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.

Answer 1

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  $Q =2.094 C$

Explanation:

From the question we are told that

    The diameter of the wire is  $d = 0.205cm = 0.00205 \ m$

     The radius of  the wire is  $r = \frac{0.00205}{2} = 0.001025 \ m$

     The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

       The electric field change is mathematically defied as

         $E (t) = 0.0004t^2 - 0.0001 +0.0004$

     

Generally the charge is  mathematically represented as

       $Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

       $A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

 So

       $\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

      $Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

      $Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

     $Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t =  5 sec

           $Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

          $Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

         $Q =2.094 C$