What determines how much electrical charge an object can hold? Does increase voltage force more electrical charge to be store in an object (Van de Graaff generator), since electric field increase as voltage increase. I don't think it is about relative permittivity in dielectric material because it just creates a bipolar.

  • $\begingroup$ i think its the surface area. $\endgroup$ – Hubble07 Feb 24 '14 at 13:42
  • $\begingroup$ So doesn't matter what material or applied voltage it is? $\endgroup$ – Kevin Feb 24 '14 at 13:44
  • $\begingroup$ the material will certainly matter but i was thinking like if we had two object of same material but different surface area at same applied potential then the one with larger surface area can hold more charge. $\endgroup$ – Hubble07 Feb 24 '14 at 13:54
  • $\begingroup$ Thanks, but what are other factors, is it electric field strength and free electron? $\endgroup$ – Kevin Feb 24 '14 at 14:00
  • $\begingroup$ regarding material this can matter en.wikipedia.org/wiki/Triboelectric_effect $\endgroup$ – Hubble07 Feb 24 '14 at 14:09

It should be associated with the work function, which is the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface, and different materials have different work functions.

Consider a very simple case, that a spherical electrical object exists in vacuum.

Considering that we move an electron from the object to the infinitely distance, if the energy of the system decrease, then it shows the electrical object is unstable. Hence, the maximum electric charge that an object can hold should make the energy unchanged in the process of removing electron.

If its radius is $R$, the work function is $W$, the maximum electric charge that it can hold is $Q$, then $$\frac{1}{4 \pi \epsilon_0} \frac{Qe}{R}=W$$ the left side is the change of electrical energy.

As for the object with other shape, you have to change the form of expression in the left side.

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  • $\begingroup$ Effective explanation. +1 $\endgroup$ – Gaurav May 27 '15 at 6:14

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