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JEE Main 2025 (28 Jan, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2025 (28 Jan, Shift 1)

Q6JEE Main 2025 · 28 JAN · SHIFT 1Difficult

Motion of Charges in an Electric Field

A particle of mass ' mm ' and charge ' qq ' is fastened to one end ' AA ' of a massless string having equilibrium length ll, whose other end is fixed at point ' OO '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the xx-axis is

Figure

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Answer: (c)

  1. The force qEqE along +x+x swings AA about OO on a circle of radius ll until the string lies on the xx-axis.
  2. Displacement along E⃗\vec{E}: from x=lcos⁡60∘=l2x = l\cos 60^\circ = \frac{l}{2} to x=lx = l, i.e. l2\frac{l}{2}.
  3. Work–energy: 12mv2=qEl2⇒v=qElm\frac{1}{2}mv^2 = qE\frac{l}{2} \Rightarrow v = \sqrt{\frac{qEl}{m}}
  4. So the answer is (c).
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