This question is about the generalization of Coulomb's law to continuous bodies of charge. The basic statement of Coulomb's Law involves two discrete charges $q_1$ an $q_2$:

$$\vec{F}_i = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_{12}} \hat{r}_i $$

Here $i$ represents the charge on which the force is exerted, and $\hat{r}_i$ represents the unit displacement vector between the other charge and the charge $i$.

Many treatment of electrostatics extend this law to the case that one charge is not discrete, but rather a continuous body. The force on the discrete charge $Q$ is then:

$$\vec{F} = \frac{Q}{4 \pi \epsilon_0} \int \frac{dq}{r^2} \hat{r} $$

Here $dq$ is the infinitesimal charge element of the continuous body, while $r$ and $\hat{r}$ represent the distance and displacement vectors between $dq$ and $Q$.

Continuing this way, we could probably propose an expression for force between two continuous bodies of charge, like so:

$$\vec{F} = \frac{1}{4 \pi \epsilon_0} \int \int \frac{dq_1 dq_2}{r^2} \hat{r}$$

However, I have not really seen this expression in the literature/treatments of electrostatics. Does anyone know why this is the case? Is the expression not useful, or are there no applications demanding the above expression?

  • 4
    $\begingroup$ This is a result of superposition principle applied to Coulomb's law. $\endgroup$
    – Ruslan
    Commented Aug 11, 2013 at 9:36

3 Answers 3


I am not quite sure which literature are you looking for, but it should be written in standard textbook. Anyway, you are right that it is not that useful except a formulation and exercise.

Usually, it is written in the following form. Given a charge distribution $\rho_{1}(\vec{\mathbf{r}_{1}})$ for object with volume $V_1$ and $\rho_{2}(\vec{\mathbf{r}_{2}})$ for object with volume $V_2$, we have the electric field: $$\vec{\mathbf{E}}(\vec{\mathbf{r}_{2}})=\frac{1}{4\pi\epsilon_{0}}\int_{V_{1}}\frac{\rho_{1}(\vec{\mathbf{r}_{1}})}{|\vec{\mathbf{r}_{2}}-\vec{\mathbf{r}_{1}}|^{3}}(\vec{\mathbf{r}_{2}}-\vec{\mathbf{r}_{1}})d\vec{\mathbf{r}_{1}} \tag{1}$$ and force $$\vec{\mathbf{F}}=\frac{1}{4\pi\epsilon_{0}}\int_{V_{2}}\int_{V_{1}}\frac{\rho_{1}(\vec{\mathbf{r}_{1}})\rho_{2}(\vec{\mathbf{r}_{2}})}{|\vec{\mathbf{r}_{2}}-\vec{\mathbf{r}_{1}}|^{3}}(\vec{\mathbf{r}_{2}}-\vec{\mathbf{r}_{1}})d\vec{\mathbf{r}_{1}}d\vec{\mathbf{r}_{2}} \tag{2}$$

Because of practicality, it is not used that often:

  1. It is hard to apply on non-fixed charge distribution. For example, for conductor, the charge itself would change as two object approaching. This internal dynamics is not captured in Eq (2).
  2. When the objects are far away, it is just like a point charge. Why do you want to do the cumbersome integration?
  3. There are better tools to handle this situation: Multipole expansion. As two objects become closer and closer, you include the monopole first (i.e. a point charge), then dipolar, then quadrupole... This expansion is very systematical and have good physical meaning.
  4. People care more about electric field rather than force, as it is more fundamental, so the Eq (1) is more emphasized than Eq (2).

That expression is fine as far as it goes but usually people express things in terms of fields and potentials. In other words, the physics behind your formula is present in standard treatments, it's just that the textbook authors have in mind where the material is going and so they don't phrase things in terms of forces between extended bodies directly.

The field due to object 1 is given by a standard formula

\begin{equation} \vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int\frac{dq_1}{|\vec{r}-\vec{r}_1|^3}(\vec{r}-\vec{r}_1) \end{equation}

The force on an extended object is given by the standard formula \begin{equation} \vec{F}=\int dq_2 \vec{E}(\vec{r}_2) \end{equation} (it's just $\vec{F}=q\vec{E}$ added up for each volume element)

If you combine these you get your formula.

The point of making this split is that it is not specific to the specific combination of the two objects. In other words, I can use the same electric field $\vec{E}_1$ for any object 2, there's no need to do that part of the integral separately every time.

Also, at a deeper level, it turns out that the fields are the real physical quantities that (1) are the best ways to understand electromagnetism in time dependent situations and (2) generalize to quantum mechanics. The concept of force between two objects ends up becoming less useful as you go on. So it's useful to learn how to think in terms of fields.


Coulomb's Law obeys Newton's third law - charges exert equal and opposite forces on each other under electrostatic conditions. The integral in the last expression you wrote down would evaluate to zero - it represents the total electrostatic force on a charge distribution, but there is no net electrostatic force on a charge distribution (under electrostatic conditions) as the action and reaction pairs cancel out. Thats why you haven't seen such expressions in the literature - it is too trivial to even point out.

  • $\begingroup$ That's not the case if the integrals over $dq_1$ and $dq_2$ have disjoint support (i.e. 1 and 2 are separate bodies and the integral over $dq_1$ is over the first body only etc.). $\endgroup$
    – Michael
    Commented Aug 11, 2013 at 13:39
  • $\begingroup$ @MichaelBrown: The issue isn't whether they have disjoint support. The reason the integral doesn't vanish identically is simply (except in cases with special symmetry) that $\hat{r}$ is (implicitly) defined asymmetrically as $\hat{r}=(r_2-r_1)/|r_2-r_1|$. $\endgroup$
    – user4552
    Commented Aug 11, 2013 at 15:29
  • 1
    $\begingroup$ @BenCrowell Right, so if both integrals run over both bodies then the forces cancel by antisymmetry... you have to define the integrals so one integral is over the first body and the other is over the second. The OP left this implicit and guru interpreted the equation in a different way which is trivial and probably not what the OP meant. You're right, "disjoint support" isn't really the issue. It was sloppy wording for what I meant in the brackets. $\endgroup$
    – Michael
    Commented Aug 11, 2013 at 15:55

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