New answers tagged

0

I am a little unsure what you are asking. You seem to think that two Majorana fermions can interact in such a way that one gets to Dirac fermions. The Majorana fermion is its own antiparticle. The charge conjugation of $\psi$ is $C\psi~=~i\psi^*$. The appearance of $\psi$ and $C\psi$ in the Lagrangian means that the Majorana field must be electrically ...


2

If electrons obeyed classical mechanics, they would rearrange in a new configuration in order to maximize the distance between them. They would not stay still because of thermal motion, as pointed out by CuriousOne, but on average they will still maximize this distance. However, electrons don't obey classical mechanics, but quantum mechanics. The behavior ...


1

Electrons in metals have states which people call Bloch states. If we want to describe these states we should bear in mind that electrons have wave like behavior. Bloch states are a subset of the many infinite wave states which electrons can have in general. They're such that when an electron is in a Bloch state, it somehow fills everywhere in the metal. If ...


2

According to Maxwell equations, steady currents and steady charge density won't produce EM waves. So if you have a steady current loop, you can calculate its magnetic field just by the Biot-Savart law, which gives an steady magnetic field in space. If a current loop would radiate energy, then it would be impossible to produce persistent currents in ...


9

If you consider that electric current is actually the flow of individual charged electrons, then as John Rennie pointed out, the radiation exists but is negligibly small. But if you were to imagine breaking the current into more and more point particles with less and less charge while holding the linear charge density $\lambda$ fixed, then the radiation ...


4

To back up John Rennie's answer, consider the Bremsstrahlung formula for velocity perpendicular to acceleration: $P= {{q^2a^2\gamma^4}\over{6\pi\epsilon_0c^3}}$. For all practical purposes $\gamma=1$, so we can simplify this to $P\approx ({q \over \mathrm{C}})^2 ({a\over \mathrm{m/s^2}})^2 {1\over{18.85\times 8.85\times 10^{-12}\times 2.7\times 10^{25}}}\...


27

Circular currents do produce EM, and indeed this is exactly how X-rays are produced by synchotrons such as the (sadly now defunct) synchotron radiation source at Daresbury. In this case the current is flowing in a vacuum not in a wire, but the principle is the same. Current flowing in loops of wire don't produce radiation in everyday life because the ...


1

I would like to add that if we do not consider the elementary particles but think of those charged spheres made of metal, they can actually break. If you keep on removing electrons from a material block and protect the discharge from the neighboring atmosphere, after a stage the repulsion among the like charges become stronger than their cohesive force of ...


1

When you put +ve charge on a conductor, you are really removing the same amount of -ve charge, because only the electrons move. Gauss' Law assumes that charge is infinitely divisible, and can be spread uniformly throughout a volume or over a surface. This is a good approximation when the charge is on the order of $1 \mu C$, corresponding to about $10^{13}$ ...


0

Quoting from the Wikipedia page on the CGS system: The e.s.u of charge, also called the franklin or statcoulomb, is the charge such that two equal $q=1\:\mathrm{statC}$ charges at a distance of $1\:\mathrm{cm}$ from each other exert an electrostatic force of $1\:\mathrm{dyn}$ on each other. The e.m.u. of current, also called the biot or abampere, is the ...


3

Another point of view is to consider the electron as quantum-mechanically smeared due to its permanent coupling to electromagnetic field oscillators. Such a construction is rather "soft" and is easy to excite - radiate and absorb soft photons. In this sense, this construction is not elementary and point-like. Point-likeness is then an inclusive picture, not ...


57

Composite particles like protons don't break apart because of the strong interaction which holds their constituents (the quarks) together. Elementary particles like electrons don't break apart because they are point-like particles, i.e. they are not made of “parts” (if the Standard Model is correct).


37

This was one of those big questions in the 19th century. It still causes some consternation. If you have a composite system, such as the nucleus of an atom, some other force is necessary. This force of course is the nuclear interaction. This keeps the protons from flying apart, though for some unstable nuclei there are transitions that eject charged ...


3

Like charges repel via the electromagnetic interaction which is mediated by exchange particles (gauge bosons) called the photon. Since the photon is massless, the electromagnetic force has infinite range, and all like charges will attempt to break apart from each other. However, those that don't are being held together by a force that is not electromagnetic ...


5

When you have a charged object, for example a charged metal sphere, of course the charges on the surface of the sphere interfere with each other. Because of these effect the charge gets distributed equally over the sphere. However, these effects are not big enough to actually break up the sphere or something like that. If your object is charged high enough ...


3

Griffith's 1987 book correctly states a totally reasonable hypothesis, that the neutron's core is positive. Here's a simple model which probably goes back to Fermi: the neutron ought to spend part of its time as a virtual proton-$\pi^-$ pair, in a strong-interaction analog to the photon spending part of its time as an electron-positron pair; since the ...


-1

According to me positive charge is present in nucleus so it will attract towards negative charge and -charge is present outward nucleus so attraction is radially outward.


3

Great Question! First of all, you can solve this problem for the Magnetic Field and the Electromagnetic Field. You won't have to do any work with the Lorentz Force Equation. Instead you will use the solutions to Maxwell's Equations. $$\pmb{\mathcal{E}}=-\nabla \phi-\frac{\partial{\pmb{A}}}{\partial{t}}$$ $$\pmb{\mathcal{B}}=\nabla \times\pmb{A}$$ where $\...


6

Initially, when first glass rods were systematically being rubbed, the "charging" phenomena was observed. The electric charges were hypothesized to be positive and negative, and the pioneer (Franklin? forgot the name...) pretty much arbitrarily decided to call one positive and the other negative. Further experiments helped him deduce that two like charges ...


16

The Hall Effect shows that negative charge is moving. In the Hall effect, one passes a current through a wide strip of metal exposed to a perpendicular magnetic field. If positive charges moved, we'd expect the positive charges to be travelling in the same direction as $\vec{I}$, and the magnetic force $q\vec{v}\times\vec{B}$ would be to the right. Thus, ...


11

Physics's don't know that only negatively-charged particles move. We can create ion currents on demand in many environments. We do know that the current flowing in a metal wire is negatively charged particles in motion. As for how to determine that, you do a Hall effect measurement. The measurement works by subjecting a current in a relatively wide bar to ...


1

When you connect them, charge can flow, and it will do so until there is no net electric field it can move along (the surfaces are the same potential). Imagine I just suddenly destroyed the wire; now the charges are stuck on the spheres. They are still under the condition that they want to equilibrate and put a constant voltage across the surface, but they ...


0

I have annotated the diagram in the Wikipedia article that you cited in your question. Note that the signs for the charges on the parallel plates were the wrong way round in the article. To simplify the derivation I assume that the condenser plate length is approximately the same as the source to aperture distance ($AS = a)$ and the angular deflections ...


1

The mathematical definition that John Rennie gave explains it well, but I'd like to give an intuitive answer to your question. Imagine a rubber sheet horizontally stretched. The height of the rubber sheet at a point is equivalent to the electric potential at that point. Now, since as of now there is no disturbance at all, the height (potential) throughout ...


1

All of your examples look perfectly consistent with each other. The thing to note here is that your first textbook's second example is working out the case for a magnesium ion, i.e. a neutral atom with a number of electrons stripped (or added in). When this is the case, it's not enough to say that it's an ion: you need to say which ion, i.e. how many ...


2

Among competing hypotheses, the one with the fewest assumptions should be selected. Some electrified objects repel, some attract. This can be explained by two kinds of charge. Nothing that cannot be explained by two charges can be explained by adding a third kind of charge. So we continue to describe electricity as occurring in two kinds.


5

There are some good answers here, but I think I want to try to abstract Franklin's work a little bit. Because Franklin found just two options - "repel" and "attract", he was forced to consider only two kinds of charges. Consider the experiment, where glass-glass repels, plastic-plastic repels, and glass-plastic attracts. If all glass is the same, the glass ...


25

I agree with DanielSank that the question is asking (wholly, not partly) about the historical development of the concept of electrical charge, not our modern description of it - "how did they know?" not "how can we know?" The latter (answered by dmckee) is the end result of more than two centuries of observation, experiment, theorising and debate, and ...


3

Let's assume we don't know how many types of charges exist. But we know that there are bodies which either attract or repel each other. Now we perform an experiment We find all such bodies that repel each other and put them in separate categories. After extensive experimentation we observe that they only belong to two piles. Furthermore we also observe ...


0

Based of the shape of the conductors and the charge distribution on one body, it is theoretically possible to find the distribution on another body using the method of images. But this may lead to difficult mathematical problems(unsolvable in some cases).


42

Get together a collection of charges. As many different ways to generate a charge as you can think of. Go ahead and invite your friends so they can think of some more. (As a practical matter you make static charges just before you use them, but still...) Now, test them pair wise to see if they attract or repel one-another. Keep careful records. Find the ...


0

Both the hollow sphere and the small metal ball are conductors, i.e. electric charge can move freely around in them. As long as they are separate the air between them acts as an insulator, so the charge stays on the hollow sphere. When the objects touch in principle the charge can move from one object to the other. But does it? One property of electric ...


