Question

An electron is placed in a uniform electric field of 5 × 105 N/C that points to the right. (e = 1.6 × 10-19 C, melectron = 9.11 × 10-31 kg). What is the magnitude of the force on the electron?

Answer 1

Force on electron due to electric field is given by:

F = q*E

q = charge on electron = -1.6*10^-19 C

E = electric field = 5*10^5 N/C

So magnitude of electric force will be:

|F| = |q|*E

|F| = 1.6*10^-19*5*10^5

|F| = 8*10^-14 N

Direction of force on electron will be towards the opposite direction of electric field, meaning towards left

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