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 involves finding the electric field at a point on the axis of a uniformly charged ring. Let's consider the provided ring of radius \( R \) with total charge \( Q \). The electric field at a point on its axis at a distance \( x \) from the center can be calculated using the principles of electrostatics.
For a point on the axis at distance \( x \), the electric field \( E \) due to a small charge element \( dq \) at a distance \( r \) from the center of the ring has only a component along the axis.

The radial components cancel out due to symmetry. 
The axial component \( dE \) due to \( dq \) is given by:
$$dE = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq \cdot x}{(R^2 + x^2)^{3/2}}$$
The total electric field \( E \) is obtained by integrating \( dE \) over the entire ring. Since each element contributes equally due to symmetry, the expression for \( E \) is:
$$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}}$$
Now, substitute \( x = \frac{R}{\sqrt{2}} \):
Calculate \( R^2 + x^2 \):
$$R^2 + \left(\frac{R}{\sqrt{2}}\right)^2 = R^2 + \frac{R^2}{2} = \frac{3R^2}{2}$$
Thus, the electric field becomes:
$$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q \cdot \frac{R}{\sqrt{2}}}{\left(\frac{3R^2}{2}\right)^{3/2}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}}$$
This matches the correct answer: \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}} \).

Answer 2

Electric Field on the Axis of a Uniformly Charged Ring

Consider a uniformly charged ring of radius \( R \) carrying a total charge \( Q \). We want to find the electric field \( E \) at a point located on the axis of the ring, at a distance \[ x = \frac{R}{\sqrt{2}} \] from the center of the ring.

Formula for the Electric Field on the Axis

The electric field due to a uniformly charged ring at a point on its axis is given by: \[ E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}} \] where:

• \( Q \) = total charge on the ring
• \( R \) = radius of the ring
• \( x \) = distance from the center along the axis
• \( \varepsilon_0 \) = permittivity of free space

 

Step-by-Step Evaluation at \( x = \frac{R}{\sqrt{2}} \)

Substitute \( x = \frac{R}{\sqrt{2}} \) into the expression for \( E \): \[ E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q \left(\frac{R}{\sqrt{2}}\right)}{\left(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2\right)^{3/2}} \]

Simplify the Denominator

Calculate the term inside the parentheses: \[ R^2 + \left(\frac{R}{\sqrt{2}}\right)^2 = R^2 + \frac{R^2}{2} = \frac{3R^2}{2} \] Thus, \[ \left(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2\right)^{3/2} = \left(\frac{3R^2}{2}\right)^{3/2} \]

Rewrite the Power

Expressing the denominator further: \[ \left(\frac{3R^2}{2}\right)^{3/2} = \left(\frac{3}{2}\right)^{3/2} \cdot R^3 = \frac{3^{3/2} \cdot R^3}{2^{3/2}} \]

Final Expression for the Electric Field

Plugging everything back into the formula: \[ E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q \cdot \frac{R}{\sqrt{2}}}{\frac{3^{3/2} \cdot R^3}{2^{3/2}}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R^2} \cdot \frac{2^{3/2}}{3^{3/2} \cdot \sqrt{2}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R^2} \cdot \frac{2}{3^{3/2}} \]

This confirms that the electric field at the point \( x = \frac{R}{\sqrt{2}} \) on the axis of the uniformly charged ring follows the standard formula: \[ { E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}} } \]

Summary

The electric field on the axis of a uniformly charged ring depends on both the radius \( R \) and the axial distance \( x \). Substituting specific values like \( x = \frac{R}{\sqrt{2}} \) helps in evaluating exact field strength and is crucial in physics problems involving ring charge distributions, electrostatics, and electric field computations