0

As Anthony B said,the number of field lines cutting any sphere surrounding a point charge is the same(because any field line which passes through a sphere of radius 1 also Passes through a sphere of radius 200) given that, the flux = E 4pir^2 should be constant. That explains the 1/r^2 dependance theoretically


1

This is a much more deeper question then it looks in first glance. The simple logic given by @Anthony B is not enough for proving the inverse square law. There are numerous experiments that have been done to verify this law. There is a collection of the experimental works in this review. In earlier days Cavendish and Coulomb have performed experiments with ...


3

As such there is no real theoretical proof to the inverse square dependence of the electric field in classical electrodynamics. It is an experimental fact famously known as the Coulomb's law. When combined with the superposition principle, it gives us the Gauss's law of classical electrodynamics: $$\nabla \cdot\mathbf E = \frac{\rho}{\epsilon_0}.$$ But, ...


2

You can prove it using the concept of electric flux. For instance. If you surround a point charge with a sphere if r=1, or a sphere with r =10, you know that the electric flux ( field strength times area) must be the same. A sphere is easy because every point is equidistant to the charge.


0

You can think of the line of charge as a collection of (very many) point charges of the same sign. Outside of the charge distribution but colinear with it, each point charge will have an electric field which points in the same direction (away from the distribution if the charges are positive, and toward the distribution if the charges are negative). These ...


0

We don't know the electirc potential of the individual plates, right? Yes, we do. Or, that depends on what we want to know. Electric potential is just the potential compared to some other point - some arbitrarily chosen reference point. If for example, you chose one of the capacitor plates as the reference, then this plate has an electric potential of $0\...


0

To know the electric potential of the plates you first need to define where and if there is a point with zero potential. This point is arbitrary. Once you defined that, you can calculate the potential of the plates. For instance, if one of the plates of the capacitor is defined to be at zero V, then the other plate will be at 2V. If you define it that way ...


0

The main question is unnecessarily complicated by alluding to a phone battery and its battery pack. Concentrating strictly on two "plain" batteries, one being charged to 5% of it capacity and the other charged to 35% of its capacity. The implication is that the one with the larger charge can charge the one with smaller charge. This is not necessarily true....


0

I may add a little bit of chemistry in the hope that it would be of some use to the physicists and engineers discussing the charging process of cell phone batteries using backup power source batteries. Betteries are devices that transform chemical energy into electrical energy and vice versa. The so-called secondary batteries operate in both directions ...


-2

Consider a positively charged particle, if it comes in upward motion its velocity increases as a result its mass decreases(according to conservation of momentum i.e. m1v1=m2v2), as mass decreases its number of atoms decrease i.e. equal number of protons and electrons are decreasing which reside in an atom.thus the charge on the particle is unaffected by its ...


5

Voltage is not any part of this explanation. The answer is that each battery pack stores a certain amount of energy. This is measured in joules. At its most basic level your phone battery has a certain capacity in joules, you external battery bank also has a capacity in joules. When you charge the battery you are transferring a certain number of joules from ...


22

Connecting your phone to the battery pack doesn't directly connect the cells in parallel. I assume this is where your guess of an equilibrium with equal voltage -> equal charge percentage comes from. Shorting lithium-ion / lithium-polymer (LiPo) cells together like that would likely cause one or both to literally catch fire from the high currents, or from ...


0

In case of the battery packs, there is much lighter weight requirement, and also much smaller development / manufacturing costs. But it is important to be bigger (in the sense of Ah). If you fill a cup of tea from a large jug, the cup will be full (100%) while the tea level in the jug decreases only a little bit. The Ah capacities of the batteries are ...


1

The sign of the force direction is just given by convention as $q \vec{v} \times \vec{B}$. Whilst the direction of the force on the particles is clearly something that can be measured, the sign of the charge and the direction of the magnetic field lines are man-made constructs. For example you would get the same direction for the force if we decided that ...


5

For an iPhone the battery voltage is a nominal 3.8 V and the battery pack would probably replicate the 5 V output voltage of a USB power supply. So the battery pack would be discharged as it was driving current into the positive terminal of the phone battery and thus recharge the phone battery. So only when the battery pack voltage was less than the ...


24

The key here is the voltage of both the batteries. The battery in the phone is generally at a voltage of 3.7V. The battery pack has a higher voltage or a circuit which gives a voltage of 5V to your phone. So, as long as the voltage with which you charge the phone is higher than that of the battery, the percentage of power in it doesn't matter and the phone ...


29

Sometimes it is easier to understand circuitry in the context of water. What you're imagining is two tanks of water of equal size linked together by a pipe that has been sealed off. If one tank holds 5% water and the other holds 35% water, when you remove the seal, the tanks equalize and you end up with 20% in both tanks. What you're forgetting is that ...


0

Well, I think that the magnetic force of the moving positive-charge which you describe is directed at the positive z-axis, if I'm not mistaken ... in other words, all 3 axes come into play when one visualizes which way the magnetic force will point .....



Top 50 recent answers are included