Cap'n Refsmmat

Yes. You'll note I said the charge effects are only cancelled out at a sufficiently long distance that the distance between the opposite charges is negligible. If they come close enough. At a distance, the positive and negative charges will exert an equal and opposite force, and the ions won't be attracted.

Indeed they do, and the equations work almost exactly the same way no matter which angle you come from. So long as the additional distance [imath]\epsilon[/imath] is small, the charges neutralize each other effectively. That's wrong. "Insulated" means the charges can't move, not that their electrostatic effect is negated. If I take an insulated sphere and embed a billion extra electrons in it, it will repel other negatively charged objects, despite being insulated. You can see this very easily by rubbing an inflated balloon on your hair. Balloons aren't conductive, but nevertheless it can carry a charge and have electrostatic effects, like exerting a force on your hair or attracting bits of paper and foam. Same thing goes with styrofoam packing pellets that get stuck to your hands with static electricity. They're insulators, but they carry a charge, and that charge exerts a force.

Perhaps it does have something to anchor it; that's irrelevant to the scenario. Right, so suppose I'm a positively charged particle sitting a distance r from a proton. I experience this force: [math]F_1 = k \frac{q e}{r^2}[/math] where q is my charge and e is the charge of an electron (which is the same magnitude as a proton's charge, of course). Now, suppose there's an electron lurking directly behind that proton at a distance [imath]\epsilon[/imath] behind it. I experience the following force from it: [math]F_2 = - k \frac{q e}{(r + \epsilon)^2}[/math] (Note the minus sign, because the charge of the electron is opposite.) Now, what's the total force I experience? [math]F = F_1 + F_2 = k \frac{q e}{r^2} - k \frac{q e}{(r + \epsilon)^2} = k q e \left(\frac{1}{r^2} - \frac{1}{(r+\epsilon)^2}\right)[/math] Now, as [imath]\epsilon[/imath] becomes negligibly small compared to the distance r... [math]F = \lim_{\epsilon \to 0} \left[ k q e \left(\frac{1}{r^2} - \frac{1}{(r+\epsilon)^2}\right) \right]= k q e \left(\frac{1}{r^2} - \frac{1}{r^2}\right) = 0[/math] And so we say that the proton and electron neutralize each other. Note, of course, this doesn't work when [imath]\epsilon \geq r[/imath]. Hence why a particle very very close to the two charged particles will still feel a net force. Nope. Electrostatic forces apply to static electric charges, as the name suggests.

They don't have to be. All that matters is that the electrostatic force from the positive charge is equal and opposite to the force from the negative charge; this can be achieved just by having a stationary electron sitting nearby. Imagine the two charges in my example as completely stationary, fixed in whatever material they're part of. For example, they might be charges fixed in an insulator that does not allow charge to move around. Now imagine bringing another charged particle to within 100r or so. It will essentially experience zero net force, because the positive fixed charge would have neutralized the negative fixed charge.

Er, no, not really. No, it isn't. Suppose I have two oppositely charged particles a small distance r apart. If I am very close to them -- say, within r -- I will feel electric forces from both charges, and I will be moved around by their electrostatic forces. However, if I move a distance away that is large compared to r -- say, 100r away -- I feel essentially zero net force, because the force from the charges are nearly equal and opposite. One charge may be slightly farther away from me than the other, but that difference is tiny compared to the overall distance, and so they essentially neutralize each other. Note that absolutely no particle motion was required.

What do you mean by "faster charge neutralization"? In an insulating material, the charges will move among individual molecules and nothing else. From any significant distance away, it does not matter how the charge moves -- the motion is insignificant compared to the distance. The motion of the charges only matters when you're sitting right next to the molecule.

Yes, but that doesn't change the overall neutrality of the substance.

No, I can't, really. Why should the electrons have to move at all to neutralize the positive charge?

This is not true. Shared-electron compounds are usually electrically neutral.

Unless, for some strange reason, women find men in back braces to be incredibly sexy.

If I were to simply take a human and make him proportionally ten times taller and wider and so on, his weight would go up as the cube of its size but his bone strength as its square, meaning the bones would be proportionally weaker and he'd be more vulnerable to fractures. Something like that. I might've gotten the powers wrong.

