Question:medium

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?

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For axis electric fields of a charged ring, use the formula provided and substitute the values of \(x\) and \(R\) to calculate the field.
Updated On: Nov 26, 2025
  • \( \frac{kQ}{R^2} \)
  • \( \frac{2kQ}{R^2} \)
  • \( \frac{3kQ}{R^2} \)
  • \( \frac{kQ}{2R^2} \)