An electron is projected with horizontal speed 105m / s in a downwardly directed 304N / C electric field. Find the vertical position (in m) after 2.0 μs.

-107 m

Explanation:

Sum of forces in the y direction:

∑F = ma

-qE = ma

a = -qE/m

a = -(1.60×10⁻¹⁹ C) (304 N/C) / (9.11×10⁻³¹ kg)

a = -53.4×10¹² m/s²

Given in the y direction:

v₀ = 0 m/s

a = -53.4×10¹² m/s²

t = 2×10⁻⁶ s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (2×10⁻⁶ s) + ½ (-53.4×10¹² m/s²) (2×10⁻⁶ s)²

Δy = -107 m