Final answer:
The current I at time t = 4.00 s for the given function of charge is calculated by differentiating the function and then substituting the time into the derived current function, which results in 0.762 A.
Step-by-step explanation:
The quantity of charge Q that flows through a conductor over time is related to the current I by the formula I = ΔQ/Δt. To find the current at a specific time, we differentiate the charge function with respect to time t to obtain the instantaneous current as a function of time. For the given charge function Q = (3.00 mC/s4)t4 − (6.00 mC/s)t + 7.00 mC, differentiating yields: I(t) = dQ/dt = 4(3.00 mC/s4)t3 − 6.00 mC/s. At time t = 4.00 s, the current I is I(4.00 s) = 4(3.00 mC/s4)(4.00 s)3 − 6.00 mC/s, which simplifies to I(4.00 s) = 4(3.00 × 10−6 C/s4)(64.00 s3) − 6.00 × 10−6 C/s. Calculating this gives I(4.00 s) = (4 × 3.00 × 64.00 × 10−6 − 6.00 × 10−6) A, which results in I(4.00 s) = 0.762 A. The quantity of charge through a conductor is modeled by the equation Q = (3.00 mC/s4)t4 - (6.00 mC/s)t + 7.00 mC. To calculate the current at time t = 4.00 s, we can find the derivative of Q with respect to t, which will give us the rate of change of charge with respect to time. Taking the derivative, we get dQ/dt = 12.00 mC/s4t3 - 6.00 mC/s. Substituting t = 4.00 s into the derivative, the current at time t = 4.00 s is: