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.

colorectal cancer may also develop from areas of abnormal cells in the lining of the colon or rectum. this area of abnormal cells is called dysplasia and is more commonly seen in people with certain inflammatory diseases of the bowel such as crohn's disease or ulcerative colitis.

true tan speed = ang vel x radius