1
$\begingroup$

Can anybody help me to understand this statement given in my book?

The electric field due to a discrete charge configuration is not defined at the locations of the discrete charges. For continuous volume charge distribution, it is defined at any point in the distribution. For a surface charge distribution, electric field is discontinuous across the surface.

Why is the field defined at all points for a volumetric distribution but not for a surface or a point charge?

$\endgroup$
  • $\begingroup$ Continue your study and you'ld get self-forces $\endgroup$ – user36790 Feb 18 '16 at 6:58
2
$\begingroup$

The field at a distance $r$ from a point charge is:

$$ E = \frac{kQ}{r^2} $$

At the point charge itself $r=0$ so the field is:

$$ E = \frac{kQ}{0} $$

and this isn't defined because we can't divide by zero. Likewise, for a 2D surface charge the field isn't defined in the surface.

This is an artefact of the model we are using to do the calculation. in reality there are no point charges because the Heisenberg uncertainty principle ensures particles are always delocalised to some extent. For the same reason there is no such thing as a purely 2D surface charge.

$\endgroup$
  • $\begingroup$ These points are known as singularities. +1. $\endgroup$ – user36790 Feb 18 '16 at 6:56
  • $\begingroup$ in reality there are no point charges because the Heisenberg uncertainty principle ensures particles are always delocalised to some extent. Uncertainty principle does not imply particles are delocalized. It limits product of standard deviations of measured coordinate and canonical momentum, but this has no implication for size of particles. Spread of wave function of an electron may be of the order of Bohr radius, but the electron is still point-like. Current upper limit on the size of electrons is something like $10^{-18}$m. $\endgroup$ – Ján Lalinský Feb 18 '16 at 7:48
  • $\begingroup$ In case of charge distributed on a surface, it is true that the electric field is discontinuous and the jump in its normal component is proportional to charge density, but actually it can be defined in the discontinuity as well. For a uniformly charged sphere, the electric field in the surface is half of the electric field outside the surface (closely above). $\endgroup$ – Ján Lalinský Feb 18 '16 at 7:52

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.