Current density in a wire that is being used to charge a capacitor

[Ugnius]

Homework Statement

Wire, with diameter of 0.7 mm2, is used to charge a capacitor. Charge on the capacitor plate is described as: Q(t):=4*t^3+5*t+3 [C]. What is current density in a wire, at given time t=2s?

Relevant Equations

Q(t):=4*t^3+5*t+3 , j=I/A , I=Q/t

Somehow this answer is incorrect , but i realize that even numbers are hipothetical , 45 coulumbs is too much charge , what is wrong in my calculations?

 

[DaveE]

Q(t) is the total charge on the capacitor, including what has accumulated in the past. The current in the wire is the rate at which new charge is added/removed.

 

[Ugnius]

Okay , but I = Q/t so that's 12.5 A in the wire at the moment t=2, but why isn't the answer correct , is there some kind of a mistake in calculation?

 

[TSny]

As @DaveE pointed out, the current I in the wire at the instant t = 2 s is the instantaneous rate at which charge is flowing through a cross-section of the wire. This must equal the instantaneous rate at which charge is arriving at the plate of the capacitor at t = 2.

How do you calculate the instantaneous rate of change of some function of time?

 

[TSny]

Homework Statement:: Wire, with diameter of 0.7 mm2,

A diameter could have units of mm, but not mm².

 

[Steve4Physics]

what is wrong in my calculations?

The equation I=Q/t can’t be used here. This equation applies to situations where current is constant or we only want the average current. T help you understand your mistake, try this problem:

An object moves along a straight line so its distance (x) from a reference point (x=o) is given by:
x(t) = 4t³ + 5t + 3
Q1. What is the object’s average speed during the interval t=0 to t=2s?
Q2. What is the object’s speed at the instant t=2s?
_____________

Also note “Wire, with diameter of 0.7 mm2” doesn’t make sense. You probably mean:
“Wire, with cross-sectional area of of 0.7 mm²” or
“Wire, with diameter of 0.7 mm”
You need to be certain which one it is.

@Edit. @TSny beat me to it!

 

[Ugnius]

Thanks guys , I managed to solve it using derivatives , your help much appreciated.

 

[DaveE]

Congratulations, good work!

 

[Steve4Physics]

Thanks guys , I managed to solve it using derivatives , your help much appreciated.
View attachment 290868

Well done. But don't forget that 0.7mm² and 2s are each specified to only 1 significant figure.