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I wonder why charges reside only on the surface on conductor?

And I read this question and the answer to it:

Why charges reside on the surface on conductor? https://physics.stackexchange.com/a/210634/374591

But unfortunately, I could not understand the answer perfectly.

My question is the following:

Consider the case in which the external electric field is extremely strong.
After all the free electrons migrate to one side of the conductor, if the electric field created inside the conductor itself due to separation of positive and negative charges is not so strong and cannot be equal to the external field, what happens?

VoulKons
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佐武五郎
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1 Answers1

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Let's take mole of free electrons in a copper plate, say $d=1$ mm thick. We can estimate the electric field energy by considering the a mole of $+1$ charges separated by $d$ from a mole of electron (it won't matter that the lattice is not all on one side of the sheet--it's a correction factor of order $\frac 1 3$).

So a mole of copper (64 g/mol) has a volume of $V=7$ cc, and an area of $A=V/d=70$ sq-cm, for a charge density:

$$ \sigma = \frac{eN_A}A $$

For a field strength of:

$$ E = \frac{\sigma}{\epsilon_0}$$

with an energy density

$$ U =\frac 1 2 \frac{E^2}{\epsilon_0}$$

which when integrate over the volume is:

$$ E' = UV $$ $$E' = \frac 12 \frac d A \frac{e^2N_A^2}{\epsilon_0^3} \approx 10^{40}\, {\rm J} $$

so... " if the electric field created inside the conductor itself due to separation of positive and negative charges is not so strong" may need a re-think.

JEB
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