An electron enters midway between parallel horizontal plates moving right at 3.600 × 10^6 m s⁻¹. The plate length is 0.092 m and separation 0.0400 m. The upper plate is 9 V above the lower plate, creating a uniform field. Use electron charge −1.602 × 10⁻¹⁹ C and mass 9.11 × 10⁻³¹ kg. Ignore gravity, fringing and relativity. Find field magnitude, vertical acceleration, transit time and upward deflection at the plate exit. Check whether the electron strikes a plate before leaving.

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Field 225 V m⁻¹ downward; acceleration 3.957 × 10^13 m s⁻² upward; time 2.556 × 10^-8 s; exit deflection 0.01292 m. It does not strike a plate. The electric field points from the higher-potential upper plate to the lower plate, downward, with magnitude E = V/d = 225 V m⁻¹. An electron's negative charge reverses the force direction. Its acceleration is upward with magnitude a = |q|E/m = 3.957 × 10^13 m s⁻². No horizontal force acts in the stated model, so horizontal speed stays constant. Transit time follows from horizontal motion, t = L/u = 2.556 × 10^-8 s. The electron begins with zero vertical velocity, giving y = ½at² = 0.01292 m upward. The starting point is halfway between plates, so the available upward clearance is d/2 = 0.0200 m. The calculated deflection is smaller; because upward displacement increases throughout transit, no earlier collision occurs. The direction is consistent with attraction toward the positive upper plate. Combining horizontal uniform motion with vertical acceleration is essential; using the vertical acceleration to alter the horizontal transit time would mix perpendicular components.

Charged-particle deflection between plates · set 6

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