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# An electron is initially at rest in a uniform electric field of intensity 1400 N/C. Find the following:

A) its acceleration in the electric field.

B) its kinetic energy after 3 nanoseconds.

A

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

4 months ago
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