Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
[tex]V=\frac{I}{nAq}[/tex]
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
[tex]\frac{I}{q} = 1.2*10^{18}[/tex]
[tex]A= 1.3*10^{-8}m^2[/tex]
[tex]n=6.3*10^{28} e/m^3[/tex]
[tex]\omicron{O} = 1.2*10^{-4}(m/s)(N/c)[/tex] Mobility
We can find the drift velocity replacing,
[tex]V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}[/tex]
[tex]V= 1.465*10^-3m/s[/tex]
The electric field is given by,
[tex]E= \frac{V}{\omicron{O}}[/tex]
[tex]E=\frac{1.465*10^-3}{1.2*10^{-4}}[/tex]
[tex]E=12.2V/m[/tex]
