The potential difference between two parallel plates 175 V. An α particle with mass of 6.5 * 10-27 kg and charge of 3.2 * 10-19 C is released from rest near the positive plate. What is the kinetic energy of the α particle when it reaches the other plate? The distance between the plates is 20 cm.

5.6 * 10⁻¹⁷ J

Explanation:

We know that the work done by an electric field, E on an electric charge, q moving a distance, d is W = qEdcosθ. From above, q = 3.2 * 10⁻¹⁹ C, d = 20 cm = 2 * 10⁻² m, E = V/d = 175V / 2 * 10⁻² m = 8750 V/m and θ = 0 since the α particle moves in the same direction as the electric field. So W = qEdcosθ = 3.2 * 10⁻¹⁹ C * 8750 V/m * 2 * 10⁻² m * cos0 = 5.6 * 10⁻¹⁷ J. We know that the work done by the electric field on the charge W = ΔK the change in kinetic energy of the charge. So, W = 1/2m (v₂² - v₁²) where v₁ = initial velocity and v₂ = final velocity. Since the charge is at rest at the positive plate, v₁ = 0. So, W = 1/2mv₂² = K which is the kinetic energy of the particle after moving the distance of 20 cm between the plates. So K = W = 5.6 * 10⁻¹⁷ J