We have asserted that, when a conductor is placed in an external E vector-field,

Question

We have asserted that, when a conductor is placed in an external E vector-field, the conduction electrons move under the influence of the field until they set up their own field that perfectly balances the external field at all points inside the conductor. This process is often described in terms of charge "pile-up" at the surface of the conductor. In this problem we examine the meaning of the term "pile-up". The figure below illustrates how this process works for a thin sheet of copper placed perpendicular to the external field. In absence of the field (left figure), the negative conduction electrons and positive copper ions fill the same volume, yielding a local charge density equal to zero at all points. When the external field is applied (right figure), the conduction electrons are displaced en masse (keeping the same volume density) until the resulting surface charge - negative on the left and positive on the right - produces a field that perfectly balances the external field. Now, according to the literature, the largest E-fields that can be produced at the surface of a conductor are no greater than 250 times 106 V/m. Beyond this value the metal breaks down due to field emission. Of course, any fields you are likely to encounter in the lab are much smaller than this. By what displacement tau, shown in the figure above, would all of the conduction electrons have to be displaced to perfectly compensate an external field of magnitude 250 times 106 V/m in copper? How does this displacement compare to the radius a0 of a hydrogen atom, a0 = 5.3 times 10-11 m? (Copper has one conduction electron per atom. Its mass density is 8.96 g/cm3, and its atomic mass is 63.546 g/mol.)

Explanation / Answer

The idea is that you know how many conduction electrons there are per unit volume, and you know how the electric field relates to the surface charge density of the electrons. You can put the two ideas together to answer the question.

number of conduction electrons per unit volume:

8.96 g/cc * (1 mol / 63.546 g) * 6.022 x 10^23 electrons / mol * 1.60 x 10^-19 coulomb/electron =

= 13586 coulomb/cc = 1.3586 x 10^10 coulomb/m^3

surface charge density:

E = density / epsilon naught so

density = E * epsilon naught = 250 x 10^6 * 8.85 x 10^-12 = 0.0022125 coulomb/m^2

Now... imagine a thin layer of charge with thickness tau, as illustrated. You can find that thickness by using the fact that

area * thickness = volume so

thickness = volume / area = (charge / area) / (charge / volume) =

= 0.0022125 coulomb/m^2 / 1.3586 x 10^10 coulomb/m^3 =

= 1.63 x 10^-13 meter

This is a tiny fraction of the radius of a hydrogen atom, i.e.

1.63 x 10^-13 / 5.3 x 10^-11 = 0.00307 times the radius of a H atom

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