A uniform electric field E=(8 m / e) V / m is created between two parallel plates of length 1 m as shown in figure, (where m= mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m / s. The angle of the deviation (θ) of the path of the electron as it comes out of the field will be _________.

Solution

Physics sir · 4 steps

Horizontal speed stays 2 m/s; the field only adds vertical speed. Divide the two at the exit.

1. Force on electron. The field points down, so the force on the negative electron points up, towards the + plate. Its size is eE = e × 8m/e = 8m, so the acceleration is a = eE/m = 8 m s⁻², upward.
2. Time between plates. Nothing acts horizontally, so vₓ = 2 m s⁻¹ throughout. The electron covers the 1 m plate length in t = 1/2 = 0.5 s.
3. Vertical speed. Starting with no vertical speed, it leaves with v_y = at = 8 × 0.5 = 4 m s⁻¹.

   tan θ = v_y/vₓ = 4/2 = 2

4. So. The path leaves the field at θ = tan⁻¹(2), which is option (b).

The idea to keep. A charge crossing a uniform field moves like a projectile: steady speed along the plates, constant acceleration across them. The exit angle comes from tan θ = v_y/vₓ.

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