Question

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 plate?_________ J

Answer 1

a) 1.28 *10^5 N/C

b)2.05 *10^{-14} N

c) 4.83 *10^{-17} J

Step-by-step explanation:

Given Data:

Distance between the plates, d = 4.67 mm

`= (4.67) *10^(-3) m`

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

Potential difference, V = 600 V

Solution:

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

`E = (V)/(d)`

`= (600 V)/(4.67 *10^(-3)) m`

`= 1.28 *10^5 V/m or 1.28 *10^5N/C`

(b) Force on electron btwn the plates is,

F = q E

`= (1.6 *10^(-19) C) (1.28 *10^5N/C`

`= 2.05 *10^(-14) N`

(c) Work done on the electron is

W = F * s

`= (2.05 *10^(-14) N) * (5.31 *10^(-3) m - 2.95 *10^(-3) m)`

`= 4.83 *10^(-17) J`