Physics Capacitance

Question

A parallel-plate capacitor has a plate separation of 0.0015 m and is charged to 700 V.

If an electron leaves the negative plate, starting from rest, how fast is it going when it hits the positive plate? 

Answer 1

The force acting on a charge in an electric field = Eq ... where E is the magnitude of the field and q = the amount of charge. The magnitude of E = V/d = 700/.0015= 4.67E5. The charge of an electron = 1.6E-19 C thus the force on the electron = .821E-14 N. This force is also equal to ma (Newton's 2nd). The electron mass = 9.11E-31 thus the acceleration = 9E15 m/s^2. Using the relationship V= sqrt(2ax), V (final) becomes 5.2E6 m/s.