0
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ What do you mean by "the electric field at source points"? $\endgroup$
    – probably_someone
    Commented Nov 23, 2018 at 14:28
  • $\begingroup$ Electric field at points inside continuous charge distribution. $\endgroup$
    – N.G.Tyson
    Commented Nov 23, 2018 at 14:31
  • $\begingroup$ By continuous do you mean equal everywhere or just varying smoothly? $\endgroup$
    – Guy Inchbald
    Commented Jun 22, 2020 at 13:49

1 Answer 1

Reset to default
1
$\begingroup$

The expression for the electric field of a point charge,

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

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):

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

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:

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

Improve this answer
$\endgroup$
4
  • $\begingroup$ What is $\vec{s}$ here? $\endgroup$
    – N.G.Tyson
    Commented Nov 23, 2018 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
    Commented Nov 23, 2018 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
    Commented Nov 23, 2018 at 16:05
  • $\begingroup$ Any idea of how to deal with singularity of line, surface or volume charges??? $\endgroup$
    – N.G.Tyson
    Commented Nov 23, 2018 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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