Equation S1
From Gauss's law,
\[\int_S^{ }E\cdot dS=\frac{Q}{\varepsilon}\ .\]
From Coulomb's law,
\(\begin{equation}
F=\frac{1}{4\pi\epsilon}\cdot \frac{Q_1Q_2}{|\bm{r}_1-\bm{r}_2|} \cdot \frac{\bm{r}_1-\bm{r}_2}{|\bm{r}_1-\bm{r}_2|^2} .
\end{equation}\)
From the model in Figure 3F,
\(\begin{align}
\overrightarrow{OA}&=(-a\sin\theta, a\cos\theta)\\
\overrightarrow{OB}&=(b\sin\theta, b\cos\theta) .
\end{align}\)
where the force received by the charges at position A from the 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 to the above, it is important to note that the moment in FAB around O is
\(\begin{equation}
\overrightarrow{\Delta N}=\overrightarrow{OA}\times \overrightarrow{F_{AB}} .
\end{equation}\)
From the total moment around O=0, we get
\(\begin{equation}
\overrightarrow{N}=\sum_a \sum_b \overrightarrow{OA}\times \overrightarrow{F_{AB}}=mg \cdot \frac{h}{2}\cdot \sin\theta .
\end{equation}\)
Substituting in equation 4, 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 foil electroscope 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, permittivity: ε, gravitational acceleration: g, foil mass: m, foil length: h, foil junction point: O
Figure S5
Photo of the fluorescent lamp flushed.