An aluminum “12 gauge” wire has a diameter d of 0.205 centimeters. The resistivity rho of aluminum is 2.75 X 10^8 ohmmeters. The electric field in the wire changes with time as E(t)=0.0004t^2 – 0.0001t + 0.0004 newtons per coulomb, where time is measured in seconds.
Find the current I through the conductor at time 5.0 seconds.
I got the answer for this part..its 1.2 A
Find the charge Q passing through a crosssection of the conductor between time 0 seconds and time 5 seconds
i cant seem to get this part right
1 Answer

You have to find an expression for the current as a function of time. You Know that current is a measure of the rate at which charge passes through a wire’s area, basically I = dQ/dt, taking this you can develop an expression for Q = ∫ I(t)*dt. From doing the first part, you must have gotten the an expression where I = EA/p? Assuming you did, you can plug that in for I, and solve taking the antiderivative.
Q = ∫ I(t)*dt = ∫ E*A/p, where A/p is a constant allowing you to pull it ouside the integral
Q = A/p ∫E, once u get to this point, you take the antiderivative of E
Q = ((pi*r^2)/p) (E antiderivative)
Q = ((3.14*0.001025^2)/(2.75 X 10^8)) * (0.0004t^3/3 – 0.0001t^2/2 +0.0004t)
Q = 120 * [(0.0004t^3/3 – 0.0001t^2/2 +0.0004t)] where t=5
Q = 2.09 C