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JEE Main 2023 (1 Feb, Shift 1) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2023 (1 Feb, Shift 1)

Q20JEE Main 2023 · 1 FEB · SHIFT 1Medium

Electric Field & Field Lines

Two equal positive point charges are separated by a distance 2a2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0q_0 becomes maximum is ax\frac{a}{\sqrt{x}}. The value of xx is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. On the bisector at distance yy: F=2kqq0 y(y2+a2)3/2F = \dfrac{2kqq_0\,y}{(y^2 + a^2)^{3/2}}.
  2. dFdy=0⇒y2+a2=3y2⇒y=a2\dfrac{dF}{dy} = 0 \Rightarrow y^2 + a^2 = 3y^2 \Rightarrow y = \dfrac{a}{\sqrt{2}}
  3. So x=2x = 2.
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