Question

Oppositely charged parallel plates are separated by 6.34 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 2.92 mm from the positive plate? J

Answer 1

(a) Magnitude of an electric field between the plates which will be given as -

we know that,   V = E d

E = [(600 V) / (6.34 x 10^-3 m)]

E = 94637.2 N/C

(b) Magnitude of the force on an electron between the plates which will be given as -

we know that,   F_e = q E

F_e = [(1.6 x 10^-19 C) (94637.2 N/C)]

F_e = 1.51 x 10^-14 N

(c) If it is initially positioned 2.92 mm from the positive plate, then the work done on an electron to move it to the negative plate which will be given as -

using a formula, we have

W = F d cos

W = (1.51 x 10^-14 N) [(6.34 x 10^-3 m) - (2.92 x 10^-3 m)] cos 0⁰

W = [(1.51 x 10^-14 N) (3.42 x 10^-3 m)]

W = 5.16 x 10^-17 J