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JEE Main 2019 (9 Jan, Shift 1) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2019 (9 Jan, Shift 1)

Q40JEE Main 2019 · 9 JAN · SHIFT 1Medium

Electric Field & Field Lines

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is:

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Answer: (b)

  1. E=kQh(h2+R2)3/2E = \dfrac{kQh}{(h^2+R^2)^{3/2}}; dEdh=0⇒(h2+R2)−3h2=0\dfrac{dE}{dh} = 0 \Rightarrow (h^2+R^2) - 3h^2 = 0.
  2. 2h2=R2⇒h=R22h^2 = R^2 \Rightarrow h = \dfrac{R}{\sqrt2}
  3. So the answer is (b) R2\dfrac{R}{\sqrt2}.
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