Imagine a sphere and wire charged to $10 \ \text{V}$ is connected to a neutral sphere. There are more charge carriers on the $10 \ \text{V}$ sphere and wire, so surely more electrons flow through the wire when the connection occurs. Why do we still use the same intrinsic value for $n$ in $I = n e A v$? Perhaps this is already encoded in $v$ being greater due to there being a greater field. If you were to talk about the charge density of the wire, it is supposedly (according to Wikipedia, for instance) $n \cdot e$, but surely $n$ is not the intrinsic value of $n$ for the material in this case. It would have to be either $n_{\text{charged material}} \cdot e$ for free charge density or $(n_{\text{charged material}} - n) \cdot e$ for the traditional definition of charge density.
Asked
Active
Viewed 40 times
1 Answers
1
Doesn't charge carrier density depend on the charge of the metal?
Yes and no.
In the example given in the Q., the charge brought to the sphere (or the number of electrons) is minuscule, compared to the charge/density of the electrons already present in the material, so it does not affect the value of the current an appreciable degree.
However, for materials with lower carrier density, controlling the amount of charge is possible... and indeed widely used - one could even say that I am using this effect while typing the answer:
Indeed, all kinds of semiconductor devices are using variation of carrier density, e.g.:
- by applying potential to an external gate (transistors, but also more ancient vacuum tubes technology)
- by photoexcitation of carriers (photodetectors, solar cells)
Roger V.
- 68,984