# Tag Info

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 .....

0

Regarding the first part, the induced charges can never be more than the charge placed and at best they will be equal in magnitude to the charge placed in the cavity( but opposite in sign). So the fields due to them will cancel out inside the conductor. Outside electric field will induce charges on the surface of the conductor but the direction of fields in ...

0

Gauss law in 2D would have to be: $$\oint \mathbf{E} \cdot \mathbf{\hat{n}} dl = 2 \pi q$$ because you are reducing your surface in 3D to a line in 2D, and keep the idea of measure of the boundary and its orthogonal direction or normal. To get the expression of the field you have to make use of the fact that the electric field is isotropic. In other words,...

4

This question has already been answered by a Randall Monroe of XKCD and Dr. Cindy Keeler of the Niels Bohr Institute. It forms a Naked Singularity. Which is an infinitely dense object, from which light can escape. Source: https://what-if.xkcd.com/140/ You Reissner–Nordström metric for this question, as opposed to the more well known Schwarzschild metric. ...

7

I think this is one of those situations where it is cleanest to think relativistically and concentrate on gravity as spacetime geometry, not gravity as a force. If the stress energy tensor's components reach a such a magnitude that an horizon forms, then the mass/energy inside the horizon is doomed to stay inside evermore. This is a question of spacetime ...

5

The strong force is responsible for neutron stars not becoming black holes. Although attractive at nuclear densities, it becomes repulsive at somewhat higher densities. This repulsion is many times greater than the electromagnetic repulsion between protons at similar densities. Yet we know that if a star is too large, it's supernova event will lead to a ...

2

The phenomenon you are talking about is called dielectric absorption. The way it works is this: Let's say you've just discharged a capacitor. An ideal capacitor would remain at zero volts after this. However, in real life, the capacitor will develop a small voltage from time-delayed dipole discharging (also known as dielectric relaxation). Dielectric ...

0

What happens during the charging process is this: During charging, an external electrical power source (the charging circuit) applies an over-voltage (a higher voltage than the battery produces, of the same polarity), forcing a charging current to flow within the battery from the positive to the negative electrode, i.e. in the reverse direction of a ...

0

In a word, yes. What you're doing is called sodium chloride electrolysis. A diagram of the experiment is shown below: Basically, because the salt has been heated until it melts, the sodium ions flow toward the negative electrode and the chlorine ions flow toward the positive electrode. So, since you are turning on one at a time, and your solution is not ...

1

In principle any acceleration of an electron causes some radiation, and an electron has to accelerate in order to leak from one plate to the other. However: the velocities, and therefore the accelerations, of electrons in electrical circuits are small. Calculating the electron drift velocity is an exercise routinely given to students and the results tend ...

10

The problem with "weak charges" is that electroweak symmetry is spontaneously broken. Before the symmetry breaking, electroweak symmetry is described by an $SU(2)_L \times U(1)_Y$ gauge group.This amounts to three charges: weak hypercharge $Y$ for $U(1)_Y$ and weak isospin (total isospin $T$ and third component $T_3$) for the $SU(2)_L$. Some examples of ...

7

You're mixing a few things up here. When you say "three" for the strong force, you're counting the number of colors of quarks, but when you guess "three" for the weak force, you're counting the number of force carriers. These are two different things. For example, if you counted the number of gluons (the force carriers for the strong force), you'd get ...

4

I would say two, which is pleasantly consistent with the $SU(2)$ structure of the weak force. One is the coupling strength with the $Z$ boson, and one is the weak isospin which is raised and lowered by the $W^\pm$.

0

If the shell is conductive in this case, one thing for sure is that the net charge on the interior surface has the opposite sign and equal amount of charge as the charge placed inside of the cavity. If there are no net charge on the conductor before the charge inside of the cavity was placed, the charge on the conductor equal to the charge placed inside the ...

2

Your statement is correct. The charge distribution is such that the hollow cavity of the conductor has a equal amount of negative charge induced on its inner part. This distribution is such that field due the cavity (including the charge inside the cavity) cancels out everywhere outside the cavity. So looking it the other way around external sources do not ...

2

How come in all my EM courses...I never had to account for this loss of energy in the form of EM waves? In the simple examples shown in introductory courses, the loss of energy of charged particle moving in external field via radiation is neglected, because radiation of EM waves is a difficult topic and even if it is not neglected, it can be shown its ...

4

If we assume that velocities are well below the relativistic region then in the absence of radiation the equation of motion of the electron is simply: $$\frac{d^2x}{dt^2} = \frac{Eq}{m} \tag{1}$$ where $E$ is the field strength and $q$ and $m$ are the electron charge and mass. As Lawrence says, the power emitted as radiation is given by the Larmour ...

3

The Larmor formula comes from the Poynting vector $$\vec S = \frac{c}{4\pi}\vec E\times\vec B$$ The electric field in a relativistic content is derived with the Lienard-Weichert potential $$\vec E = q\frac{\hat n - (\vec v/c)}{\gamma^2(1 - \vec v\cdot\hat n)^3r^2} + \frac{q}{c}\frac{\hat n \times((\hat n - (\vec v/c)))\times\vec v_t/c}{\gamma^2(1 - \vec v\... 0 Charge is conserved, so the equation of continuity should be applied, . It states that the divergence of the current density J (in amperes per square meter) is equal to the negative rate of change of the charge density ρ (in coulombs per cubic metre), Current is the flow of electric charge. So if the divergence of J is positive, then more charge is ... 1 Long ago somebody decided that the direction of "conventional" current flow was the same direction as the direction of flow of positive charges. In that convention the flow of negative charge in one direction is equivalent to the flow of positive charge (and hence the conventional current) in the opposite direction. When introduced electricity usually ... 0 Nothing "flows" actually. Electrons transfer the electrical energy by hitting each other. And even if you consider flowing, only electrons free. Protons cannot because they're held strongly in nucleus. About charge, textbooks usually refers it as positive. That means, we just take the opposite direction of electron flow as +ve charge (because electrons are ... 2 Start by noting that the electrical potential is an energy per unit charge. In an electric field E the field produces a force on a charge Q of:$$ F=EQ $$so if we move the charge a distance dr the work done is just force times distance or:$$ W=EQ\,dr  The work done per unit charge is $E\,dr$, and this is what we mean by the change in the ...

0

The Electric field outside a point of the shell=KQ/r^2 assuming Q=Charge on the shell and r=distance of the point from the Centre of the Sphere.This can be easily derived if you draw another Gaussian sphere of radius r enclosing the given Sphere.By Gauss Law, Closed integral of E.dA=Charge enclosed/Epsilon. And You can easily find out the Electric Field. ...

0

If you know Gauss's Law, apply it. Within the boundary, charge enclosed is zero. So, closed integral of E.dS is zero(i.e. flux). Now, we know that the field and area vectors are parallel, and area vector is non-zero. What is indeed zero is the magnitude of field vector, and there you go. Field inside the spherical shell is zero at any point. On drawing a ...

2

Charged particles can't have Majorana masses of any type because they would violate the charge conservation law. The Majorana mass is really a term that is converting a particle into its antiparticle. It implies that the particle must be considered "physically indistinguishable" from its antiparticle. The Majorana mass term violates the lepton number or its ...

0

Start off by considering what the charge distribution would be like without the battery being connected. The charge distribution must be such that there is no electric field inside either of the plates. The consider which of the charges will change when the battery is connected.

Top 50 recent answers are included