# Tag Info

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Electric charge is not special. Every charge is replaced by its opposite under the C conjugation. For instance, electric charge goes from positive to negative and vice versa. Color charge goes from blue to antiblue and vice versa. Mutatis mutandis with green/red color charge. Every charge is sent to its anti-charge.

3

It is the way one reads/writes Feynman diagrams, a particle going backwards in time is the antiparticle. The electron radiates a gamma, and continues to meet the positron , annihilating charge with another photon. Two real particles are needed for momentum conservation in the center of mass, and two photon vertices are the simplest case within the standard ...

1

Can you tell from the image below if Q1 and Q2 are attracted or repelled? No, you do not have enough information. Will Q2 only be attracted to the sphere if Q2 is enough bigger than Q1? For any nonzero values of Q1 and Q2 you can compute the distance at which there is no net force. Will the positive charge inside the shell attract electrons interior to ...

0

Two electrons when they move experience these forces $$F_{electrostatic repulsion } = \frac{ke^2}{r^2}$$ And, $$F_{magnetic attraction} = \frac{μ_0 . e^2 v^2}{4 \pi . r^2}$$ As you can see from the formulae for attraction there must be a velocity. For the two forces to be the same the speed of the electrons must as fast as light, practically these two ...

0

you can check the discussion here. There is a certain case in which a phonon mediates attraction between two electrons. Indeed, acoustic phonons correspond to a slowly varying in-space displacement of atoms which produces a charge. This charge, in turn, results in an electric potential for the electrons. This means that the electron distorts the crystal ...

0

In certain scenarios there can be a magnetic attraction, but the electrostatic replusion will greatly overpower it.

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While Gauss' law does imply that the electric field terminates only on charges, it certainly does not imply that it cannot form closed paths. Gauss' law may be expressed as follows: $$\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$ From the Helmholtz theorem of vector calculus, as any vector can be expressed in terms of two parts ($\phi$ and $\mathbf{A}$ ...

0

Gauss law states that electric flux through a closed surface equals $\frac{q}{\epsilon_0}$, where $\epsilon_0$ is a constant and $q$ is the charge contained in the surface. So, if we corner a 'point charge' using a spherical closed surface, with the charge at the center of the sphere, all we have to do is to start reducing the radius of the sphere. Because ...

1

To do this you must use the electrostatic image method : The problem with two spheres is that you will have image charges of the image charges Here is a diagram of what it will look like after two iterations : Using the method of images we have the image charges inside the spheres: $Q_1$ has an image $q'_1$ located at $O_2 - ( \frac{R_2^2}{D} )$ with ...

2

Can we have electronics with charge carriers OTHER than electrons? Yes, see what Sebastian said above. And see the physicsworld article Taming light at the nanoscale: "Look around, and you will probably see numerous electronic and optical gadgets, such as mobile phones, personal digital assistants, laptops, TVs and digital cameras. These may all do ...

4

Depending on your view, there is electronics with other charge carriers. It is commonplace to have semiconductor devices where the relevant carriers are holes! Furthermore, batteries and electrolysis relies heavily on ions as charge carriers (but hardly count as electronics). I guess genuine electronics with ions will be difficult as charge carrier mobility ...

0

You need something that can be conducted along the wire to power electronics; if you were to get protons, rather than spreading from atom to atom you'd just end up changing the element of the atom or splitting it. The closest thing that you can do other than add electrons is chemically charge it, as in replace the batteries.

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Appending to the answer given above, since I cannot post comments. Integrating in the onion peel manner, charge enclosed in the gaussian surface of a 'thin' shell taking care of limits (inner radius and beyond to whatever distance so desired) should yield you the answer to your question. Note the exponent of the distance parameter in your 'distance - ...

0

In reality, "point charges" don't exist. When you get close enough, you find that charge is spread out in a little cloud - just think of the way electrons behave around the nucleus of an atom, or about the Heisenberg uncertainty principle. Once you accept that there is no such thing as a "point source", but instead everything is a "charge cloud" of finite ...

0

Let's measure mass in $MeV$ here. The mass of an electron is about $0.5\,MeV$. So if you add one electron the mass of the sphere should increase by $0.5\,MeV$. But at the same time you also change the electric field surrounding the sphere. Due to the mass-energy equivalence the energy contained in the electric field contributes to the mass of the sphere. If ...

1

I'm assuming you mean a solid metallic sphere, not a shell. The analysis is slightly different for a shell but rests on the same principles. It is actually not correct that the total charge induced on the sphere in the case of grounding is $-q$. The basic concepts you need to know about electrostatics with conductors is that the entirety of the conductor ...

0

The mass of the object in concern is only dependent on the way you choose to 'charge' it. If you add charged particles to it, its mass would obviously increase, and if you take away charged particles from it, the mass would increase. (Both these changes can happen by extremely negligible amounts.)

