Imagine we have two parallel wires with a potential difference of $V$ volts that form the opposite sides of a square of size $\lambda$.
Any virtual electron-positron pairs that form between the wires pick up an acceleration $a$ given by
$$a\sim\frac{eV}{m_e\lambda}.\tag{1}$$
According to the Unruh effect the accelerating electron and positron are immersed in a heat bath with temperature $T$ given by
$$T \sim \frac{ha}{ck_B}.\tag{2}$$
For simplicity let us assume that we just have photons of wavelength $\lambda$ such that
$$k_BT\sim\frac{hc}{\lambda}.\tag{3}$$
Combining Equations $(1\hbox{-}3)$ we find that the voltage $V$ must be given by
$$eV\sim m_ec^2.\tag{4}$$
An inertial observer sees the electron-positron pairs radiating at temperature $T$. According to the Stefan-Boltzmann law the radiation flux $j$ through the square is given by
$$j=\sigma T^4\tag{5}$$
where
$$\sigma\sim \frac{k_B^4}{h^3c^2}.\tag{6}$$
If we assume the radiation consists of photons with wavelength $\lambda$ as described by Equation $(3)$ then the number of photons per second, $N$, through the square is given by
$$N \sim \frac{j\lambda^2}{k_BT}\sim\frac{\sigma T^3 \lambda^2}{k_B}.\tag{7}$$
Combining Equations $(3)$,$(6)$ and $(7)$ we find that the number of photons going through the square per second, $N$, is given by
$$N\sim\frac{c}{\lambda}\sim \nu\tag{8}$$
where $\nu$ is the frequency of the radiation of wavelength $\lambda$.
It seems to me that these real photons are generated from the vacuum itself. However as I have not used any quantum field theory in this argument my conclusions are almost certainly completely false!
Reference post: Does an electric field cause the vacuum to emit photons?
