If the electric field and boundary conditions are known exactly for a region of space, is it true that there exists only one charge distribution in that region of space that could have produced it?

My understanding of the uniqueness theorem in electrostatics is that for a given charge distribution and boundary conditions for a volume, there exists only one (unique) solution to Poisson's equation, and thus the electric field in that volume is uniquely determined. Does the arrow point the other way, too? If we know the field and boundary conditions, is the charge distribution uniquely determined in the volume? Is there a simple example that illustrates why or why not?

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    $\begingroup$ To know the field $\phi$ is enough: up to constant factors depending on the adopted units $\rho(x) = \Delta \phi(x)$. Also $\rho = \nabla \cdot \vec{E}$ $\endgroup$ Commented Nov 17, 2014 at 19:48
  • $\begingroup$ For both of the equations you gave, is $\rho$ required to be unique from a mathematical standpoint? (i.e. Do the divergence and gradient always have unique solutions? I'm inclined to say "yes", but I'll leave it as a question.) $\endgroup$
    – higgy
    Commented Nov 17, 2014 at 19:59
  • $\begingroup$ I do not understand well. If you know $\vec{E}$ in a neighborhood of $x$, the charge density at $x$ is simply $\nabla_x \cdot \vec{E}$, there are no issues regarding uniqueness and all that... $\endgroup$ Commented Nov 17, 2014 at 20:08
  • $\begingroup$ @Valter By "gradient", I meant "Laplacian". Perhaps we could discuss this in chat? $\endgroup$
    – higgy
    Commented Nov 17, 2014 at 20:08
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    $\begingroup$ Given the field in a region R, are you asking whether (1) the charge density is uniquely determined in R, or (2) whether it's uniquely determined everywhere? If 1, then Valter Moretti's comment answers your question. If 2, then TZDZ's answer. $\endgroup$
    – user4552
    Commented Nov 17, 2014 at 21:41

3 Answers 3


Distribution of charge within the region where the field is located, is obviously uniquely defined, because it's just

$$\rho=\epsilon_0\nabla \vec{E}$$

However, if you cut out a region of space, and want to predict the contents of this region based only on the field outside, you can't do it in a unique way.

The reason is that you have too many degrees of freedom. If you decompose the field outside into the inhomogeneous contribution of the charges outside, plus the extra that was supposed to be caused by the cut out region (and obeys the Laplace equation outside the region), then the second contribution (the one with $\nabla \vec{E}=0$ outside the region) can be exactly reproduced simply by putting the correct charges on the surface of the boundary! That's what many call the holographic principle. You see now that there is one unique surface solution for every bulk distribution - but however differently you choose to distribute the charges inside the region, you can always rearrange the surface charge to accomodate the same external field. That's how conductive objects "mask" whatever you put inside them.

So... charges that occupy the space where the field is known, are uniquely defined. Charges in regions where the field is unknown, are arbitrary. Charges on the boundary of the known region are uniquely defined for each particular choice of distribution inside - assuming no charge is inside, and all field is caused by the surface, is simply one of the infinite number of solutions.


Since your question states "for a region of space", I'd say no... The electric field of a big sphere and the field of a smaller sphere are the same outside the radius of the biggest sphere (if the total charge is the same).


Charge distributed on a sphere should produce the same electric field regardless of the size of the sphere. For example, if you measure the electric field at a point 1 m from X you cant tell if the the field you observe results from a sphere of 1 cm or 99 cm radius centered at X.


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