Curvature of electrostatic potential is zero Could you please expound upon this claim? I found such claim on Zangwill's Classical Electrodynamics, which states that constraint coming from Laplacian equation implies 


*

*electrostatic potential has zero curvature, and 

*is not bounded in at least one direction.
What kind of curvature are we talking about here? 
Please explain both points for me.
 A: The electric potential $\phi:\mathbb{R}^3\to\mathbb{R}$ is the solution to Laplace's equation and therefore a harmonic function. Harmonic functions enjoy several nice properties, some of them listed on the Wikipedia page. 
Concerning OP's second point, let us mention that there is a theorem similar to Liouville's theorem from complex analysis that a bounded harmonic function defined on the whole $\mathbb{R}^3$ is a constant function .
Concerning OP's first point, Zangwill is looking at the graph 
$$ {\rm graph}(\phi)~=~ \{({\bf r}, \phi({\bf r}))~\in~ \mathbb{R}^4\mid {\bf r}\in\mathbb{R}^3 \} ~\subset~ \mathbb{R}^4 .$$
The graph of $\phi$ is a 3-dimensional submanifold with possible curvature embedded in $\mathbb{R}^4$. The metric on the graph is induced from the standard metric on $\mathbb{R}^4$.  
A: There are several definitions of the term curvature in mathematics, but I think the simplest explanation of the first point is in regards to the second derivative of a function.  The Laplacian is really just the second derivative of the electric potential equal to zero.  Think of a straight line.  Such an object has a finite first derivative but zero second derivative.  In contrast, a parabola will have finite first and second derivatives, and thus a finite curvature.
I think this is consistent with @Qmechanic's post, but he/she is much more adept at math than I so I will let them clarify.
Side Note:  The Poisson equation for an electrostatic potential, for comparison, can have curvature under the definition I used.
