# Tag Info

## Hot answers tagged electrostatics

52

Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but this is currently the most fundamental explanation known. It's ...

26

The reason is that you have a boundary layer on the surface of the blade of the fan. On the frame of the blade (the blade moves with some velocity, but at the frame of the blade the air moves) the boundary layer starts from the surface of the blade where the fluids velocity is zero and as you move away from the blade, the velocity increases up to the value ...

24

Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law). You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or ...

18

What about this hypothesis: Dust sticks everywhere, but since the propeller cuts through a lot of air, it meets more dust particles. Thus, more dust sticks to the propeller than elsewhere. Evidence I (Mark) took photos my the fan my room to support Damien's hypothesis. The first photo is of the leading edge of the fan blade, which impacts a lot of air, ...

17

Gauss's law is always fine. It is one of the tenets of electromagnetism, as one of Maxwell's equations, and as far as we can tell they always agree with experiment. The problem you've uncovered is simply that "a uniform charge density of infinite extent" is not actually physically possible, and it turns out that (i) it is not possible to express it as the ...

16

There does seem to be a lot of mythology around about the "grape in a microwave" experiment. I have never see any publications on the subject in a respectable journal, however from chatting to other scientists there seems to be a consensus about what happens. It's all rather boring really. The grape is the right size (about a quarter wavelength) and shape ...

16

Charge is a fundamental conserved property of particles. It is, if you like, a measure of how much a particle interacts with electromagnetic fields. A particle with charge can produce and be affected by electromagnetic fields. This is what we mean when we say a particle has electric charge. It might help to think of it as a simple quantised way to measure ...

15

Electric field lines are a visualization of the electrical vector field. At each point, the direction (tangent) of the field line is in the direction of the electric field. At each point in space (in the absence of any charge), the electric field has a single direction, whereas crossing field lines would somehow indicate the electric field pointing in two ...

15

Wind doesn't actually touch the surface. You can see the same effect on a car: even if you move at speeds beyond 70mph, the dust doesn't get blown away. If you look closely, there is a boundary layer between the matter of the fan and the air around the fan. When you get closer to the fan blades, the air starts to move with the fan (the blade pulls it ...

13

The electric and magnetic fields are real things: they can store energy and transfer momentum. "Field lines" or "lines of force" are a visualization tool suitable for drawing vector fields. They are maps of the fields and the fields are real things. Is that good enough for you? And, yes, the electromagnetic interaction can be described in another (more ...

13

Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form. In electric circuit theory, the across variable is voltage while the through variable is current. The analogous quantities in mechanics are force and velocity. Note that ...

13

The effect in which two objects get charged by rubbing and remain charged is called the triboelectric effect, http://en.wikipedia.org/wiki/Triboelectric_effect where the root "tribo" means friction in Greek (The Greek word $\tau\rho\iota\beta\omega$ means 'to rub'). Friction is actually unnecessary: contact is enough in principle. This effect ...

12

This is a good example of a procedure that happens in many areas of physics. In general, physical laws - and particularly conservation laws - tend to be most naturally phrased in integral form, or even in mixed integro-differential form. For an example of the latter, consider the integral form of Faraday's law: $$\oint_{\partial S}\mathbf{E}\cdot\text ... 12 If there was a closed field line a particle following that line would eventually return to the same place but having a different energy so the field would not be conservative. 12 The short answer is that there's no wind near the blade. This is called no-slip condition in hydrodynamics of viscous fluids. [Concession] It is actually more than that. There's minor van der waals sticking which contributes to this otherwise purely hydrodynamic phenomenon. 12 Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ... 12 The maximum charge a capacitor stores depends on the voltage V_0 you've used to charge it according to the formula:$$ Q_0=CV_0 $$However, a real capacitor will only work for voltages up to the breakdown voltage of the dielectric medium in the capacitor. So in reality, for every capacitor there is a maximum possible charge Q_{max} given by:$$ ...

11

The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law. More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law ... 11 If you put your rod in a ultra high vacuum it will stay charged almost forever, but since you probably keep it exposed to air, this is where the electron excess slowly migrates (and the same for the electron defect in the silk). Since the charge exchange requires an hit between an air molecule and a spot of the rod where an electron excess is present, and ... 11 Of course you can define such a quantity, but the question is: does it mean anything physically? Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great ... 11 You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy. 11 The statement "electric field inside a conductor is zero" is true only after charges have distributed themselves in the most optimal way on the surface - it is an electrostatic result. Starting with an arbitrary charge distribution, there will be forces that cause a redistribution of the charge until, for a sphere, they are distributed uniformly. At that ... 11 You want a gas so you don't need to expend energy vaporising the propellant. You also want the gas to be as dense as possible so you can get as much impulse per unit volume of propellant as possible. It's also nice if the gas is inert and non-corrosive so you don't need to worry about it degrading or corroding whatever you're storing it in. Finally it's nice ... 10 This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law $$\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}$$ If we integrate this over some volume V and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} ... 10 If you want to avoid factors of \pi in the more fundamental equations like \nabla . E = \rho / \epsilon_0, you have to accept them where they belong, for instance in: E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}. As remarked by others, Newton failed to put a factor 4 \pi into his gravitation equation (he stipulated g = G \frac{M}{r^2}, instead of ... 10 Although this is quite an old question, I have to disagree with answer by Luboš. First, Pauli exclusion principle says that no two fermions can share the same state. But, if the electrons have different spins (i.e. are in so called spin-singlet state), then they can be in the same positional state. Next, indeed, in classical case, two charged particles ... 10 Coulomb's law becomes invalid at distances of the order of the electron Compton wavelength and smaller, due to vacuum polarization. To first order in the fine structure constant, the electric potential due to a charge q at the origin is given by:V(r) = \frac{q u(r)}{r}$$where$$u(r) = 1 +\frac{2\alpha}{3\pi}\int_1^{\infty}du ...

10

Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will". If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction. Regarding your equal-and-opposite situation, the closest ...

10

James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential $\Phi$ relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is: \Phi_{inner} = \Phi \,q\, ...

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