A particle of charge q and mass m is subjected to an electric field
E = E₀ (1 – ax²) in the x-direction, where a and E₀ are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

Solution

Physics sir · 4 steps

Kinetic energy is zero again when the total work done by the field adds back up to zero.

1. Work-energy theorem. The particle starts from rest, so its kinetic energy at any position x equals the work done on it by the field. Kinetic energy is zero again exactly when this work is zero.

   W = ∫₀ˣ qE₀(1 - a x²)dx = qE₀(x - (a x³)/3)

2. Set work to zero. qE₀ x(1 - (a x²)/3) = 0. The root x = 0 is the starting point, so the other root comes from 1 - (a x²)/3 = 0, which gives x² = 3/a.
3. Why it turns back. Up to x = √(1/a) the force pushes forward and the particle speeds up. Beyond that point the field reverses and slows it down, until all the gained energy is taken back at x = √(3/a).
4. So. The kinetic energy becomes zero again at a distance √(3/a) from the origin, which is option (c).

The idea to keep. When a force depends on position, integrate it to get the work. The particle stops where the total work is zero, not where the force is zero.

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