Question

A uniformly charged ring of radius \( R \) carries total charge \( Q \). Find the electric field at a point on the axis at a distance \( x = \frac{R}{\sqrt{2}} \) from the center.

• A. \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}} \)
• B. \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{QR}{(R^2 + x^2)^{3/2}} \)
• C. \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R^2} \)
• D. \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{(2R^2)^{3/2}} \)

Hint:

\textbf{Tip:} For field on the axis of a ring, only the axial components add; radial components cancel due to symmetry.

Answer 1

The Correct Option is A

The problem requires determining the electric field at a specific point on the axis of a uniformly charged ring. Given a ring with radius \( R \) and total charge \( Q \), the electric field at a point on its axis, located a distance \( x \) from the center, can be ascertained using electrostatics principles.
For a point on the axis at distance \( x \), the electric field \( E \) originating from a small charge element \( dq \) at a distance \( r \) from the ring's center is exclusively along the axis.

Due to symmetry, the radial components of the electric field cancel out.
The axial component \( dE \) attributed to \( dq \) is expressed as:
$$dE = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq \cdot x}{(R^2 + x^2)^{3/2}}$$
The total electric field \( E \) is found by integrating \( dE \) across the entire ring. Given that symmetry ensures equal contributions from each element, the expression for \( E \) is:
$$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}}$$
Now, we substitute \( x = \frac{R}{\sqrt{2}} \):
The term \( R^2 + x^2 \) is calculated as:
$$R^2 + \left(\frac{R}{\sqrt{2}}\right)^2 = R^2 + \frac{R^2}{2} = \frac{3R^2}{2}$$
Consequently, the electric field is derived as:
$$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q \cdot \frac{R}{\sqrt{2}}}{\left(\frac{3R^2}{2}\right)^{3/2}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}}$$
This result aligns with the correct answer: \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}} \).