Suppose I have a conducting sphere with a charge of $$1.6 * 10^{-19} C. $$ What will happen when i touch this sphere with an identical neutral sphere? On which sphere will the charge reside?


I don't know solid state physics very well, so I am maybe wrong here. But since your charge is only one electron, its actual position on the first sphere is given by a probability distribution. If you now touch the first sphere with another one, The over all space on which the electron can be is enhanced by the second sphere and thus its position is defined by the probability distribution over all spheres. So you result with one object (out of two spheres) with the charge 1.6*10⁻¹⁹C. However if you now separate the two spheres again, You will again only have one sphere with charge 1.6*10⁻¹⁹C and one with charge zero, as the charge is determined by the electron, which can not split up. If the electron is truly free on the sphere, then the probability distribution on the one, as well as on both spheres is constant. If know there would be areas on any sphere attracting the electron, while not binding it, the probability distribution might be more complex, but would still reside on both spheres, if they touch.

Note: This however assumes, that the electron is not bound by any means. If the electron is bound it cannot move, while still contributing its charge.

  • $\begingroup$ how about conducting cubes, which would make this a classic QM particle in a box problem, in which the box size is suddenly doubled. The $x$-basis wave function is the same, but the energy basis eigenstates change "under" it. (Of course I'm ignoring things like lattices, temperature, and Boltzmann factors). $\endgroup$ – JEB Feb 28 at 17:34

When two conducing spheres are put in contact, the charge of the spheres is redistributed so that the potential over the two bodies is the same. If there was a potential difference between the two bodies, a charge current would flow between the two spheres until equilirium was reached.

If two conducing spheres (which are far enough from one another so that we can assume that the surface charge density is constant in each sphere), are connected with a wire, the charge will flow through the wire until the potential in both spheres is the same. The potential of a conducting sphere with charge $Q$ and radius $r$ is

$$V=\frac{Q}{4\pi\epsilon_0 r}$$

If one of the spheres has an initial charge $Q_1$ and the other one doesn't have any charge ($Q_2=0$), then, half the charge will flow to the second sphere. In your case, however, $Q_1=1.6\cdot 10^{-19}$ is the elementary charge. The charge of any object is an integer multiple of the elementary charge so you can't divide the charge between both spheres.

In a realistic case, a single electron wouldn't make much of a difference in the electric poperties of a macroscopic body. The potential difference between both spheres would be so small compared to other interactions that the electron wouldn't feel it.


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