# How to calculate static electric field produced by multiple point charges at a point (simple...)

[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!

• Your equation is a priori wrong, as equating a vector to a sum of scalars. Commented Jan 27, 2017 at 19:37
• @Frobenius Shouldn't I just change $r^2_{ij}$ to $\vec{r}^2_{ij}$ ? Commented Jan 27, 2017 at 19:39
• See CR Drost answer. Commented Jan 27, 2017 at 19:45

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.
• 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 Commented Jan 27, 2017 at 19:55
• 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 $$E_x(x,y,z) = k \sum_i \frac{q_i (x - x_i)}{[(x - x_i)^2 + (y - y_i)^2 + (z - z_i)^2]^{3/2}}.$$ Commented Jan 27, 2017 at 20:44
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.