0
$\begingroup$

[I've seen several related questions but I ask for a confirmation here.]

I have a set of point charges to model atoms, say:

$q_1$ at $(x_1,y_1,z_1)$

$q_2$ at $(x_2,y_2,z_2)$

...

$q_n$ at $(x_n,y_n,z_n)$

Is the electric field at a given point $\vec{r}_j$ simply:

$$\vec{E}_{j} = k\sum_{i=1}^n \frac{q_i}{\vec{r}_{ij}^2} $$ ?

This remains the same even if $\vec{r}_j$ has a point charge itself, right? If not how does it change?

Thanks!

$\endgroup$
3
  • 1
    $\begingroup$ Your equation is a priori wrong, as equating a vector to a sum of scalars. $\endgroup$
    – Frobenius
    Jan 27 '17 at 19:37
  • $\begingroup$ @Frobenius Shouldn't I just change $r^2_{ij}$ to $\vec{r}^2_{ij}$ ? $\endgroup$
    – khaverim
    Jan 27 '17 at 19:39
  • 1
    $\begingroup$ See CR Drost answer. $\endgroup$
    – Frobenius
    Jan 27 '17 at 19:45
3
$\begingroup$

Almost correct; the problem is that electric fields are vectors (replacing your $1/r_{ij}^2$ with $1/r_{ij}$ you'd get a valid expression for the electric potential, which is a scalar).

Instead the electric field at a point $\vec r$ is:$$\vec E(\vec r) = \sum_{i=1}^n\frac{q_i (\vec r - \vec r_i)}{4\pi\epsilon_0|\vec r - \vec r_i|^3}.$$This does remain the same even if $\vec r$ has a point charge at it, i.e. $\vec r = \vec r_i,$ but you will notice that the expression diverges. This basically says "you cannot take two point charges (of like sign) and put them infinitely close together without considerable effort." Note that the $\vec E$ field is not the appropriate way to calculate the force on any of the $\vec r_i$ particles themselves, but rather you have to subtract its contribution from the $\vec E$ field to get a suitable field that will push on that charge. In other words, the $\vec E$ field is dispositional, it says "here is how something would be pushed at this point, if there were something here."

You can also eliminate the discontinuities by "smearing out" the charge so that $q_i = dq = \rho~dV.$ Then you get $$\vec E(\vec r) = \iiint d^3 r'~ \frac{\rho(r')}{4\pi\epsilon_0}~\frac{\vec r - \vec r'}{|\vec r - \vec r'|^3}.$$ This can potentially be well-defined everywhere.

$\endgroup$
4
  • $\begingroup$ Right on, tyvm. $\endgroup$
    – khaverim
    Jan 27 '17 at 19:45
  • $\begingroup$ But if I split the field into its components I can reduce to the form, e.g. $E(x) = k \sum_{i=1}^n \frac{q_i}{(x-x_i)^2}$, no? Specifically applied to computer simulation, where I compute $E_x$, $E_y$, $E_z$ for each point $\endgroup$
    – khaverim
    Jan 27 '17 at 19:55
  • 1
    $\begingroup$ Nonono. You are being very sloppy and assuming that things like the inverse cubic function have properties like the linearity property $f(x + y) = f(x) + f(y)$ when they don't. Instead the proper formula would have $$E_x(x,y,z) = k \sum_i \frac{q_i (x - x_i)}{[(x - x_i)^2 + (y - y_i)^2 + (z - z_i)^2]^{3/2}}.$$ $\endgroup$
    – CR Drost
    Jan 27 '17 at 20:44
  • $\begingroup$ Ah shoot. Of course. This is clear to me now, thanks again $\endgroup$
    – khaverim
    Jan 27 '17 at 20:46
1
$\begingroup$

The expression fails if $\vec r_j$ is a point where there is a charge, i.e. if $\vec r_i-\vec r_j=0$. Basically, the field becomes technically infinite there. In some cases, it is possible to find a workaround using Gauss' law to find the field of charge distributions where there is charge, but for a discrete distribution there isn't much you can technically do.

$\endgroup$

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.