Instantaneous current through an area

[Smazmbazm]

Homework Statement

Help with this question would be greatly appreciated as I have attempted it many times without success and I don't know where I'm going wrong.

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 = 8t^3 + 7t + 6, where t is in seconds.

What is the instantaneous current through the surface at t = 0.900 s?

Homework Equations

I = Q/t
dQ/dt

The Attempt at a Solution

I would think that you find dQ/dt which would give 24t^2 + 7, substitute 0.9 into t which equals the charge and then divide that by the time to find instantaneous current. Current was = 20.15A, which was incorrect

I also tried substituting 0.9 into 8t^3 + 7t + 6 and then dividing by time. Here the current was = 29.37A. Incorrect as well.

 

[berkeman]

Homework Statement

Help with this question would be greatly appreciated as I have attempted it many times without success and I don't know where I'm going wrong.

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 = 8t^3 + 7t + 6, where t is in seconds.

What is the instantaneous current through the surface at t = 0.900 s?

Homework Equations

I = Q/t
dQ/dt

The Attempt at a Solution

I would think that you find dQ/dt which would give 24t^2 + 7, substitute 0.9 into t which equals the charge and then divide that by the time to find instantaneous current. Current was = 20.15A, which was incorrect

I also tried substituting 0.9 into 8t^3 + 7t + 6 and then dividing by time. Here the current was = 29.37A. Incorrect as well.

When you find dQ/dt and substitute in t=0.9s, that is the answer. You do not need to divide by time or anything else.

 

[gneill]

Why divide by time? If you take the derivative w.r.t. time, that's already dQ/dt.