Tag Info

Hot answers tagged

3

I think that electrons do escape. For example, Electrostatic Ion Thruster: Electrons are emitted from a separate cathode placed near the ion beam, called the neutralizer, towards the ion beam to ensure that equal amounts of positive and negative charge are ejected. Neutralizing is needed to prevent the spacecraft from gaining a net negative charge. I ...


3

What keeps electrons on a negatively-charged conductor from leaving? It is a quantum mechanical phenomenon. Wherever there exists an electric field potential there exist energy levels , i.e. stable orbital locations which can be occupied by an electron. How does this happen? Even the simple Hydrogen atom has a negative ion state, an anion. This is ...


3

If the valence band maximum and the conduction band minimum are on the same position in k-space, this means that you have a direct gap semiconductor (GaAs, InAs, ...). If the CB minimum is at a finite k-value, it would be an indirect gap semiconductor (like Si, Ge, AlAs, ...) The bands appear as parabolas due to the dispersion of a quasi-free electron/hole ...


3

It is because of the Lorentz force, a basic law of electromagnetism. Its expression is $\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})$ What this means is that an electrostatic charge is only influenced by a magnetic field if it is moving (the first term is just the electrostatic force). The other is that the force is perpendicular to both the magnetic field ...


3

Why are electrons attracted to a magnetic field? Perhaps nobody knows. You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... So let's look at the bird and see what it's doing — that's what counts. I learned very early the difference between knowing the ...


2

Why do not you try this? The shell is always an equipotential object. When the charge is inside the shell, the potential of the shell is $kq/R$ with $R$ the radius of the shell, so the energy of the charge is $kq^2/R$. When the charge moves to infinity, the potential energy of it is zero. Therefore, the work is $kq^2/R$.


2

The path of charged particles in electrostatic field is completely indepedent of mass and charge; the path depends on the kinetic energy. From another point of view the time taken to travel along a path does depend on mass and charge. It is a classic result in the motion of charged particles in electric and magnetic fields that in magnetic fields the ...


2

Yes, a force is still acting on the particle. Moving perpendicular to the lines of force may result in no work being done ($W = \mathbf{F}\cdot \mathbf{r}$) , but since the particle is charged, it will be experiencing a force when moving through any electric field. If the particle is (initially) moving in the $y$ direction and the electric field is in the ...


1

Thanks laying out your work so neatly in the question. I think the solution is the following $$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$ where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge). so, for example, if the point charge goes from $r$ to $2r$ we have two positive ...


1

Electrons do leave surfaces either due to electrical potentials pulling them as you describe or as a result of a combination of heat plus electrical potentials. THe electrons are held in because the Fermi level is lower than the vacuum level. Furthermore, the difference in potential between the Fermi level and the vacuum level is 'felt' by the electrons ...


1

You seem to be asking about mode transformers; these are extensively used in antennas as they connect to waveguide feeds. The waveguides usually employ TE10 (rectangular) or TE11 (circular) modes, but if you want to feed a horn, say, then you have to shape the field properly to avoid reflection, reduce sidelobes, and reduce cross-polarization coupling, etc. ...


1

The tangential component due to the locally flat piece of surface, is indeed zero, at the surface. But, the total electric field is the field due to the locally flat patch, plus the rest of the surface. Hence, the total tangential component need not be zero.


1

One way to think of a vector field like $\mathbf{E}$ is to separate it into a divergent part and a curling part. Roughly, at a given location in space, the divergent part spreads out from that location and the curling part curls in a closed contour around that location. The divergent part (also called "irrotational") has $\mathbf{\nabla}\times\mathbf{E} = ...


1

"Vacuum" actually means the ground state of space. When there are real particles in space, "it" is in an excited state, and no longer the vacuum. But we usually think of a small region of empty space as approximating a vacuum, for reasons of locality, etc. If you have two real particles in space, and talk about their mutual interaction, then you necessarily ...


1

Your boundaries are at $r=a$ and $r=b$. Notice that the potentials at these two surfaces are independent of $\theta$ (they are spherically symmetric). Look at a list of the first few Legendre Polynomials $P_{l}(\cos{\theta})$. For what value of $l$ does $P_{l}(\cos{\theta})$ not depend on $\theta$? Further, notice that $V(a) = V(r=a,\theta) = V_{0}$, and ...


1

The original system as described in inherently unstable (or at best metastable). According to Earnshaw's Theorem, "a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges" http://en.wikipedia.org/wiki/Earnshaw's_theorem So the introduction of the slightest ...


1

Electrons are able to accelerate more freely then protons within a conductor. Therefore when a leaf electroscope is negatively induced the charged will move outwards while in a positively charged electroscope the leafs will stay hanging.


1

I guess different authors use different definitions. For me, it is that the E- and B-fields do not have time derivatives, hence curl free, conservative E-fields and B-fields that can depend only on steady currents. The condition that the divergence of $\partial {\bf E}/\partial t = 0$ is not the same thing. The E-field could be time variable and have this ...


1

Static electromagnetic fields implies: $$ \frac{\partial\mathbf E}{\partial t} = 0 \quad\mbox{ and }\quad \frac{\partial\mathbf B}{\partial t} = 0 $$ This means for electrostatics: $$ \nabla\cdot\mathbf E = \frac{\rho}{\epsilon_0}, \quad \nabla\times\mathbf E = 0 \quad $$ And for magnetostatics: $$ \nabla\cdot\mathbf B = 0, \quad \nabla\times\mathbf B = ...


1

Force per unit length would be given by $$F=\alpha E.$$ For an infinite line charge the electric field at a distance $d$ is, by Gauss' Law, $$E=\frac{\alpha}{2\pi \epsilon_{0} d}.$$ The dielectric is made of dipoles, so you should be able to figure out why it makes no difference to the field outside the wire. And the two wires are far enough apart that we ...



Only top voted, non community-wiki answers of a minimum length are eligible