Electric field just near the surface of a surface charge is $\dfrac{\sigma}{2 \epsilon_0} (\hat{n})$ on one side and $-\dfrac{\sigma}{2 \epsilon_0} (\hat{n})$ on the other side.

What would be the electric field on the surface of a surface charge?

  • $\begingroup$ How are you getting two different signs? If the surface is positively charged the field will be + on both sides (pointing away from the surface). If it is negatively charged, the field will be - on both sides (pointing into the surface). $\endgroup$ – Bob D Apr 24 '19 at 13:03
  • $\begingroup$ w.r.t. our coordinate system, the unit vector has been inverted in the latter. $\endgroup$ – N.G.Tyson Apr 24 '19 at 13:24
  • $\begingroup$ Is the surface positively or negatively charged? $\endgroup$ – Bob D Apr 24 '19 at 13:41
  • $\begingroup$ well...if you consider the electric field the direction a test charge would move to... and you are in the plane of symmetry... $\endgroup$ – mikuszefski Apr 24 '19 at 14:17
  • $\begingroup$ well, there is no on the surface. You are above or in the surface. Where would a test charge move to if it is in the surface? $\endgroup$ – mikuszefski Apr 25 '19 at 8:10

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