Solution:
E = 1400 N/C,
The electron’s initial speed (V1) = 0ms,
M = 9.11×10^-31 kg,
Q = 1.60×10^-19 C.
1. F (electric force exerted on the electron) is expressed as:
F = Eq
A (magnitude of the acceleration of the electron in the electric field) is expressed as:
A = FM (and F = Eq), so A = Eqm
A = (1400N/C) x (1.6×10^-19C) x (9.11×10^-31) x 1 kg⋅ms21 N)
A = 2.5×10^14 ms²
2. V2 (the final speed of the electron after an interval, t = 3×10^-9 s) is expressed as:
v = u+at,
Where u = 0 = Eqmt
Therefore, K (the electron’s final kinetic energy) is expressed as:
K =12mv2 =12 m(Eqmt)2 =(Eqt)22 m =[1400 NCx (1.60×10−19 C x (3×10−9 s)]22x(9.11×10−31 kg) x 1 kg⋅ms21 N 2×1 J1 kg⋅m2s2 ≈ 2.5×20−19 J
3 years ago
131 Views