Having read Mr. Benjamin's book, I can say that he certainly does not use the standard calculations -- he uses mental methods that often result in getting the leftmost digit first. He does not memorize the computations in advance apart from basic things like single-digit squares.

Learning the fundamentals of OOP with PHP isn't a bad start; PHP is fairly easy to learn and very easy to tinker with. However, you'll probably be required to learn other languages for your degree. Object-oriented patterns in PHP translate well to other languages, but you will have to learn new syntax and things like static typing when you transition to C or Java. My advice: Go ahead and learn PHP. Tinker with it and write some code. When it comes time for your degree and they ask you to write Python or Java or C, you'll understand the basics of programming and learning will be much easier. (Although if you hadn't started learning PHP already, I'd say to learn Python.)

Parallel data transfer systems aren't directly analogous to parallel processors. Parallel processing enables higher efficiency, since a single-core processor with sufficient speed to match a dual-core processor would be so fast as to melt itself with heat. Pipelining is also a common efficiency-boosting trick in processors.

Ah. Then run with FAT32. It'll get you basic compatibility between Linux, Mac, and Windows.

It's definitely not, in your case. Domains refer to an XP install in a large corporate network, where logins are managed by a central server. If it's your personal PC, it's not in a domain. There's instructions on how to get the Security tab visible later in the article. Probably not. FAT32 is usually assumed, so FAT=FAT32 in this case. However, as I said, FAT32 is inefficient with larger disks. Do you intend to use this drive under both Linux and Windows? If so, FAT may be a good choice, even if it's less efficient; NTFS can be problematic under Linux.

I removed the link, since it's an Amazon affiliate link that gets the poster a share of the profits.

Here's the tutorial on the fancy mathematics: http://www.scienceforums.net/topic/3751-quick-latex-tutorial/ You can click on any equation you see to learn how it was done. There's a popup with the appropriate code.

How do you jump to this conclusion? Going from "the protests would be unlikely to occur without external cause" to "the external cause was Western insurgents" requires evidence.

Formatting FAT32 on a large disk is probably not a good idea, as FAT32 doesn't handle large disks very well. (It was designed in the days when 2GB was big.) Reformatting will likely cause the same problems. You need to do something like this: http://support.microsoft.com/kb/308419 See the instructions under "Set, view, change, or remove special permissions for files and folders". You want to make sure the "Permissions" box has granted you the "Write Attributes" permission, although you should just grant yourself everything, really.

Because Iraq's government made an easy transition to democracy, right? No, really, could you please elucidate your hypothesis?

I think that's to be expected, but also the scale of the events is not the same. Libya was in all-out civil war, not domestic protest. And then of course there's the illegitimate regime in Ivory Coast which has been resisting outside criticism for a long time now, but that's so far away from anyone's oil or strategic basing interests that it's not worth even thinking about military action to restore democracy there! What do you propose the true motivation is? I do not see the economic benefit of the current military action.

Perhaps I should have noted that the UN resolution also authorizes military action to protect civilians, such as airstrikes on tanks or military command centers.

And?

It's my understanding the Libyan average income is actually fairly good for the region, although unemployment is particularly high. Why do corruption and invasions matter? The stated goal is to prevent Gaddafi from killing his own civilians, which he most certainly was doing. The fact that the UN vacillates so easily does not make its decisive action here wrong; it merely implies that UN members believed there was no other option. Certainly diplomacy does not work for Gaddafi, since he's been under sanctions and international scorn for years. Right, we're in this for our personal gain, but we got France, England, Spain, Italy, Canada, Qatar, Denmark and the UAE to join us on a whim. That must be why our top military leaders are warning politicians that we should do nothing more than a no-fly zone: http://www.telegraph.co.uk/news/worldnews/africaandindianocean/libya/8394442/Libya-It-would-be-unwise-to-directly-target-Col-Gaddafi-Robert-Gates-warns.html And, of course, the Arab League endorsement just seals the deal. The oil-producing nations of the world would clearly endorse an effort to steal oil from their neighbors, right?