Question

Oppositely charged parallel plates are separated by 4.49 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? N/C (b) What is the magnitude of the force on an electron between the plates? N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 3.14 mm from the positive plate?

Answer 1

A. Using

E=V/d

= 600/4.49*10^-3

= 1.336 x10^5 N/C

b) F = E*q = 1.33610^5 x 1.6*10^-19

= 2.17 x 10^-14 N

c) Work = Fs distance = 2.17 x 10^-14 N (4.49-3.14)*.001= 1.35 x 10^-17 J