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What is the maximum charge in Coulombs that can be stored in a copper wire? Lets say I have a thin copper wire (0.1 mm wide) that is 1 meter long. I'm asking about what could I practically achieve, not the theoretical maximum. For example how would I calculate the total charge of the wire after connecting it to a 100kV source, waiting a few seconds and then disconnecting it?

Would the largest charge in Coulombs be positive (by removing as many electrons from the wire as possible) or negative (by cramming as many extra electrons onto the wire as possible?

Additionally, if I wanted to build an object that would hold as much charge as possible for the least mass how would I do so? Would I use a conductor like copper or somehow statically charging an insulator? Would I use a long 1 dimensional shape like a wire, or a thin plate or a solid sphere or something else?

I have read some related questions How much electric charge can a copper sphere, made of N atoms, contain? and What limits the maximum sustainable surface charge density of a sphere in space? , but I am looking for what could be practically achieved, not a theoretical maximum. Even a rough order of magnitude estimation would be much appreciated.

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The objects which are designed to hold the most charge for the least mass are the capacitors which you can buy very cheap from any electronics store, such as RS Components. They do this by storing an equal amount of +ve and -ve charge in close proximity yet separated by an insulator with a high dielectric constant, which increases the charge-storing storage capacity by orders of magnitude compared with having air between the stored charges. Wires will not be anywhere near as effective.

The specification of particular capacitors will tell you the maximum voltage $V$ which can be applied. This is related to charge stored $Q$ and capacitance $C$ by $Q=CV$.


Conducting wire can be modelled as a coaxial capacitor with the outer plate at infinity. In practice the outer plate will be any of several nearby conducting surfaces, which could be anything between a few mm to a few m away.

The capacitance of a wire of length $1m$ and diameter $0.1mm$ is about $10pF$ (pF=$10^{-12}$ Farads). Typical electrolytic capacitors are between $0.1\mu F$ and $0.1mF$, a factor of $10^4$ to $10^7$ times larger - ie for the same applied voltage they can store around a million times as much charge.

Charge storage capacity is limited by the breakdown strength of the insulating material between the capacitor plates. In the case of bare wire this is air, which breaks down at a field strength of $E=3kV/mm$. (Mica, used in old capacitors, breaks down at $118kV/mm$.) Using the relation $E=\frac{\sigma}{\epsilon_0}$ this requires a surface charge density $\sigma$ of about $30pC/mm^2$. Your $1m$ wire would have an area of about $600mm^2$. Assuming a uniform charge density, it could hold at most $20nC$. With capacitance of $10pF$ the breakdown voltage would be $V=Q/C=20nC/10pF=2kV$.

The weak points for wire, at which air will break down will occur, are the sharp ends, where surface charge density and electric field are highest. So with a more rigorous calculation the maximum voltage and charge stored will be significantly lower.

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  • $\begingroup$ Thank you for the detailed answer. I am confused about how you got 30pC/mm^2. I tried to obtain it from 10pF, 3kV/mm and 600 mm^2, but I can not. Is it just a property inherent to wires?Also is it possible to separate the negative (or positive) half of the capacitor, to get something overall highly charged? Capacitors have large amounts of both positive and negative charge so overall they are electrically neutral. $\endgroup$ – Andrew Aug 14 '16 at 2:26
  • $\begingroup$ My calculation is only very approximate and uses $E=\sigma/\epsilon_0$ which applies close to the surface of any charged conductor. I don't understand what you are asking about separating the -ve or +ve half of the capacitor, because they are already separated. For parallel plates, using $C=\epsilon A/d$, if you separate the plates (increase d) this reduces C. This is (partly) why wires are poor capacitors - the 2nd 'plate' is so far away. $\endgroup$ – sammy gerbil Aug 14 '16 at 15:28
  • $\begingroup$ Thank you for the clarification about $ϵ_0$. About the separation, I mean to just have 1 of the plates in isolation, just to have a very negatively (or positively) charged object. If the plates are together (separated by a very small distance) than the overall charge of the capacitor is zero (it is electrically neutral) because there is equal positive charge on one plate balancing the negative charge on the other plate. $\endgroup$ – Andrew Aug 14 '16 at 17:19

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