Editing Openai/68713b82-90e0-8006-8973-ea152bf13a2c (section)

====== - You trap free electrons inside — say n=1014n = 10^{14}n=1014 electrons (this is realistic for lab-scale static charge buildup). ======
* Each electron carries 1.6×10−19 C1.6 \times 10^{-19} \, C1.6×10−19C.
* Total charge Q=n⋅e=1014⋅1.6×10−19=1.6×10−5 CQ = n \cdot e = 10^{14} \cdot 1.6 \times 10^{-19} = 1.6 \times 10^{-5} \, CQ=n⋅e=1014⋅1.6×10−19=1.6×10−5C

Let’s assume effective capacitance of the box is similar to a sphere of radius ≈1 cm:

C=4πε0r=4π(8.85×10−12)(0.01)≈1.1×10−12 FC = 4 \pi \varepsilon_0 r = 4 \pi (8.85 \times 10^{-12}) (0.01) \approx 1.1 \times 10^{-12} \, FC=4πε0r=4π(8.85×10−12)(0.01)≈1.1×10−12F
Then:

V=QC=1.6×10−51.1×10−12≈1.45×107 volts (14.5 MV)V = \frac{Q}{C} = \frac{1.6 \times 10^{-5}}{1.1 \times 10^{-12}} \approx 1.45 \times 10^{7} \, \text{volts} \, (14.5 \, \text{MV})V=CQ=1.1×10−121.6×10−5≈1.45×107volts(14.5MV)
⚠️ That’s a huge voltage — not practically achievable this way without dielectric breakdown (air breaks down at ~30kV/cm).

So the max voltage across a 2cm box in air is:

30 kV/cm×2 cm=60 kV30\, \text{kV/cm} \times 2 \, \text{cm} = 60\, \text{kV}30kV/cm×2cm=60kV
So only a much smaller amount of charge is stable before dielectric breakdown occurs (unless in vacuum or insulation).