0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 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 at 13:24
  • $\begingroup$ Is the surface positively or negatively charged? $\endgroup$ – Bob D Apr 24 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 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 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.