An electron is released from rest in a uniform electric field. the electron accelerates, travelling 5.50 m in 4.00 µs after it is released. what is the magnitude of the electric field in n/c?

3.91 N/C

Explanation:

u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s

Let a be the acceleration.

Use second equation of motion

s = u t + 1/2 a t^2

5.5 = 0 + 1/2 a (4 x 10^-6)^2

a = 6.875 x 10^11 m/s^2

F = m a

The electrostatic force, Fe = q E

Where E be the strength of electric field.

So, q E = m a

E = m a / q

E = (9.1 x 10^-31 x 6.875 x 10^11) / ( 1.6 x 10^-19)

E = 3.91 N/C

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