Respuesta :
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is [tex]Q =2.094 C[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 0.205cm = 0.00205 \ m[/tex]
The radius of the wire is [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]
The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]
The electric field change is mathematically defied as
[tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]
Generally the charge is mathematically represented as
[tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]
Where A is the area which is mathematically represented as
[tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]
So
[tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]
Therefore
[tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]
substituting values
[tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]
[tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]
From the question we are told that t = 5 sec
[tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]
[tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]
[tex]Q =2.094 C[/tex]
The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.
Given the following data:
- Diameter of wire = 0.205 centimeters.
- Resistivity of aluminum = [tex]2.75\times 10^{-8}[/tex] Ohm-meters.
- [tex]E(t)=0.0004t^2-0.0001t+0.0004[/tex] Newton per coulomb.
Conversion:
Diameter of wire = 0.205 cm to m = 0.00205 meter.
Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]
To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:
First of all, we would find the area of the wire.
[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]
Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:
[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]
Where:
- A is the area of a conductor.
- [tex]\rho[/tex] is the resistivity of a conductor.
- t is the time.
- E is the electric field.
Substituting the given parameters into the formula, we have;
[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]
[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]
When t = 5 seconds:
[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]
Q = 2.094 Coulomb.
Find more information: https://brainly.com/question/18214726
