I want to understand how exactly $$ \nabla^2 V = - \frac{\rho}{\epsilon_0}$$ turns into $$ \nabla^2 V = 0.$$

Of course it is by setting $ \rho$ equal to $0$ but what does setting $ \rho$ equal to $0$ mean?

$$ \int_{S} \vec E \cdot d \vec a = \int_{V} \nabla \cdot \vec E \, d \tau = \frac {Q}{\epsilon_0} = \frac {1}{\epsilon_0} \int_{V} \rho \,d\tau$$

$\rho$ is a function of $ \vec r'$ which is a position vector to the source coordinates, and $\nabla \cdot \vec E$ operates with respect to the general coordinates, we get the equation

$$ \nabla \cdot \vec E = \frac { \rho(\vec r')}{\epsilon_0},$$

which gives us

$$ \nabla^2 V = - \frac{\rho(\vec r')}{\epsilon_0}.$$

I don't know what it means to set $\rho$ equal to $0$ because I'm picturing a point charge at the origin, or let's say some configuration of charges, and I want to calculate the divergence of $ \vec E$ away from this configuration ('outside the configuration'), does that mean that I have to set $\rho =0$? And then the divergence becomes $0$? But how does that make sense?

  • $\begingroup$ Well, $\rho$ might happen to be zero in some region, no? $\endgroup$ – thedude Aug 25 '19 at 12:48
  • $\begingroup$ Yes, but does that mean that outside any charge configuration, the divergence is zero? $\endgroup$ – khaled014z Aug 25 '19 at 12:55
  • 2
    $\begingroup$ yes, it does... $\endgroup$ – thedude Aug 25 '19 at 13:16

Generally, setting $\rho$ to zero means setting it to zero everywhere in the region of interest, i.e. $\rho(\vec r) \equiv 0$.

Typically, though, we only say that the governing equation is Laplace's equation, $\nabla^2 V \equiv 0$, if there really aren't any charges in the region, and the only sources for the electrostatic field come from the boundary conditions. Thus, you might have one or more conducting surfaces that are held at a constant potential (corresponding to Dirichlet boundary conditions); those do hold charge, but you don't care about the effect of those charges directly, as they're not held constant $-$ you only care about the constant-potential boundary condition.

In contrast, if you do have sources in some of the region of interest, then we normally say that the governing equation is Poisson's equation, even if those sources only cover a fraction of the region of interest. Thus, you might have a solid sphere of charge, $$ \rho(\vec r) = \begin{cases} \rho_0 & |\vec r|\leq R \\ 0 & |\vec r|>R,\end{cases} $$ with vanishing charge density outside of a given radius, and we'd still say that you're dealing with a Poisson's-equation problem, even though for $|\vec r|>R$ the equation reads $\nabla^2 V \equiv 0$. This is primarily because the methods of solution are different as soon as you have any sources inside your region of interest.

In the particular case of a point charge, for example, we typically classify the equation as a Poisson's-equation problem, with a charge density given by $$ \rho(\vec r) = q\, \delta(\vec r) $$ in terms of a Dirac delta. (Why? because this is the only density that will give $\iiint_\Omega \rho(\vec r)\mathrm d^3\vec r = q$ if $\vec 0\in\Omega$ and $\iiint_\Omega \rho(\vec r)\mathrm d^3\vec r = 0$ if $\vec 0\notin\Omega$.) At all points of space other than the point where the point charge sits the electric-field divergence is zero, $\nabla\cdot\vec E = -\nabla^2 V =0$, but the behaviour at the origin forces a nonzero solution (which nevertheless has a vanishing laplacian) at all those other points.


Following specification should help: In electrostatics, Laplace's equation holds at any point where the charge density is zero, i.e. $\rho = 0$, even if $\rho \neq 0$ elsewhere. It informs us about how potential behaves locally at a point of consideration.


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