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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2019
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(B) \(\frac{R}{{\sqrt 2 }}\)

Step-by-step Solution

Detailed explanation

Electric field \(E=\frac{k Q x}{\left(x^{2}+R^{2}\right)^{3 / 2}}\)

For maxima \(\frac{d E}{d x}=0\)

After solving we get, \(\left(x \pm \frac{\mathrm{R}}{\sqrt{2}}\right)\)

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