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If the electric field on a positive charge is non-zero, then the charge accelerates in the direction of the field.

The field at the surface of a conductor is perpendicular to the surface. Why don't the surface charges accelerate away from the surface of the conductor then?

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    $\begingroup$ I know that you have asked this question before...such as physics.stackexchange.com/q/95889. $\endgroup$ Mar 12 '14 at 1:52
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    $\begingroup$ Also look again at my answer to you previous question physics.stackexchange.com/q/102345 ... the conduction electrons have sufficient energy to be free of individual nuclei, but are still bound in the collective energy well of the bulk material. $\endgroup$ Mar 12 '14 at 2:14
  • $\begingroup$ @dmckee You are absolutely right, I'm sorry. I completely forgot about that question. Could you delete this for me? $\endgroup$
    – dfg
    Mar 12 '14 at 2:14
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In a conductor the electric field within itself is always equal to 0 and therefore all of the charge is found on the surfaces of the conductor, forming an equipotential. We are left with a surface charge density $\sigma$ and an electric field (close to the surface):

$$\vec{E}\approx \frac{\sigma}{2\varepsilon_{0}}\hat{n}$$

We are left with the natural question: Why do the surface charges not try to escape the boundary of the material they are contained in?

The answer is that they do, but there is a potential barrier (dependent upon the work function $\Phi$). A good way to think about this is imagining that an electron does manage to momentarily leave the surface of the object. The object will then be temporarily positively charged and the electron will experience an acceleration back towards the object it left (you can think of this using the method of image charges). Therefore, unless the electron leaves the surface with sufficient kinetic energy (as in the photoelectric effect) it will remain trapped within the confines of the conductor.

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  • $\begingroup$ But if another field is causing it to be held back (from the positive charge is leaves behind), doesn't it mean the net field is zero? $\endgroup$
    – dfg
    Mar 11 '14 at 19:41

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