Equation S1
From Gauss's law, we get
\[\int_S^{ }E\cdot dS=\frac{Q}{\varepsilon}\ .\]
From Coulomb's law, we get
\(\begin{equation}
F=\frac{1}{4\pi\epsilon}\cdot \frac{Q_1Q_2}{|\bm{r}_1-\bm{r}_2|} \cdot \frac{\bm{r}_1-\bm{r}_2}{|\bm{r}_1-\bm{r}_2|^2} .
\end{equation}\)
From the model in Figure 3F, we get
\(\begin{align}
\overrightarrow{OA}&=(-a\sin\theta, a\cos\theta)\\
\overrightarrow{OB}&=(b\sin\theta, b\cos\theta) .
\end{align}\)
where the force experienced by the charge at position A from that at position B is
\(\begin{equation}
F_{AB}=\frac{1}{4\pi\varepsilon}\cdot \frac{Q_1Q_2}{|\overrightarrow{OA}-\overrightarrow{OB}|} \cdot \frac{\overrightarrow{OA}-\overrightarrow{OB}}{|\overrightarrow{OA}-\overrightarrow{OB}|^2} .
\end{equation}\)
In addition, it is essential to note that the moment resulting from FAB around O is
\(\begin{equation}
\overrightarrow{\Delta N}=\overrightarrow{OA}\times \overrightarrow{F_{AB}} .
\end{equation}\)
By setting the total moment around O=0, we get
\(\begin{equation}
\overrightarrow{N}=\sum_a \sum_b \overrightarrow{OA}\times \overrightarrow{F_{AB}}=mg \cdot \frac{h}{2}\cdot \sin\theta .
\end{equation}\)
On substituting this in Equation 4, we get , we get
\(\)\(\begin{align}
\sum_a \sum_b \overrightarrow{OA} \times \frac{\overrightarrow{OA}-\overrightarrow{OB}}{|\overrightarrow{OA}-\overrightarrow{OB}|^3} \cdot \frac{\rho \Delta r \cdot \rho \Delta r}{4\pi\varepsilon}=mg\cdot \frac{h}{2}\cdot \sin\theta.
\end{align}\)
Therefore, the relationship between the angle of the electroscope leaves and the charge density is as follows:
\(\begin{equation}
\rho =\sqrt{\frac{\sin\theta}{\sum_a \sum_b \overrightarrow{OA} \times \frac{\overrightarrow{OA}-\overrightarrow{OB}}{|\overrightarrow{OA}-\overrightarrow{OB}|^3}} \cdot \frac{mgh}{2} \cdot \frac{4\pi\varepsilon}{\Delta r^2}}
\end{equation}\)
where, ε is the permittivity of air, g is the acceleration due to gravity, m is the leaf mass, h is the leaf length, and O is the leaf junction point.
Figure S6
(A) Photograph of fluorescent lamp lit using SEG: (i) before being driven by SEG and (ii) after being driven by SEG. (B) Inside the fluorescent lamp: Electrons supplied by SEG are emitted from the electrodes of the lamp. The mercury vapor reflects the electrons that produce ultraviolet light, which is emitted by the luminescent material coated on the inner surface of the glass.