While studying Electrostatics, I was wondering about whether a charged sphere gains or loses mass while just adding electrons? If it is possible then the negatively charged sphere will have more mass than positively charged sphere, but according to me I think that since the mass of electron is so negligible then there must be no change in mass just because of electrons. So is it possible or not ?
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1$\begingroup$ It depends on what you mean by "negligible". $\endgroup$– hftCommented May 13, 2015 at 16:23
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1$\begingroup$ This is somewhat off-point, but interesting. W.B. Bonnor wrote a paper in 1960 that argued an electric charge can contribute to the gravitational mass of a charged sphere. As the radius of the sphere tends to zero, it becomes a point charge with non-zero mass. The charge itself contributes to mass. Here is the first part of the paper: link.springer.com/article/10.1007/BF01337478#page-1. Unfortunately, the paper is behind a pay wall. $\endgroup$– ErnieCommented May 13, 2015 at 16:51
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$\begingroup$ @ernie: see Why is spacetime curved by mass but not charge? and Do electromagnetic fields gravitate? and possibly Do objects have energy because of their charge?. $\endgroup$– John RennieCommented May 13, 2015 at 17:03
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5 Answers
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We can easily do this calculation. The capacitance of a sphere is:
$$ C = 4\pi\varepsilon_0r $$
and the charge is given by:
$$ Q = CV = 4\pi\varepsilon_0r V $$
The number of extra electrons is:
$$ n_e = \frac{Q}{e} = \frac {4\pi\varepsilon_0r V}{e} $$
And finally the mass of the extra electrons is:
$$ M = m_e n_e = \frac{Q}{e} = \frac {4\pi\varepsilon_0r Vm_e}{e} \tag{1} $$
Let's take a copper sphere with a radius of 10cm charged to a million volts. If we put $r = 0.1$m and $V = 10^6$V into equation (1) we get:
$$ M \approx 6.3 \times 10^{-17} \,\text{kg} $$
Pretty negligable.
answered May 13, 2015 at 16:34
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$\begingroup$ So we can say that it increases slightly as coming value is so small. Thanks ! $\endgroup$– ShashankCommented May 13, 2015 at 16:58
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Let's measure mass in $MeV$ here. The mass of an electron is about $0.5\,MeV$. So if you add one electron the mass of the sphere should increase by $0.5\,MeV$. But at the same time you also change the electric field surrounding the sphere. Due to the mass-energy equivalence the energy contained in the electric field contributes to the mass of the sphere. If for example the sphere has a potential of +1 million volt and you add one electron you will reduce that energy by $1\,MeV$. So the total change in mass of that sphere should be $0.5\,MeV$(electron mass) $-\,V_{sphere} * 1e$. Example: $$ V_{sphere} = +1 MV $$ $$ d_{mass} = 0.5MeV - 1MeV = - 0.5MeV = -8.9 * 10^{-37}\,\text{kg} $$
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The mass of the object in concern is only dependent on the way you choose to 'charge' it. If you add charged particles to it, its mass would obviously increase, and if you take away charged particles from it, the mass would increase. (Both these changes can happen by extremely negligible amounts.)
answered May 13, 2015 at 16:36
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The charges on the sphere can able exert their pressure on it ,due to the pressure the volume can get decreased and so the mass of the sphere can decrease
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$\begingroup$ The volume of an object does not actually change its mass. Only a change in the matter that constitutes the object can change the mass. You are talking about a change in density! Make sure that your answer is useful to the question. $\endgroup$ Commented Mar 22, 2016 at 10:11
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Now, try doing the same experiment by charging the sphere from the inside with a Tesla coil on the inside. The sphere is hollow a Faraday cage has formed and the high voltage Corona of the now equivalent ball lightning doesn't allow any external forms of emg penetrating it. However another Tesla coil is pumping even more electrostatic charge into it from the inside. Does ball lightning overcome gravity?
