View Question
Q: Electrical Discharge at Zero Gravity ( No Answer,   3 Comments )
 Question
 Subject: Electrical Discharge at Zero Gravity Category: Science > Physics Asked by: rhodescollegephysics-ga List Price: \$35.00 Posted: 20 Jul 2006 12:09 PDT Expires: 19 Aug 2006 12:09 PDT Question ID: 748064
 ```Our team from Rhodes College will be performing an experiment in microgravity aboard NASA's "Vomit Comet" in August. For our experiment to come off successfully, we need to know how well electrical charge persists on a metallic, spherical object (radius of about 1.5 cm) at high altitudes. To answer this question, we would specifically like quantitative data showing how the following affect the dielectric strength (i.e. the electric field at which breakdown occurs) of air: 1. air pressure 2. humidity 3. cosmic ray intensity We suspect that low humidity may increase dielectric strength and high cosmic ray intensity may decrease it, but we would like more specific information about all three of the above factors. Any suggestions about how to prevent electrical discharge at high altitudes would also be appreciated. Note that we need this information by the first week of August at the latest.``` Request for Question Clarification by hedgie-ga on 21 Jul 2006 04:52 PDT ```This is a fairly complex problem. You would have to estimate ion density - and the 'cosmic rays' may have to be better defined. For price shown, you can get couple references, like this one: http://www.blazelabs.com/l-vacuum.asp not a complete answer. Are you interested?``` Clarification of Question by rhodescollegephysics-ga on 26 Jul 2006 11:57 PDT ```Let me clear up two issues: the amount of charge on each of the two metal spheres we'll use, and the definition of cosmic rays. 1. We will use two different-sized spheres (1.5 and 6.5 cm radii), both of which will be at a voltage of 30 kV. Using V = k*q/R to calculate the charge q, we expect 217 nC on the large sphere and 50 nC on the small sphere. Dividing by the surface area of each, we obtain 0.4 and 1.8 nC/m^2 for the large and small spheres. So there you have the surface charge densities. These distributions aren't exactly uniform in practice, but for the purposes of this question we can assume they are. 2. We were warned by another college that cosmic rays are more intense at high altitudes (at least 25,000 feet) and could hinder us from keeping charge on our spheres. By cosmic rays, I mean radiation (mainly protons) streaming from space into Earth's atmosphere.```
 There is no answer at this time.

 ```One of the most significant contributors to electrical discharge is the relative humidity. Generally speaking, the lower the RH, the more likely it will be to have an electrostatic discharge (ESD). In many labs where ESD events would be an issue (i.e. computer chip manufacturing plants), RH levels are closely monitored at all times. If they drop below some threshold (usually around 30%), work ceases immediately. So, if you don't want ESD events, humidify the heck out of that cabin. Hard to do, since most airplane cabins are drier than...well, you can finish that yourself.```
 ```There is a wealth of knowledge available on ionization chambers used for radiation detection and measurement. You are in some ways creating a giant radiation detector. Study how they work with respect to humidity, voltage and air pressure and you may have you answer.```
 ```If this information isn't already in 'the literature,' I'd be surprised. There are probably oodles of experimental measures of air's dielectric constant as a function of (pressure, humidity). That should be pretty sufficient phenomenological in roads. There may be not be as many as a function of "cosmic rays." I haven't thought much about it, but I'd physically intuit that this means (since you mentioned protons) that there's some charge flux density in the surrounding air and well that's proportional to the dielectric constant... There are a lot of ways you could solve these problems: isobaric chamber, EM shielding (like faraday cage or something), (de)humidifiers-- all of various cost and effectiveness. Or just take pressure, humidity, and cosmic ray measurements... and then compensate for them in the calculations based on whatever dependence of air's dielectric constant you find. But sorry, I have no specifics not for \$35, heh. Hope this helps, though...```