Is it possible to use the Van de Graaff generator to obtain the electron mass to charge ratio? If the sphere has mass $M_{0}$ while uncharged and $M_{1}$ after charged with electrons, should that differential of mass be revealed while the measurement of the sphere's charge is easy to do?
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asked Sep 23, 2021 at 21:21
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3 Answers
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Suppose the dome of the generator to be a sphere of radius $0.25\,\rm m$, thickness $1\,\rm mm$ and made of aluminium density $2700\,\rm kg\,m^{-3}$.
The mass of such a dome is $4\,\pi\,r^2\,\delta r\,\rho \approx 2.1\,\rm kg$.
Assume the dome to be an isolated sphere then the capacitance of the dome is $4\,\pi \,\epsilon_0\, r \approx 28 \,\rm pF$ and if the potential of the dome is $100\,\rm kV$ then the charge on the dome is $C\,V \approx 28\,\rm \mu C$.
Given that the breakdown potential gradient of air is $3\,\rm MV\,m^{-1}$ and electric field at the surface of the sphere is $\dfrac {q}{4\pi\epsilon_0 r^2}\approx 400\,\rm kV\,m^{-1}$ such a potential will not cause the air to become a conductor.
$28\,\rm \mu C$ is equivalent to $\approx 1.7 \times 10 ^{13}$ electrons which have a mass of $\approx 1.6\times 10^{-17}\,\rm kg$.
So the answer to the question is "no" as one could not measure the increase in mass even without thinking about how to configure the apparatus to actually measure the mass of the charged dome or another isolated sphere connected to the dome.
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In principle, but the problem is that electrons are really really tiny. Let's consider a $r = 15$ cm Van de Graaff generator, which can be charged up to about $V = 450$ kV before the air around it breaks down. The charge $Q$ on the sphere is then related to $V$ and $r$ by $$ V = \frac{Q}{4 \pi \epsilon_0 r} \quad \Rightarrow \quad Q = 4 \pi \epsilon_0 r V, $$ and given the numbers, this means that the charge on the sphere would be about $7 \times 10^{-6}$ C. Given what we know about the electron charge and mass, this would be a deficit of about $4.5 \times 10^{13}$ electrons, which would have a mass of $4 \times 10^{-17}$ kg. This is comparable to the mass of a bacterium, and it would require heroic efforts to measure such a small mass difference in the mass of the generator, which would presumably be on the order of a few kilograms.
answered Sep 23, 2021 at 21:37
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$\begingroup$ How much difference would it make if the generator was inside a vacuum chamber? $\endgroup$– Xfce4Commented Sep 23, 2021 at 22:08
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$\begingroup$ @Xfce4: Practically speaking there's no such thing as a perfect vacuum; so to answer your question one would need to know how high the vacuum was to figure out its dielectric strength. (I say "one" because I'm not an expert in vacuum chambers; I get the sense that the dependence of dielectric strength on air pressure is a complicated subject.) ... $\endgroup$ Commented Sep 24, 2021 at 0:55
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1$\begingroup$ ... But let's assume a perfect vacuum (unattainable on Earth, but never mind.) If the electric field gets up to around $10^{18}$ V/m, the electric field is so strong that it can spontaneously create electrons & positrons (the Schwinger effect), so it's doubtful that you could get much larger than that. That would require an electric field $10^{12}$ times larger, a potential larger by the same factor, and would bump up the mass deficit to almost 0.1 grams. So... maybe? $\endgroup$ Commented Sep 24, 2021 at 0:55
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$\begingroup$ The practical way of increasing the maximum potential of the dome is to increase the pressure eg dry and purified sulfur hexafluoride (SF 6) gas or mixtures of nitrogen and carbon dioxide (80% N2 and 20% CO2) are used with pressures up to 20 bar. $\endgroup$– FarcherCommented Sep 24, 2021 at 6:58
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$\begingroup$ So it would work only for really gigantic spheres.... $\endgroup$ Commented Sep 25, 2021 at 18:42
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That is a very impractical approach. As a student I measured this ratio with a table top setup. The technique of mass spectrometry should be used. Note the e/m is 2000*A times larger for an electron than for an atom with mass number A, so you need to tweak the ration B/E quite a bit. Perhaps use an old fashioned CRT?
