Question

The electric field at a point on the axis of a uniformly charged ring of radius R at a distance x from its center is given by: \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2\pi kQx}{(x^2 + R^2)^{3/2}}. \] If x = 2R, what is the magnitude of the electric field?

• A. \( \frac{kQ}{R^2} \)
• B. \( \frac{2kQ}{R^2} \)
• C. \( \frac{3kQ}{R^2} \)
• D. \( \frac{kQ}{2R^2} \)

Hint:

For axis electric fields of a charged ring, use the formula provided and substitute the values of \(x\) and \(R\) to calculate the field.

Answer 1

The Correct Option is A

Given the formula for the electric field at a point on the axis of a uniformly charged ring: \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2\pi kQx}{(x^2 + R^2)^{3/2}}. \] Substitute \( x = 2R \): \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2\pi kQ \cdot 2R}{(4R^2 + R^2)^{3/2}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{4\pi kQR}{(5R^2)^{3/2}}. \] Simplifying: \[ E = \frac{kQ}{R^2}. \] Thus, the electric field is: \[ \boxed{\frac{kQ}{R^2}}. \]