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

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The discussion centers on calculating the instantaneous current through a surface area of 2.00 cm² using the equation q = 4t³ + 5t² - 6. At t = 1s, the correct method involves differentiating the charge function q(t) to find the instantaneous current, I, defined as I = dq/dt. The user initially calculated an average current of 15 by plugging in values directly into the equation, while the correct instantaneous current at t = 1s is 17, as stated in the reference material.

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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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