In classical electromagnetism, is there a way to find the electric field at source points of a line, surface or volume charge? If there is a way, please explain in detail

My idea: I don't really know; but it seems to me we can proceed by making an infinitesimal $-$ linear, circular or spherical $-$ cavity and then find the electric field inside the cavity. By the way, I do not even have an approach to find the electric field inside the cavity we made.

  • $\begingroup$ What do you mean by "the electric field at source points"? $\endgroup$ – probably_someone Nov 23 '18 at 14:28
  • $\begingroup$ Electric field at points inside continuous charge distribution. $\endgroup$ – N.G.Tyson Nov 23 '18 at 14:31

The expression for the electric field of a point charge,

$$\vec{E}=\frac{q}{4\pi\epsilon_0 r^2}$$

can be extended to a continuous charge distribution divided into parcels with approximately constant density $\rho$ with volume $dV$ (as $dV\to 0$, $\rho$ becomes exactly constant):

$$\vec{E}=\frac{\rho}{4\pi\epsilon_0 r^2}dV$$

To get the total electric field at a point $\vec{r}$, we must add up the infinitesimal contributions from each parcel of continuous charge. This is accomplished using an integral:

$$\vec{E}(\vec{r})=\int \frac{\rho(\vec{s})}{4\pi\epsilon_0 |\vec{r}-\vec{s}|^2}\; d^3\vec{s}$$

  • $\begingroup$ What is $\vec{s}$ here? $\endgroup$ – N.G.Tyson Nov 23 '18 at 15:54
  • $\begingroup$ The position of a particular point emitting electric field within the continuous charge distribution. When you integrate, $\vec{s}$ ranges over every point in the distribution (well, really, it ranges over all space, but these are the same because $\rho=0$ in the vacuum). $\endgroup$ – probably_someone Nov 23 '18 at 15:58
  • $\begingroup$ $\vec{s}$ ranges over every point in the distribution...Even at the point where $\vec{s}=\vec{r}$?? Won't it be a singularity.... I do not understand this. Please explain $\endgroup$ – N.G.Tyson Nov 23 '18 at 16:05
  • $\begingroup$ Any idea of how to deal with singularity of line, surface or volume charges??? $\endgroup$ – N.G.Tyson Nov 23 '18 at 16:56

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