One of the Laplace equation's property says that the maxima and minima can only occur at the boundaries. Okay so lets take 2 positive charges, one at the origin and the other $d$ distance apart on the $x$-axis. So the potential between them would be somewhat like what I have drawn in the image. enter image description here

Now lets take a region between $x=d/3$ and $x=2d/3$. Now apply the Laplace equation here in the region, (as there is no charge in this region), and so the potential's maxima and minima should occur at the boundary. But its maxima occurs at $x=d/2$ ??

  • $\begingroup$ @AlmostClueless, that's why I am taking the region of interest to from $d/3$ to $2d/3$ $\endgroup$ Aug 26, 2020 at 9:27
  • $\begingroup$ As pointed out, Coulomb's law applies to 3 dimensions. If we extend those charges into the y- and z- directions, we get plates of infinite size. This is a capacitor. If both plates carry the same charge density, then the potential is constant everywhere between the plates. In other words, it satisfies the Laplace equation property. $\endgroup$ Aug 26, 2020 at 20:28

4 Answers 4


The issue is that you have applied a 3-dimensional concept (i.e. Laplace's equation and potential falling with $\frac{1}{r}$) to a 1 dimensional boundary I think. The only solution to Laplace's equation in 1-dimension are solutions of the form $f = ax+b$ which obviously satisfies the conditions. If you consider the potential in 3 dimensions, then you'd actually realise it's a saddle point at $x = \frac{d}{2}$ (or a saddle point in four dimensions, since you have $V = V(x,y,z)$ if you get me).

I'm a bit rusty on electromag, so I could be mistaken here but I think this is it.

  • 1
    $\begingroup$ This might not be a question connected to the topic, but how can you see that in 1D the function $f = ax+b$ satisfies the conditions? (this is probably my misunderstood of the physical and mathematical meaning of being solution to laplace equation) $\endgroup$
    – Sagigever
    Aug 26, 2020 at 22:28
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    $\begingroup$ @sagigever In 1 dimension Laplace's equation is just $d^2y/dx^2=0$. $\endgroup$
    – Carmeister
    Aug 26, 2020 at 23:28
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    $\begingroup$ @Sagigever The condition I was referring to was the extrema being located on the boundaries for solutions to Laplace's equation, which the function $f$ described does in 1 dimension $\endgroup$
    – FizzKicks
    Aug 27, 2020 at 13:02
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    $\begingroup$ @Sagigever if $ax$ is positive, then $ax+b$ increases linearly so the minimum must necessarily be the left-most boundary and the maximum the right-most. If $ax$ is negative, the opposite holds. $\endgroup$
    – Andy
    Aug 27, 2020 at 20:29

(Assuming that you're referring to the $3$D case and not the $1$D case).

In your example, $V(r)$ is not an extremum. Let's place two unit charges at $x = \pm d/2$, and look at the resulting potential $V(x,y)$ (taking $V(r \to \infty) = 0$).

At $x = 0$, the potential is $V(0,0) = \frac{2}{d} + \frac{2}{d} = \frac{4}{d}$

Along the $x$ axis (close to $x = 0$), we have

\begin{align} V(x=\varepsilon, 0) &= \frac{1}{\frac{d}{2} + \varepsilon} + \frac{1}{\frac{d}{2} - \varepsilon}\\ &= \frac{2}{d} \left(\frac{1}{1 + \frac{2 \varepsilon}{d}} + \frac{1}{1 - \frac{2 \varepsilon}{d}}\right)\\ &= \frac{4}{d} \left(1 + 4 \frac{\varepsilon^2}{d^2} + o\left(\frac{\varepsilon^2}{d^2}\right) \right) > V(0, 0). \end{align}

But along the y axis, we have:

\begin{align} V(x=0, y=\varepsilon) &= \frac{1}{\sqrt{\frac{d^2}{4} + \varepsilon^2}} + \frac{1}{\sqrt{\frac{d^2}{4} + \varepsilon^2}}\\ &= \frac{4}{d} \frac{1}{\sqrt{1 + \frac{4 \varepsilon^2}{d^2}}}\\ &= \frac{4}{d} \left(1 - 2 \frac{\varepsilon^2}{d^2} + o\left(\frac{\varepsilon}{d}^2\right) \right) < V(0, 0). \end{align}

So while the partial derivative along $x$ and $y$ is indeed zero, this corresponds neither to a maximum nor a minimum of $V$ but rather to a saddle point.


I was confused about this for an embarrassingly long time until I actually plotted it out, and so to supplement @MC2k's excellent answer, here is a small plot of the potential of two positive point charges, placed at $(0,0)$, and $(5,0)$.

As you can see, while there exists what might appear to be a minimum along the $x-$axis (the line joining the charges) there is no minimum along the $y-$axis. Clearly, $V(x,y)$ has no local maxima or minima, and the point in question is a saddle point.

enter image description here


Other answers have shown the point in question is not a maximum of the potential as a function in three dimensions and hinted that the reason things went wrong has something to do with dimension. While it is true that the subtle mathematical point of confusion has to do with dimension, I'd like to clarify that the relevant theorem, known as the maximum principle, holds just fine. It states:

If $U \subset \mathbb{R}^n$ is open and $u: U \to \mathbb{R}$ is a (classical) solution to the $n$-dimensional Laplace equation $\Delta u = 0$, then for any precompact set $V \subset U$, the restriction of $u$ to the closure of $V$ achieves its maximum and minimum values on (and only on, provided $u$ isn't constant) the boundary $\partial V$.

You've observed that the potential $u$ generated by the two charges on the open set $U = \mathbb{R}^3 \backslash \{(0,0,0),(d,0,0)\}$ is indeed a (classical) solution to Laplace's equation in three dimensions on its domain, and you're trying to make a conclusion about the extrema of $u$ when restricted to the compact set $V= \{(x,0,0) \; | \; d/3 \leq x \leq 2d/3\} \subset U$.

The theorem then applies, and its conclusion that the maximum of the restriction of $u$ to $V$ is achieved on $\partial V$ must hold. What gives? Notice that we've applied the theorem in the ambient space with $n=3$, and so $\partial V$ refers to the boundary of the set in $\mathbb{R}^3$. Mathematically, the boundary is defined as $\overline{V} \backslash V^\circ$, the closure of $V$ remove its $3$-dimensional interior. But $V$ is closed and, being a line segment, $V$ has no $3$-dimensional interior, so in fact $\partial V = \overline{V} = V$. Hence the theorem, while applicable and true, is giving us no information because we've chosen too small of a set to apply it to-- it just states that the largest value $u$ takes on $V$ is achieved somewhere in $V$. The set of interest must have an interior, some "wiggle room", in order for the maximum principle to yield nontrivial information.

One might take an alternative perspective and say we're identifying the $x$-axis with $\mathbb{R}$ and considering the potential $u$ as a function on $U = \mathbb{R} \backslash \{0,d\}$. Now the set of interest $V = [d/3,2d/3] \subset U$ is again a compact subset of this one-dimensional space, and its boundary $\partial V$ is legitimately the two points $\{d/3, 2d/3\}$. The hypotheses of the theorem, however, are no longer satisfied (and hence the conclusion need not apply) because we are now working in $n=1$, but $u$ is not a solution of the $1$-dimensional Laplace equation $\frac{d^2 u}{dx^2} = 0$.


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