The Reissner Nordstrom metric

$$ds^2 = - \bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)dt^2 + \dfrac{1}{\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)} dr^2 + r^2 d\Omega\_2^2 $$

is a solution of Einstein equations if the non-trivial components of the stress-energy tensor are the following: 

$$T\_{tt} = \dfrac{Q^2\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)}{r^4}$$

$$T\_{rr} = \dfrac{-Q^2}{r^4\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)}$$

$$T\_{\theta\theta} = \dfrac{Q^2}{2 r^2}$$

Here, the origin of these energy densities can be easily explained if we identify the parameter $Q$ as an electric charge. The energy densities will then be just the energy densities of the electric field of such an electric charge as obtained by solving general relativistic Maxwell's equations. And thus, the Reissner Nordstrom metric is generally regarded as a solution of coupled Einstein-Maxwell equations and is also regarded to represent a charged black hole with charge $Q$ and mass $M$. 

Now, clearly, the metric of a charged black hole will be a Reissner Nordstrom metric, but my doubt is that can this metric also represent something else? I mean can the energy densities to which the metric is a solution of Einstein equations have a different origin? I cannot think of a reason why the same gravitational situation should imply the same electromagnetic situation. Is there any argument which dictates that these energy densities can only arise out of a radially falling static electric field?