# when does $\frac{dq}{dT} = \frac{Q}{T}$

If I have a current going through a wire, I can say say that:

$$i = \frac{dq}{dT} = \frac{Q}{T}$$

If there is some sort of symmetry or homogeneity. If I were to solve this Differential Equation, I would find:

$$\int \frac{dq}{Q}=\int \frac{dT}{T} \implies Q = \pm e^{c + ln|T|} = CT$$

This is similar to having a uniform (homogenous) mass density, so that I may write density as:

$$\rho = \frac{dm}{dv} = \frac{m}{v}$$

So is the necessary mathematical condition that Q = CT for me to write current as $i = \frac QT$ ? Is this a necessary condition for homogeneity?

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How are $q$ and $Q$ related? –  John Moeller Mar 14 '13 at 23:04
Assuming that Q is just some total charge accumulated over time, I think you're on the right track. The diff. eq. should work basically the same as the homogeneous mass density one. –  John Moeller Mar 14 '13 at 23:17

In general, you only know that $i = \frac{dQ}{dT}$ - that is what the physical law says.
If you are working on a special case, and someone has told you $i = \frac{Q}{T}$, then yes, you could conclude that $Q = CT$. This information might be stated slightly differently, e.g. they might say "in this case, there is a steady (constant) flow of charge through the wire", in which case you can interpret that to mean $i = \frac{Q}{T}$, from which you can conclude $Q = CT$. (using the proof you give in your question)
Likewise, if you were told ahead of time "in this wire, $Q = CT$", then you could differentiate to get $\frac{dQ}{dT} = C$. Since you know from physical law that $i = \frac{dQ}{dT}$, you can say $i = C$. Then, going back to $Q = CT$ and dividing both sides by $T$ shows $\frac{Q}{T} = C$ as well, finally allowing you to conclude $i = \frac{Q}{T}$.