# Electric field at source points of any continuous charge distribution

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.

• What do you mean by "the electric field at source points"? – probably_someone Nov 23 '18 at 14:28
• Electric field at points inside continuous charge distribution. – 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:

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

• What is $\vec{s}$ here? – N.G.Tyson Nov 23 '18 at 15:54
• 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). – probably_someone Nov 23 '18 at 15:58
• $\vec{s}$ 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 – N.G.Tyson Nov 23 '18 at 16:05
• Any idea of how to deal with singularity of line, surface or volume charges??? – N.G.Tyson Nov 23 '18 at 16:56