# The quantity of charge q (in coulombs) that has passed through a surface of area 1.95 cm^2 varies with time according to the equation q = 3t^3 + 6t + 6, where t is in seconds.

The quantity of charge q (in coulombs) that has passed through a surface of area 1.95 cm2 varies with time according to the equation q = 3t3 + 6t + 6, where t is in seconds.

(a) What is the instantaneous current through the surface at t = 0.900 s?
A

(b) What is the value of the current density?
kA/m2
asked Feb 2, 2012 in Physics

The formula for current is:

i = dq/dt  (where i=current, q=charge, t=time)

the equation for q is given in the questions as:

q = 3t3 + 6t + 6,

and t = 0.9 s

To find the current we take the derivative of q with respect to time, then plug in t.

so

i = 9t^2 + 6

Now plug in t = 0.9s

i = 9(0.9s)^2 + 6 = 13.29 A

i = 13.29 A

Part B, the value of current density is equal to the current divided by the area

Area = 1.95 cm^2 = 1.95*10^-4 m^2

current = 13.29 A

So current density is:

current density = 13.29 A / 1.95*10^-4 m^2   = 68153 A/m^2   = 68.15 kA/m^2

So current density is   68.15 kA/m^2

answered Feb 2, 2012 by ~Expert~ (3,856 points)
selected Feb 2, 2012 by yoshi