The quantity of charge q (in coulombs) that has passed through a surface of area 2.08 cm2 varies with time according to the equation q = 4t3 + 5t + 6, where t is in seconds.

(a) What is the instantaneous current through the surface at t = 1.08 s? (in A)

(b) What is the value of the current density? (with units ka/m^2)

(A)

Current is flow rate of charge. (Amp=Coulomb/Second)
Take the derivative and evaluate at the time.

I(t) = dq(t)/dt
I(t) = d(4t^3 + 5t + 6)/dt
I(t) = 12t^2 + 5
I(1.08) = 12(1.08)^2 + 5
-------

(B) Divide this current by the area to get the current density.
Note: kA/m^2 = (1/10) A/cm^2

To find the instantaneous current through the surface at t = 1.08 s, we need to calculate the derivative of the charge with respect to time, dq/dt, and evaluate it at t = 1.08 s.

Given: q = 4t^3 + 5t + 6

(a) First, calculate the derivative of q with respect to t to find dq/dt:
dq/dt = d/dt (4t^3 + 5t + 6)
= 12t^2 + 5

Now, substitute t = 1.08 s into the derivative equation to find the instantaneous current:
dq/dt at t = 1.08 s = 12(1.08)^2 + 5
= 12(1.1664) + 5
= 13.997 + 5
= 18.997 A

Therefore, the instantaneous current through the surface at t = 1.08 s is approximately 18.997 A.

(b) To find the current density, we need to divide the current by the area of the surface. Given that the surface area is 2.08 cm^2, we first need to convert it to square meters.

1 cm^2 = (1/100)^2 m^2 [converting from cm^2 to m^2]
= 0.0001 m^2

So, the area of the surface in square meters is:
2.08 cm^2 = 2.08 * 0.0001 m^2
= 0.000208 m^2

Now, divide the instantaneous current (18.997 A) by the surface area (0.000208 m^2) to find the current density:
Current density = Instantaneous current / Surface area
= 18.997 A / 0.000208 m^2
= 91182.692 A/m^2

Therefore, the value of the current density is approximately 91182.692 A/m^2.