# Tag Info

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The electron and proton aren't like pool balls. The electron is normally considered to be pointlike, i.e. has no size, but what this really means is that any apparent size we measure is a function of our probe energy and as we take the probe energy to infinity the measured size falls without limit. The proton has a size (about 1fm) but only because it's made ...

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Electro-magnetism is a good guess, simply because it's the only force you commonly see that's powerful enough. It's not very useful as an explanation, though, because almost everything you see around you is due to electro-magnetism (e.g. the way the spoon holds together in the first place, or the light that allows you to see the sugar, or the way the water ...

28

In general the answer is "yes it is possible" - but in your case the answer is "that is not a Faraday cage". Radio waves are (partially) reflected by any discontinuity in dielectric constant of the medium they propagate through. The ones that propagate (through walls etc) will also experience attenuation. A faraday cage is a continuous conducting structure ...

21

This was a big mystery before quantum mechanics was discovered. Not only are electrons attracted to protons, electrons radiate away energy when accelerated. A classical electron in orbit around a proton should spiral into the nucleus in a small fraction of a second. The "explanation" is that classical physics doesn't work on a small scale. Quantum ...

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You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...

17

First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...

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The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.

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To add to ACuriousMind's answer on the Liénard-Weichert potentials, you can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula from them for the radiation from a moving charge: $$\vec{E} = ... 16 Just to add to what Floris has said. It is frequent (in the UK) that institutional settings would have toughened glass in windows, particularly in bathrooms, gyms etc. that would have the form of a wire mesh (of order 1cm grid) embedded in the glass. That would do a particularly good job of blocking phone signals that would otherwise penetrate the glass. 15 Not magnetism but static electricity - the other side of the electromagnetic force. Sorry this answer is short, but I think the link will give you all the information you need. 15 I have reconstructed your diagram computationally and numerically computed the potential energy as a function of the orientation of the ring: As you can see, when we neglect friction (something that you can't do in the real world) the machine is indeed capable of perpetual motion. However, some important points: The total change in energy of the system ... 15 Then will every field line originate from the +q and end up to -6q No, every field line won't end to negative charge. will there be some extra lines coming to -6q from infinity because of higher charge to get 6 times the number of field lines? Yes, many extra lines will come. I think it should be that every line will originate from the ... 12 This is really just a footnote to Anubhav's answer so accept his not this one! Anubhav mentions Gauss's law, and this states: The net electric flux through any closed surface is equal to 1/\varepsilon times the net electric charge within that closed surface. So if you consider a spherical surface around the +q charge the total flux through this ... 11 You are right - potential is a scalar. But a dipole moment is a vector - it has magnitude and direction. When sodium channels open up, charge flows. Lots of charge moving a little bit causes a change in the dipole moment of the heart. This in turn induces charge to move elsewhere in response (the dielectric properties of tissue cause a propagation of the ... 10 In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ... 9 What the picture shows is a corona discharge (see also Wikipedia). It isn't a circuit in the usual sense of the word. It happens because the voltage is so high that it raises the electron energy to above the work function and the electrons just leak off. In effect the coil is charging the air around it. The charge will end up on the furniture, walls, floor, ... 8 If you are talking about point charges then, as explained above, the answer is no. But in the case of non-uniform charge distributions, it is possible for same-charge particles to attract, if they are sufficiently close. As an example, the following two particles are identical, each having a net charge of -1. Plotted below them is their potential energy as ... 8 You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner ... 8 This may help your, it comes from Rutherford scattering by which they determined that the atom has a hard core. It is positive alphas against positive nucleus, but the math is the same. Determining the closest approach to the nucleus amounts to calculating the minimum distance for the hyperbolic orbit which is produced by the coulomb repulsive force. ... 7 There have been lots of experimental attempts to test the validity of Coulomb's r^{-2} law. Many of these are reviewed by Tu & Luo (2004), and is where I am getting the numbers quoted below. Somewhat equivalently, experiments have looked at trying to set an upper limit to the photon mass, which is testing the hypothesis that rather than a r^{-1} ... 6 Perhaps a gravitational analogy will help? The positive charges are railway trucks on a circular track in a frictionless environment. The positive plate is the top of a hill and the negative plate is the bottom of the hill. Set the railway trucks moving The trucks going down the hill lose gravitational potential energy whilst there are trucks going up the ... 6 This type of model, a classical model, led to the Bohr model and quantum mechanics for the atom, as it is an experimental fact that the Hydrogen atom exists and does not turn into a neutron. For the large distances you illustrate the classical trajectory would have to be exactly centered otherwise, even classically there will be lateral motion that will ... 6 The reason why the method of the images is easily applicable in the case of the sphere or the plane is that it uses the symmetries of the Laplace operator$$\Delta\Phi=\frac{\partial^2\Phi}{\partial x^2}+\frac{\partial^2\Phi}{\partial y^2}+\frac{\partial^2\Phi}{\partial z^2}$$or in the spherical coordinates$$\Delta\Phi=\frac{1}{r}\frac{\partial^2}{\partial ...

6

Both gravity and electrostatic forces depend on distance ($r$) like $1/r^2$. So changing the separation between 2 atoms changes both forces equally. So whichever force is stronger initially (at any distance) will always be stronger. To determine which is stronger consider the ratio of gravitational to electric force. $$F_g/F_e = 4\pi \epsilon_0 G ... 5 Just a coincidence. There are too many quantities and not enough letters. It probably does make a difference that the fields in which these two equations exist (material science and electromagnetism) are well enough separated that you typically won't see them both in the same papers or textbooks; if that weren't the case, people would start using different ... 5 Physics does not answer why questions, except with how from postulates and mathematical models one can describe the data. The how is Coulombs law. Physics is about fitting experimental observations with mathematical models. The answer to the "why attraction" in this case, is, data dictates so. There is no other answer except that Coulomb's law fits the ... 5 The answer to your revised question is that your object 2b does exist, is correctly described as an electron stuck to a proton via Coulomb attraction, and is what you get (most of the time) if you take a single electron and a single proton and place them in an otherwise empty universe, initially at rest in the center-of-mass frame. The initial separation ... 5 I actually agree with Ruslan's comment. You cannot say that the integral blows up when \textbf{r} = \textbf{r}', with \textbf{r}' spanning the integration domain where \rho \neq 0. The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function. To see this you ... 5 The E field due to each plate is E/2 and hence the total field between the plate is E. But a plate won't exert force on itself, so the E field experienced by a plate is E/2 only. Multiplying by charge gives you the force, hence 1/2 QE. 5 The energy of the capacitor is U= \frac{\epsilon_0}{2} S\,\mathrm d E^2 where S is the area of a plate. If we increase of \Delta d the distance of, say, the right plate from the left one, keeping fixed the charge Q on each plate, E does not change and we find a variation of energy$$\Delta U = \frac{\epsilon_0}{2} S E^2 \Delta d = ...

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