The electric field strength is 20,000 n/c inside a parallel-platecapacitor with a 1.0 mm spacing. an electron is released fromrest at the negative plate. what is the electron’s speed when itreaches the positive plate?

Explanation:

It is given that,

Electric field strength, E = 20,000 N/C

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

Also,

Using third equation of motion as :

v = 2649528.2599 m/s

or

So, the velocity of the electron when it reaches the positive plate is . Hence, this is the required solution.

answer: good job it's all correct !

explanation: