# Tag Info

37

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 ...

22

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 ...

19

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 ...

14

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, ...

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: $$... 11 Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form. In electric circuit theory, the across variable is voltage while the through variable is current. The analogous quantities in mechanics are force and velocity. Note that ... 10 The effect in which two objects get charged by rubbing and remain charged is called the triboelectric effect, http://en.wikipedia.org/wiki/Triboelectric_effect where the root "tribo" means friction in Greek (The Greek word \tau\rho\iota\beta\omega means 'to rub'). Friction is actually unnecessary: contact is enough in principle. This effect ... 10 Wind doesn't actually touch the surface. You can see the same effect on a car: even if you move at speeds beyond 70mph, the dust doesn't get blown away. If you look closely, there is a boundary layer between the matter of the fan and the air around the fan. When you get closer to the fan blades, the air starts to move with the fan (the blade pulls it ... 10 Loosely speaking, as we walk away from a sphere it looks smaller, as we walk away from a cylinder just the radius looks smaller, but not the infinite length, and finally as we walk away from an infinite sheet of charge it never looks any smaller (we can never 'get away' from an infinite sheet). At more mathematical level I would say the best way to see ... 10 Short answer: this is a textbook example of the limitations of ideal circuit theory. There seems to be a paradox until the underlying premises are examined closely. The fact is that, if we assume ideal capacitors and ideal superconductors, i.e., ideal short circuits, there appears to be unexplained missing energy. What's not being considered is the ... 10 I suppose you mean k_e=\frac1{4\pi\epsilon_0}. That comes from the fact that Coulomb's law can be stated as :$$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2 $$Now, \epsilon_0 is the electric constant, or the permittivity of free space, and it essentially scales the force. The 4\pi r^2 comes from the surface ... 9 Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ... 9 There is no "center of charge" that simplifies calculations completely because there is no center gravity that does this, either. Two rigid bodies feel gravity that is quite similar to that of point masses, but not exactly the same. Unless the objects are perfect spheres, they feel tidal forces, which depend in second-order and higher derivatives of the ... 9 Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will". If you take q|\vec v| instead of q\vec v, you get something related to current (generally current times a factor). Current is conserved at a junction. Regarding your equal-and-opposite situation, the closest ... 9 If you want to avoid factors of \pi in the more fundamental equations like \nabla . E = \rho / \epsilon_0, you have to accept them where they belong, for instance in: E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}. As remarked by others, Newton failed to put a factor 4 \pi into his gravitation equation (he stipulated g = G \frac{M}{r^2}, instead of ... 9 James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential \Phi relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is:$$\Phi_{inner} = \Phi \,q\, ...

8

Coulomb repulsion it is. Specifically, if a black hole has a lot of charge, then particles with a high charge-to-mass ratio will be repelled. Anything that falls in will contribute "more mass than charge," heuristically, keeping the charge-to-mass ratio of the black hole from getting too big.

8

In the same way that bodies that are far away from you can be approximated as point masses, charge distributions that are sufficiently far away can be approximated as point charges, with total charge $Q$, and then you would, indeed, compute something like the center of charge. This is the first part of the so-called multipole expansion: You begin by ...

8

A battery generates a voltage by a chemical reaction. There is a class of chemical reactions called redox reactions that involve the transport of electrons, and you can use the reaction to drive electrons through an external circuit. This is the basis of a battery. The battery will continue to provide power until all the reagents have been used up and the ...

8

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 ...

7

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.

7

While it may be possible to derive a violation of energy conservation due to intersecting equipotentials, there is a much more intuitive and in my opinion a more fundamental reason that equipotentials cannot intersect: Potential is a single-valued function. A good analogy for potential in this case is a map of the ground elevation of the earth; a ...

7

Yes. The delta function always has the same dimensions as the inverse of its argument. You can read this from its definition, your first equation. So in one dimension $\delta (x)$ has dimensions of inverse of length, in three spatial dimensions $\delta^{3}(\vec x)$ or simply $\delta(\vec x)$ has dimension of inverse of volume, and in n spatial dimension ...

7

This problem with $N$ point charges on a sphere is a famous problem in electrostatics known as the Thomson problem. For large $N$, it is in general an open problem still under active research. References: Wikipedia.org Mathworld.wolfram.com Mathpages.com

7

In an attempt to be brief: The big thing to remember is that the flux is also proportional to the area (technically, the surface integral of the field over the area). Crudely speaking, the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases ...

7

+1, Good question,. While I don't think your idea has much of a physical implications, it is a good analogy (in my opinion, at least). A fair approximation to General Relativity is Newtonian Gravity. A better one is Newtonian Gravity with some special relativistic corrections (I mean a modification to Newton's gravity where the masses $m$ are replaced ...

7

The mistake you made is in the way you stated Coloumb's law. It's either $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}}$$ OR $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}}$$ but definitely NOT $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$

7

Freely-moving charges placed on a line will tend to fly away from each other - with no equilibrium position possible - unless there is some potential that confines them to a specific region. Enforcing the charges to lie within an interval $[0,L]$ will always mean one charge is at either end, so you might as well consider $n-2$ charges confined by the ...

7

This problem has been solved by Griffiths in Charge density of a conducting needle. David J. Griffiths and Ye Li. Am. J. Phys. 64 no. 6 (1996), p. 706. PDF from colorado.edu. The problem is nontrivial.

Only top voted, non community-wiki answers of a minimum length are eligible