A charged particle (mass m and charge q)
moves along X-axis with velocity V₀. When it
passes through the origin it enters a region having uniform electric field
E⃗ = - Eĵ which extends upto x = d.
Equation of path of electron in the region x > d is

Solution

Physics sir · 4 steps

Inside the field it moves like a projectile; after x = d it flies straight along its exit velocity.

1. Inside the field. The force on the charge is qE⃗ = -qE ĵ, so it has a constant acceleration a = qE/m along -y. The x-velocity stays V₀, so it crosses the field region in time t = d/V₀.
2. Exit point and velocity. At x = d the displacement is y₁ = -1/2qE/md²/V₀² = -qEd²/2mV₀². The vertical velocity is v_y = -qE/md/V₀, so the path makes slope v_y/V₀ = -qEd/mV₀².
3. Straight line after. For x > d there is no force, so the path is a straight line through (d, y₁) with that slope.

   y = -qEd²/2mV₀² - qEd/mV₀²(x-d) = qEd/mV₀²(d/2 - x)

4. So. The path for x > d is y = qEd/mV₀²(d/2 - x), which is option (b).

The idea to keep. A charge leaving a uniform deflecting field moves in a straight line. Trace that line back and it meets the original path at the middle of the field region, x = d/2.

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