# Why do the electrons on a charged surface not accelerate away from each other?

This question was asked to me by a 'backyard scientist' and I'm having a frustrating difficulty answering it satisfactorily.

As an example of a charged object to focus attention on, I'll use this bench-top Van de Graaff generator (VDG): The metal sphere is $$0.25\:\mathrm{m}$$ diameter, rated maximum voltage $$375,000\:\mathrm{V}$$ and maximum charge $$5.2\:\mathrm{\mu C}$$.

Based on this, I made a few 'back of an envelope' calculations and estimates for this VDG, at max. charge:

Number of electrons on the sphere $$= \frac{5.2 \times 10^{-6}}{1.6\times 10^{-19}}=3.25\times 10^{13}$$

Surface area of the sphere = $$\pi D^2=\pi \times (0.25)^2=0.20\:\mathrm{m^2}$$

Estimated distance between electrons (square packing) = $$\sqrt\frac{0.2}{3.25\times 10^{13}}=7.8\times10^{-8}\:\mathrm{m}$$

Repulsive force between two adjacent electrons (Coulomb's Law)$$=9\times10^{9}\times\frac{(1.6\times 10^{-19})^2}{(7.8\times10^{-8})^2}=3.8\times10^{-14}\:\mathrm{N}$$

Electron acceleration due to repulsive force: $$F=ma$$:

$$a=\frac{3.8\times10^{-14}}{9.1\times10^{-31}}=4.2\times10^{16}\:\mathrm{ms^{-2}}$$

That's quite a whopper! Hard to see how these electrons would not fly away from each other (and reduce their potential energy).

I found this P.S answer to a very similar question but don't find it very satisfying.

What am I missing?

• They're pulled back in by the atomic nuclei. – knzhou Dec 4 '18 at 14:49
• Can you clarify why "attraction by protons" and "work function" from the other answer don't satisfy you? – Jasper Dec 4 '18 at 14:51
• You're also forgetting that the force that an electron feels is due to the sum of the electric fields from all of the electrons, not just one. – probably_someone Dec 4 '18 at 14:55
• @knzhou what?? electrons – Sourabh Dec 4 '18 at 15:03
• @Jasper: for one, steel contains almost no protons. – Gert Dec 4 '18 at 15:04