Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The...

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ= 1.9 × 10-7 C/m2, and the plates are separated by a distance of 1.8 × 10-2 m. How fast is the electron moving just before it reaches the positive plate?

(please put answer in m/s)

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Answer #1

Given:

q = magnitude of charge on electron =

m = mass of electron =

The electric field between the plates is uniform and its magnitude is calculated as

The force on the electron is

F = qE

Also, from Newton's 2nd law, the acceleration of the electron is

a = F/m.

The initial speed of the electron is "u = 0", is accelerated by the acceleration "a" travels a distance "d" and reaches the positive plate with speed (say "v").

Applying kinematic equation:

[answer]

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