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JEE Main 2025 (2 Apr, Shift 1) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2025 (2 Apr, Shift 1)

Q6JEE Main 2025 · 2 APR · SHIFT 1Easy

Electric Field & Field Lines

A point charge +q+q is placed at the origin. A second point charge +9q+9q is placed at (d,0,0d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

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Answer: (b)

  1. Let the null point be at distance xx from +q+q: kqx2=k(9q)(d−x)2\dfrac{kq}{x^2} = \dfrac{k(9q)}{(d-x)^2}.
  2. d−xx=3⇒x=d4\dfrac{d-x}{x} = 3 \Rightarrow x = \dfrac{d}{4}
  3. So the answer is (b) (d/4,0,0)(d/4, 0, 0).
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