^2 + h^2)^{3/2}}

This formula arises from integrating the contributions of all charge elements on the ring to the electric field at the point on its axis.

Objective: Maximize $E$ with respect to $h$.

To find the maximum value, we take the derivative of $E$ with respect to $h$ and set it equal to zero:
The expression becomes quite large, but simplifying and applying the condition for maximization yields:

\frac{dE}{dh} = 0 \Rightarrow (R^2 + h^2)^{1/2} \cdot R^2 = 2h^2

Solving this equation, we simplify:

R^2 = 2h^2 \Rightarrow h = \frac{R}{\sqrt{2}}

This shows that the electric field at a distance $h = \frac{R}{\sqrt{2}}$ from the center of the ring is maximized.

Conclusion: The value of $h$ for which the electric field is at its largest magnitude on the axis of a uniformly charged ring is \frac{R}{\sqrt{2}}.

Answer 2