JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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
- A \(\frac{R}{{\sqrt 5 }}\)
- B \(\frac{R}{{\sqrt 2 }}\)
- C \(R\)
- D \(R\sqrt 2 \)
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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