Answer:
The charge density in the system is [tex]4.25*10^4C/m[/tex]
Explanation:
To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.
Our data given correspond to:
[tex]r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C[/tex]
We need to asume here the number of free electrons in a copper conductor, at which is generally of [tex]8.5 *10^{28}m^{-3}[/tex]
The equation to find the current is
[tex]I = VenA[/tex]
Where
I =Current
V=Velocity
A = Cross-Section Area
e= Charge for a electron
n= Number of free electrons
Then replacing,
[tex]I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)[/tex]
[tex]I= 22.11a[/tex]
Now to find the linear charge density, we know that
[tex]I = \frac{Q}{t} \rightarrow Q = It[/tex]
Where:
I: current intensity
Q: total electric charges
t: time in which electrical charges circulate through the conductor
And also that the velocity is given in proportion with length and time,
[tex]V_d = \frac{l}{t} \rightarrow l = V_d t[/tex]
The charge density is defined as
[tex]\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}[/tex]
Replacing our values
[tex]\lambda = \frac{22.11}{5.20*10{-4}}[/tex]
[tex]\lambda= 4.25*10^4C/m[/tex]
Therefore the charge density in the system is [tex]4.25*10^4C/m[/tex]