4

We can easily do this calculation. The capacitance of a sphere is: $$C = 4\pi\varepsilon_0r$$ and the charge is given by: $$Q = CV = 4\pi\varepsilon_0r V$$ The number of extra electrons is: $$n_e = \frac{Q}{e} = \frac {4\pi\varepsilon_0r V}{e}$$ And finally the mass of the extra electrons is: $$M = m_e n_e = \frac{Q}{e} = \frac ... 1 The "work" is the useful rotational pull you get from the motor, and excludes any wasted heat. If only 75% of the energy going in comes out as work, then you need to put in 90/0.75J of energy to get 90J of work out. 1 If you have an isolated capacitor, so that there is no conducting path for charge to flow from one plate to the other, then the charge on the plates will be conserved as you change the geometry. Since$$Q = CV,$$a drop in the capacitance C is matched by an increase in the potential V. Note that the stored energy U in the capacitor,$$ U = \frac12 ...

1

The existence of quarks is not seriously in dispute at this point AFAICT. If you want to make something meaningful out of quarks and only quarks having fraction-of-$e$ charges, I think you pretty much have to postulate that electrons are composite. For instance, the rishon model proposes that all the "fundamental particles" of the Standard Model are ...

1

Begin by analyzing the forces on the two axons, x and y. You can prove that on the y axon the sum of the forces is 0. Now on x we have, for the force from the charge on +y : $$F_{1x}=F_1 cos u$$ where u the angle between the x axon and the distance d between q and Q It is:  F_{1x}=F_1 cos u ={ k Qq \over d^2}cosu = {k Qq \over \sqrt{x^2 + y_0 ...

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That is a good question but I think you might be a bit confused. The quark charges are quantised as they are fractional values of the electron charges, so when you refer to 2/3 and -1/3 these mean 2/3 of the electron charge and -1/3 of the electron charge respectively. As such, a Hydrogen atom with a proton in the nucleus and an electron in the shell, is ...

23

Quarks do not violate quantization of charge, it's simply that $\frac{1}{3}e$ instead of the electron charge $e$ is the smallest unit of electric charge.

0

I assume you are conducting a thought experiment in which the electron is stationary in a perfect vacuum. In this case, in the equation from Amey Joshi, the electron's velocity is zero so the net force on it from the magnetic field is zero. The electron therefore feels no force and doesn't move. However, this is an unstable equilibrium. The slightest motion ...

0

Magnetic fields do not attract or repel charges the way they do so for magnetic poles. They exert a force $q(\vec{v} \times \vec{B})$ on the charges. The picture of an electric field and a magnetic field as separate entities is not entirely true. In reality, there is just an electromagnetic field which affects charges, currents, dipoles and such other ...

1

To use rules without knowing what is the reason is boring. See my paper about vector product for Lorentz force, for generators and for electric drives, in a reduced form for perpendicular vectors only. If one isn't sure that this equations could be derived see this answer from mathematicans. See my answer Why does one call $B$ the magnetic induction? too. ...

1

Let us imagine that the charge carriers in the rod are electrons (negatively charged). An electron moving to the right is equivalent to a (conventional) current to the left. Alternatively, you can use a "left hand rule" for electrons (since the current is to the left when the motion is to the right, you can represent electron motion with the thumb of your ...

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Have a look at the Wikipedia article on the left hand rule. It says: The direction of the electric current is that of conventional current: from positive to negative.

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See explanations on Quarks and Gell-Mann's Quark Model.

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It is because they only lose there energy when accelerating. A moving electron or proton with a constant velocity won't emit EM radiation. The electronmagnetic radiation only takes away the aceleration, slowly the proton or electron down to a constant velocity. The reason they have charges is much more complicated. For the proton example it is because ...

0

The significance of Gauss law is to calculate electric field at a point. This is easily done by making a gaussian surface to pass through the point at which field is required and simultaneously enclosing the charge due to which the field is required inside the gaussian surface. If the gaussian surface is made to pass through a discrete charge itself then the ...

3

A capacitor stores electric energy in a static electric field between two conductors. The conductors are separated by an insulator called a dielectric, which allows the conductors to be very close to each other without contact. The closer together the conductors, the greater the storage capacity of the capacitor. This storage capacity also is proportional ...

2

Electrons flow through the wire from the battery's negative terminal to the battery's positive terminal. If said wire is actually a capacitor, the electrons still flow the same way - from the battery's negative terminal. But since it's a capacitor, the electrons are pushed into the capacitor's negative plate instead of making it all the way to the battery's ...

2

No. Step number two already mentions electrons from the battery flowing into the negative terminal of the capacitor, giving the negative terminal a negative charge. Step number three is talking about electrons flowing out of the positive terminal of the capacitor, giving the positive terminal a net positive charge.

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$n$ is the particle density: Number of particles per cubic meter in SI units. $e$ would be the charge on a particle (Coulombs), here equal to the elementary charge (they could be protons). With that I think that you will see the the dimensions are correct.

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