Answer:
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]
[/tex]= 4.67 *10^{-3} m[/tex]
Potential difference, V = 600 V
Solution:
(a) The magnitude of the electric field between the plates is,
[tex]E = \frac{V}{d}[/tex]
[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]
[tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]
(b) Force on electron btwn the plates is,
F = q E
[tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]
[tex]= 2.05 *10^{-14} N[/tex]
(c) Work done on the electron is
W = F * s
[tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]
[tex]= 4.83 *10^{-17} J[/tex]
