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Proof that current is the rate of change of charge with respect to time

I've been looking for a proof of this but its impossible to find one.
Ive started with the fact that E=QV
therefore dQ/dt=d(E/V)/dt
dE/dt= power = P= IV
but I'm stuck with derivative of voltage.
I know quotient rule will help.

thanks
Original post by kennz
I've been looking for a proof of this but its impossible to find one.
Ive started with the fact that E=QV
therefore dQ/dt=d(E/V)/dt
dE/dt= power = P= IV
but I'm stuck with derivative of voltage.
I know quotient rule will help.

thanks


Instead of using calculus, you could just use normal equations:
E=QVE=QV and Et=P=IV\frac{E}{t}=P=IV
Equate the two equations for E:

Et=QVt=P=IV\frac{E}{t}=\frac{QV}{t}=P=IV

Cancel the voltages off both sides after combining:
QVt=IVQt=I\frac{QV}{t}=IV \rightarrow \frac{Q}{t}=I

So current is change in charge, respect to time.
Reply 2
Original post by The-Spartan
Instead of using calculus, you could just use normal equations:
E=QVE=QV and Et=P=IV\frac{E}{t}=P=IV
Equate the two equations for E:

Et=QVt=P=IV\frac{E}{t}=\frac{QV}{t}=P=IV

Cancel the voltages off both sides after combining:
QVt=IVQt=I\frac{QV}{t}=IV \rightarrow \frac{Q}{t}=I

So current is change in charge, respect to time.


Thankk you for that, is there not a way to do it with calculus though?

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