Solving q = 4t³ + 5t² - 6 for t = 1s - Instantaneous Current

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To find the instantaneous current through a surface at t=1s, the charge function q=4t³+5t²-6 must be differentiated to obtain the current I. The initial calculation of q at t=1s gives 15, which represents the average current rather than the instantaneous current. The correct approach is to differentiate q with respect to time (dq/dt) and then evaluate this derivative at t=1s. This method reveals that the instantaneous current is 17, aligning with the book's answer. Understanding the difference between average and instantaneous current is crucial for solving this problem correctly.
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the problem reads th quantity of charge q that has passed through a surface area of 2.00cm^2 varies with time accoding to the equation
q=4t^3+5t=6 where t is in seconds. what is the instantenaus current through the surface at t=1s?

What I did was plug in 1 in the equation and got 15 as an answer, then to get current (I) i did I= q/t so that gave me 15 the answer in the book is 17. how did they get this?

thanks in adavance
 
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No,no,u need to differentiate first and then evaluate the new function (the electric current) at the moment t=1s...

Daniel.
 
Exactly...you evaluated the average current over one second, by dividing the total charge that had passed by by the total time that had elapsed. This is not what the question was asking for. It asked for instantaneous current, which is defined as the time rate of change (time derivative) of the electric charge passing by a point (or in this case, passing through that surface area).

I = dq/dt, where q is of course a function of time q(t), and so I = dq(t) / dt, to write things out in full.

q(t) has been given to you.
 
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