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Is there a material with fairly high resistivity (at least semi-conductor level), but also allows the flow of charge through it (and subsequently to the ground)? The flow of charge does not need to be fast, it can be very slow if necessary. The higher resistivity, the better.

So ideally, if that material is left alone on the ground, its steady-state should have very little charge, and thus have negligible or zero electric field (even if you initially applied some charge to it). Basically, the material is able to be discharged in finite-time, regardless of its fairly high resistivity.

The speed of discharge is the property I'm particularly interested in, but based on what I've seen, this property might be independent of resistivity.

I am not sure if such properties are documented, so if you know what the property is called, please tell.

Thanks

Edit: My wording is apparently confusing, so let me try to put it in other words. Here is a phenomena I have observed. There is a piece of rubber and a piece of glass on the ground, both equal in size and resistivity. I apply a static charge to both, and the rubber ends up discharging (to ground presumably) much faster than glass. What is that material property called?

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    \$\begingroup\$ you appear to be talking about a standard resistor, maybe a mixture of a resistor and a capacitor ..... very unclear \$\endgroup\$
    – jsotola
    Commented May 23, 2018 at 2:13
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    \$\begingroup\$ @ jsotola "very unclear". More like enigmatic. Can you explain what it is you want to accomplish, in a real world application. In fewer well thought out words? \$\endgroup\$
    – Sara Heart
    Commented May 23, 2018 at 2:21
  • \$\begingroup\$ What parts are you unclear about exactly? I wish to discharge static electricity via a material that can also block current (i.e. has high resistivity). \$\endgroup\$ Commented May 23, 2018 at 2:42
  • \$\begingroup\$ Please see edit \$\endgroup\$ Commented May 23, 2018 at 2:57
  • \$\begingroup\$ I think what you want is a capacitor with a bleed resistor on it. So it can accept some charge, but the charge bleeds off over time. \$\endgroup\$
    – user57037
    Commented May 23, 2018 at 3:54

2 Answers 2

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To oversimplify a bit, the discharge will take place with a time constant (measured in seconds, hours, microseconds etc.) that depends on the product of the capacitance and the resistance.

The capacitance of an conductive object in air is mostly dependent on its surface area. An isolated sphere of radius R has a capacitance of \$4\pi\epsilon_0R\$.

The resistance is dependent on the surface area of contact and the resistivity \$\rho\$ of the material.

So, though geometry certainly enters into it, one can say that for a given geometry, in air, the discharge time constant is inversely proportional to the resistivity of the material. Resistivity is a fundamental material property that you can look up in a chart or on the internet.

Remember that discharge is an exponential decay. It does not make a lot of sense to say it is completely discharged, but after 10 time constants maybe 50 parts in 1,000,000 of the original charge remains (\$e^{-10}\$ if you want to work it out).

Take care that materials which are decent insulators tend to have a wide range of possible resistivities.

Eg.

Glass: \$10^{11} \text { to } 10^{15}\$ Ω·m

Hard Rubber: ~\$10^{13} \$ Ω·m

Acrylic: \$2\cdot 10^{15} \text { to } 1.4\cdot 10^{16} \$ Ω·m

So it's quite easy to get an order of magnitude or two difference in resistivity. Resistivity of insulators tends to be very dependent on temperature. Glass at 1500'C is about as conductive as damp wood, about 250 billion times more conductive than at room temperature.


The above assumes that volume resistivity dominates. If the surface is coated with something that is relatively conductive then things change dramatically. There are various types of anti-static spray that are designed to leave such a residue. Some of them are closer to metals, and others are so high in resistivity that they cannot be measured with an ordinary multimeter.

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  • \$\begingroup\$ Thanks for the response. I'm going to need to refresh my memory on capacitance and get back to you \$\endgroup\$ Commented May 23, 2018 at 6:18
  • \$\begingroup\$ Regarding your first paragraph, could you point me to the exact formula? \$\endgroup\$ Commented May 23, 2018 at 6:31
  • \$\begingroup\$ \$ v(t) = V_O(1-e^{t/\tau})\$ where \$\tau = R\cdot C\$, making all kinds of possibly inappropriate assumptions. If the bulk resistivity is the main factor the charge near the contact will discharge earlier than those further away, and there won't be any easy closed form formula. \$\endgroup\$ Commented May 23, 2018 at 15:28
  • \$\begingroup\$ Thanks, I am satisfied with your answer. One last thing, could you point me to a formula for resistance? I would like to see all the factors that affect resistance \$\endgroup\$ Commented May 24, 2018 at 3:21
  • \$\begingroup\$ Simplest: physics.udel.edu/~watson/scen103/cylinder.html \$\endgroup\$ Commented May 24, 2018 at 3:29
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There is a piece of rubber and a piece of glass on the ground, both equal in size and resistivity. I apply a static charge to both, and the rubber ends up discharging (to ground presumably) much faster than glass. What is that material property called?

The rubber has higher conductivity (or lower resistivity) than the glass.

There's a huge range of conductivities exhibited by materials ranging from very good (metals) to very bad (aka good insulators) like most plastics, glasses, ceramics, and a range of poor ones in the middle (wood, rubber, undoped semiconductors, pure water, antistatic-bags).

'Rubber', because of the way it's made, could have a large range all of its own, as it's often made more black by adding carbon. I once had to reject a bunch of custom-moulded mains connector parts, because an inexperienced worker at the moulding company had thought the rubber 'wasn't black enough', and heaved some carbon into the mix to make it look better.

It's sometimes necessary to make a distinction between volume conductivity, and surface conductivity. A high quality insulator can leak charge across the surface if dirt or the salts from fingerprints attract moisture to the surface. A plastic bag may be made 'antistatic' by coating it with a conductive layer. In high frequency circuits, current tends to travel on the surface of metal conductors, without penetrating into the bulk.

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  • \$\begingroup\$ Thanks for the response. But I measured the resistivity of both and it was the same \$\endgroup\$ Commented May 23, 2018 at 6:15
  • \$\begingroup\$ You interpret your experimental results incorrectly. You made two measurements of resistivity. One was to charge them and time the loss of charge, which showed rubber was lower resistivity than glass. You made another measurement by some other method, which shows them to be the same. Without saying what that second measurement was, method, apparatus, numerical results, we can't comment on the discrepancy between the two. \$\endgroup\$
    – Neil_UK
    Commented May 23, 2018 at 6:19
  • \$\begingroup\$ My bad, I 'calculated' the resistivity based on the data sheet and it was the same to a high level of accuracy. However you seem to imply that the method of measuring resistivity is to time loss of charge. I did not know that. \$\endgroup\$ Commented May 23, 2018 at 6:24
  • \$\begingroup\$ I think I understand what you're saying, but I think capacitance may be the concept I'm looking for. I forgot everything I learnt about it though \$\endgroup\$ Commented May 23, 2018 at 6:33
  • \$\begingroup\$ If they're the same geometry, they're the same capacitance. When you're trying to measure resistivity of things like rubber, glass, even antistatic bags with meter probes and a meter, it's usual to get zero, or bad answers, it's a very difficult measurement to make. It's a method that OK for metals and resistor components up to 10M or so. Loss of charge from a capacitor is less direct, but the method is more sensitive to the 1Gohm/1Tohm sort of resistances you encounter with insulating materials, it's a lousy method for metals (difficult to time nS!) \$\endgroup\$
    – Neil_UK
    Commented May 23, 2018 at 7:41

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