Oppositely charged parallel plates are separated by 4.67 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 2.00 mm from the positive j

a)  1.28 *10^5 N/C

b)2.05 *10^{-14} N

c) 4.83 *10^{-17} J

Explanation:

Given Data:

Distance between the plates, d = 4.67 mm

[/tex]= 4.67 *10^{-3} m[/tex]

Potential difference, V = 600 V

Solution:

(a) The  magnitude of the electric field between the plates is,

(b) Force on electron btwn the plates is,

   F = q E

(c) Work done on the electron is

   W = F * s

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