# Tag Info

48

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 ...

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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 charge. Its a simple quantised way to measure the coupling strength of particles with ...

13

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 ...

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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 ...

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 ... 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 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. 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 ... 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} ... 9 In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where \mathbf{E}=\mathbf{0}. Letting the charges be located at (\cos(2\pi k/N),\sin(2\pi k/N)) for k\in\{1,2,...,N\}, we can generate plots of |\mathbf{E}|^{-1} for various N. The zeros of ... 9 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 ... 9 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 ...

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 One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose V(\infty)=0, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ... 8 You have to realize that the system is invariant under rotations about the normal to the plane. Then then electric field must also be invariant under these rotations. An electric field component in the plane does change under such a rotation, so such a component must not exist if we have this invariance. Thus the electric field is purely along the normal to ... 8 The answer by @NowIGetToLearnWhatAHeadIs is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation. Start with your charge distribution and a "guess" for the direction of the electric field. As you can see, I made the guess have a component upward. We'll see shortly why ... 7 It doesn't hold for arbitrary shapes. The reason it works for spheres is that when you have a spherical charge distribution and a concentric spherical Gaussian surface, the whole system is invariant under rotations around the center of the spheres. If the electric field were different at different points on the Gaussian sphere, you could rotate the whole ... 7 Does the fact that every rain drop falls in their respective straight lines all parallel to one another imply that those lines are physically real? No. It is just the tendency of gravity to act between two massive objects--a straight line is simply the least inaccurate way to describe this interaction. You can also draw additional curved lines linking the ... 7 It is a misconception to think that just because the 29th electron is outside the gaussian surface, it will not have an effect on the electric field inside it. The total flux through the surface is indeed zero, but that doesn't mean there is no influence: imagine a point charge and draw up a sphere next to it. The electric field goes in on one face and out ... 7 Heavy clouds have condensed to the point of large droplet formation, failing the Rayleigh criterion for visible light and so no longer scatter them. It is a case of absorption being higher than reflection/scattering that causes clouds to look dark. 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. 7 This is a more down-to-earth answer as opposed to the fancy mathematics in the other one. This problem is easily solved numerically. The equations are easily stated: inverse-square forces to the right from the particles to the left and to the left from the particles to the right. Thus, for a system of n+2 charges where the first and last are fixed at x=0 ... 6 Field lines draw all of their validity from Gauss's law for the electrostatic field,$$ \nabla\cdot \mathbf{E}=\frac1{\epsilon_0}\rho,\ \text{or equivalently}\ \oint_{\partial\Omega}\mathbf{E}\cdot\text d\mathbf{S}=\frac1{\epsilon_0}Q_\Omega, $$where Q_\Omega=\int_\Omega\rho\,\text d\mathbf{r} is the electric charge in a volume \Omega whose surface is ... 6 Charge is a quantity which arises from Noether's theorem, due to continuuous global symmetries (up to a total derivative) of an action, and as such we have many types of charge, other than electric. For example, consider the Dirac Lagrangian,$$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi$$which describes fermions. It is invariant by a ... 6 I) Right, the differential form of Gauss's law$$\tag{1} {\bf\nabla} \cdot{\bf E}~=~ \frac{\rho}{\varepsilon_0} $$uses the relatively advanced mathematical concept of Dirac delta distributions in case of point charges$$\tag{2} \rho({\bf r})~=~\sum_{i=1}^n q_i\delta^3({\bf r}-{\bf r}_i).$$Note in particular, that it is technically wrong to claim (as ... 5 In physics and engineering, we often abstract and idealize a physical problem to gain insight into the physics, e.g., infinite plane of charge, infinite line of charge, point charge, etc. Now, it goes without saying that if these idealizations didn't represent good approximations of relevant physical systems, they wouldn't be used. With regards to your ... 5 The only property of metals used in deriving C=\varepsilon A/d is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter. But if you use something other than metal, then it will of course change the capacitance. 5$$A=\pi r^2\frac{dA}{dr}=\pi\cdot2rdA=2\pi rdr Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$ You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit ...

5

It will never form a stable structure as both electrostatic and gravitation forces decreases at the same rate with distance. And if one dominates at some distance, then it will continue to dominate forever, as the charge to mass ratio of the star will still be same as that of the electron.

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