# physics

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)

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1. (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

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posted by Graham

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