Combining Gauss's law and $\vec{E} = -\nabla \phi$ gives Poisson's equation
\begin{equation} -\nabla^2 \phi = \frac{\rho}{\epsilon_0} \end{equation}
for which the motivation is clear as to why the minus sign is there from a modeling standpoint. It is standard though that when you look at a Helmholtz equation with a forcing term that you look at
\begin{equation} -(\nabla^2 + k^2) \phi = f \end{equation}
I understand why this is from a variational calculus point of view. When you go through and obtain a weak form of this equation, it pops out naturally that Green's theorem turns the minus sign to a plus sign and so you have both sides positive. But is there physical motivation for this the way that there is in my example of Poisson's equation?
Here's another motivating example for my question because in this case, it doesn't seem natural why you place a minus sign. The standard derivation of the Helmholtz equation in electromagnetics starts with Maxwell's equations and with the assumptions (1) permittivity constants $\mu$ and $\epsilon$ are constant, (2) zero current density $\vec{J}$, and (3) constant charge density $\rho$. I will THROW out assumption (3) so that I can get a forcing function. From there, you apply a curl to both sides of the Maxwell-Faraday equation and apply the "curl of the curl" identity. Plug Ampere's law into the right hand side, Gauss's law into left hand side, and assumptions (1) and (2). This gives
\begin{equation} \nabla(\frac{\rho}{\epsilon}) = \nabla^2 \vec{E} -\mu\epsilon \frac{\partial^2 \vec{E}}{\partial t^2} \end{equation}
Take the Fourier transform of both sides in time to get
\begin{equation} \nabla(\frac{\rho}{\epsilon}) = \nabla^2\vec{E}+\mu\epsilon\omega^2\vec{E} \end{equation}
Now EVERYTHING has a plus sign, including the forcing function on the left. So for this example, the minus sign seems unnecessary...I haven't passed to a potential equation like $\vec{E} = -\nabla \phi$. If I pass to a weak form for this, I will now get one negative term and one positive term on the right hand side.