Suppose that the ionosphere is in thermal equilibrium at temperature $T$. It consists of electrons (mass $m$ ) and singly charged ions (mass $M$ ). If there were no large-scale electric field, the electrons and ions would have very different scale heights, $k T / m g$ and $k T / M g$ respectively. This would produce a large electric field, thereby modifying the density vs. height relation.

Taking into account the electrostatic field, find the numerical value of $E$ if the ions are protons.
(You may ignore the curvature of the Earth's surface. it may be helpful to know that the electric field turns out to be uniform.)
Use a coordinate system with origin at the Earth's surface and the $z$-axis vertically upward. The self-consistent electric field $\mathbf{E}$ is along the $z$ direction:

$$
E=-\frac{\partial V}{\partial z},
$$

or

$$
V=-E z
$$

assuming $E$ to be uniform. Then an electron and an ion at height $z$ will respectively have energies

$$
\begin{aligned}
&\varepsilon_{\mathrm{e}}=m g z+e E z \\
&\varepsilon_{\mathrm{i}}=M g z-e E z
\end{aligned}
$$

and number densities

$$
\begin{aligned}
&n_{\mathrm{e}}=n_{\mathrm{e} 0} e^{-\varepsilon_{\mathrm{e}} / k T}=n_{\mathrm{e} 0} e^{-(m g+e E) z / k T}, \\
&n_{\mathrm{i}}=n_{\mathrm{i} 0} e^{-\varepsilon_{\mathrm{i}} / k T}=n_{\mathrm{i} 0} e^{-(M g-e E) z / k T}
\end{aligned}
$$

as functions of height.Since

$$
\Delta \cdot \mathbf{E}=\frac{\rho}{\epsilon_{0}}=0
$$

for a uniform field, the net charge density $\rho=0$ everywhere. This means that

$$
-e n_{\mathrm{e}}+e n_{\mathrm{i}}=0,
$$

or

$$
n_{\mathrm{e} 0}=n_{\mathrm{i} 0}, \quad m g+e E=M g-e E .
$$

The last equation gives

$$
E=\frac{(M-m) g}{2 e} \approx \frac{M g}{2 e}
$$

as $M \gg m$.If the ions are protons, we have $M=1.672 \times 10^{-27} \mathrm{~kg}$, $e=$ $1.602 \times 10^{-19} \mathrm{C}$, and

$$
E=\frac{1.672 \times 10^{-27} \times 9.807}{2 \times 1.602 \times 10^{-19}}=5.12 \times 10^{-8} \mathrm{~V} \mathrm{~m}^{-2} .
$$