A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is ___________ (in degree).

Solution

Physics sir · 4 steps

The exit angle comes from the velocity: compare sideways speed with forward speed at the plate edge.

1. Split the motion. Along the plates there is no force, so the electron keeps its entry speed vₓ. Across the plates the force eE gives a steady acceleration a = eE/m towards the positive plate.
2. Sideways speed. The time spent between the plates is t = L/vₓ. In that time the electron gains a sideways speed v_y = at = eEL/(m vₓ).

   tan θ = v_y/vₓ = eEL/(m vₓ²) = eEL/2K

3. Put in numbers. eEL = e × 10 V m⁻¹ × 0.1 m = 1 eV, and 2K = 2 × 0.5 eV = 1 eV. So tan θ = 1/1 = 1.
4. So. tan θ = 1 gives θ = 45°, so the answer is 45.

The common slip: 26.6. Using the position angle tan θ = y/L = eEL/4K = 0.5 instead of the velocity angle gives about 26.6°. The exit direction is set by v_y/vₓ, not by y/L.

The idea to keep. Deflection angle at the exit is tan θ = qEL/2K. Keep energies in eV: then qEL in eV is just E × L in volts.

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