A charged particle of mass m and charge q enters symmetrically between two parallel plates of length 2 m with a horizontal speed of 4 m s ⁻¹ . A uniform transverse electric field E = ( 4m q ) V m ⁻¹ is maintained between the plates. After exiting the plates, the particle travels through a field-free region for a horizontal distance of 3 m before striking a vertical screen. The total vertical displacement of the parti

Step-by-step solution

The transverse acceleration of the particle inside the plates is a_y = qE m = q m ( 4m q ) = 4 m s ⁻² . The time spent inside the plates is t₁ = L v_x = 2 4 = 0.5 s . The vertical displacement inside the plates is y₁ = 1 2 a_y t₁^2 = 1 2 (4)(0.5)^2 = 0.5 m . The vertical velocity upon exiting the plates is v_y = a_y t₁ = 4(0.5) = 2 m s ⁻¹ . In the field-free region, the particle travels a horizontal distance D = 3 m . The time taken is t₂ = D v_x = 3 4 = 0.75 s . The vertical displacement in the field-free region i