Answer:
6 x 10⁻¹⁵ J
Explanation:
d = distance between the plates = 1.5 cm = 0.015 m
E = magnitude of electric field between the plates of the capacitor = 2.5 x 10⁶ V/m
q = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C
Force on the electron due to electric field is given as
F = q E
F = (1.6 x 10⁻¹⁹) (2.5 x 10⁶)
F = 4 x 10⁻¹³ N
KE₀ = initial kinetic energy of electron at negative plate = 0 J
KE = final kinetic energy of electron at positive plate = ?
Using work-change in kinetic energy
F d = KE - KE₀
(4 x 10⁻¹³) (0.015) = KE - 0
KE = 6 x 10⁻¹⁵ J
