Question

The quantity of charge q (in coulombs) that has passed through a surface of area 1.90 cm² varies with time according to the equation q=6t³+5t+6, where t is in seconds. (a) What is the instantaneous current through the surface at t=1.05 s ?

Answer 1

To find the instantaneous current at t=1.05 s, we need to find the derivative of the charge with respect to time. Plugging in t=1.05 s into the velocity function gives us the instantaneous current at that time: I=19.335 A.

Step-by-step explanation:

To find the instantaneous current at t=1.05 s, we need to find the derivative of the charge with respect to time. Taking the derivative of q=6t³+5t+6 gives us the velocity function: v=18t²+5. Now, the current is given by the derivative of charge with respect to time: I=dq/dt. Plugging in t=1.05 s into the velocity function gives us the instantaneous current at that time.

Let's find the value of I:

I=dq/dt = d(6t³+5t+6)/dt = 18t²+5

Plugging in t=1.05 s into the velocity function:

I=18(1.05)²+5 = 19.335 A