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The field near the surface of a conductor is given by

$$\bf{E}=\frac{\sigma}{\epsilon_0} \hat{\bf{n}}$$

Where $\sigma$ is the local surface charge density.

My question is: is this field $\bf{E}$ always outgoing from the conductor surface (as in picture)? What is the reason why it must be outgoing and it cannot be ingoing in the conductor?

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    $\begingroup$ is $\sigma$ unsigned? because in general the direction depends on the sign of the charge of the conductor... $\endgroup$ – AccidentalFourierTransform Jan 3 '17 at 19:29
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As @AccidentalFourierTransform pointed out in the comments, in general the direction of the electric field is related to the sign of the charge (density) generating it:

  • outgoing if the charge is positive
  • ingoing if the charge is negative

So the answer to your question is: "depends on the sign of the surface charge on the conductor". In the image you posted, as you can see, the surface charge density is positive and this leads to an electric field pointing away from the surface.

Just to be clear, as you ask whether E can be ingoing in the conductor: if you mean, as I suppose, towards the conductor, then the answer is above; if you on the other hand mean inside the conductor, then the answer is no: there cannot be an electrif field inside a conductor (as pointed out in the image in your question).

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The E field near a smooth ideal conductor is always perpendicular to the surface element, because conduction nulls the lateral voltage gradient, and the E field is in the direction of the voltage gradient. This means that those field lines around the pear-shape can be annotated with a little right-angle symbol, if you wish, to the pear-shape's surface.

And, it's why one's hair stands up on end when a large electric charge is applied, Van de Graaff hair and why the dust on a rubbed balloon looks like little hairs sprouting from the surface.

The polarity of the charge determines the field sign, whether it is into or out of the surface.

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