A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 10⁵ NC⁻¹. If the charge on the particle is 40 μC and the initial velocity is 200 ms⁻¹, how much distance it will travel before coming to the rest momentarily :

Solution

Physics sir · 3 steps

The field pushes back on the charge, so treat it as constant deceleration and use v² = u² − 2as.

1. Force on charge. The charge is positive and moves against the field, so the force qE acts opposite to its velocity and slows it down. F = qE = 40 × 10⁻⁶ × 10⁵ = 4 N.
2. Deceleration. Convert the mass first: 100 mg = 10⁻⁴ kg. Then a = F/m = 4/10⁻⁴ = 4 × 10⁴ m s⁻².

   0 = u² - 2as ⇒ s = u²/2a = (200)²/(2 × 4 × 10⁴)

3. So. s = (4 × 10⁴)/(8 × 10⁴) = 0.5 m, which is option (d).

The idea to keep. A uniform field gives a constant force qE, so the motion is plain constant-acceleration kinematics. Convert mg to kg carefully before dividing.

Answer from the NTA final answer key. Solution by PrepWiser.