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JEE Main 2022 (29 Jun, Shift 2) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2022 (29 Jun, Shift 2)

Q24JEE Main 2022 · 29 JUN · SHIFT 2Medium

Electric Field & Field Lines

Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

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Answer: (d)

  1. On the bisector the horizontal components cancel: F=2kQq x(x2+d2/4)3/2F = \dfrac{2kQq\,x}{\left(x^2 + d^2/4\right)^{3/2}}.
  2. Set dFdx=0\dfrac{dF}{dx} = 0: (x2+d24)−3x2=0⇒x2=d28\left(x^2 + \frac{d^2}{4}\right) - 3x^2 = 0 \Rightarrow x^2 = \dfrac{d^2}{8}
  3. x=d22x = \dfrac{d}{2\sqrt{2}}
  4. So the answer is (d) x=d22x = \dfrac{d}{2\sqrt{2}}.
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