# Tag Info

8

In a microwave the EMW energy is transferred to the water molecules, but, since they are in immediate contact with other molecules (as in any food), the whole volume gets heated. You will not have a two-temperature mixture.

8

At sufficiently high voltages almost everything conducts due in part to quantum tunneling of electrons. An insulator has a breakdown voltage which is the field strength required before it will start conducting. Related to the breakdown voltage is the dielectric strength which is the minimum voltage over distance ($\mathrm{V}/\mathrm{m}$) before a material ...

5

In a liquid mixture such as ethanol-water, both components vaporize to some extent. If the combined vapor pressure of the two equals the external pressure, say 1 atm, the mixture will boil. The components DO NOT boil separately. Further, the composition of the vapor and the composition of the liquid will be different from each other. This is the basic ...

5

You are correct: there is no free charge so $\vec{D}=0$ which means $$\vec{E}=-\frac{1}{\epsilon_0}\vec{P}=-\frac{k}{\epsilon_0r}\hat{r}$$ But this is for $R_1\leq r\leq R_2$. Inside the shell, $r<R_1$, there are no enclosed charges, so $\vec{E}=0$ there. Outside the shell, there is also no charge. Recall that the total charge for dielectrics can be ...

5

The only property of metals used in deriving $C=\varepsilon A/d$ is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter. But if you use something other than metal, then it will of course change the capacitance.

4

The energy is used to polarize the dielectric, i.e.: Moving charges inside the dielectric.

4

Use a setup that looks like this: The level of the water in the fine tube changes with the average density of the ice/water mixture so as the ice mets it will go down and as water freezes to ice it will go up.

4

The product of the permittivity and permeability is encoded into the geometry of spacetime because the product $\varepsilon_0\mu_0 = 1/c^2$ and the speed of light is special. So the value of the product is telling us about the geometry of spacetime. The relative values of $\varepsilon_0$ and $\mu_0$ tell us about the relative strengths of the electric and ...

3

From plasma physics perspective, those branches are called streamers. What happens here is that the pointed conductive object creates a high electric field because of it is pointy geometry. Of course it is connected to external power supply so it is biased at a certain voltage. The high electric field at the edge of the nail causes the loosely confined ...

3

It is all about wavelength versus tunnel diameter. The wavelength of GPS is about 20cm it would happily propagate in any normal tunnel if it could get in but the earth and other structures absorb it. AM radio (600kHz - 1500kHz) cannot propagate in any normal tunnel because the wavelength is too long (500m-200m) relative to the diameter, and thus gets ...

3

On point 2: While vacuum itself, being composed of nothing at all, is not expensive, a capacitor structure able to maintain a vacuum when surrounded by air is impractically expensive. On point 3: Don't think of higher-value capacitors as requiring less voltage. Rather, a higher-value capacitor allows us to "store" more charge at the same voltage. In a ...

3

About the autoionization of water ... Wikipedia (http://en.wikipedia.org/wiki/Debye_length) gives a formula for water $$\text{debye length in nm} = \frac{0.304}{\sqrt{I\text{ in molar}}}$$ where $I$ is ionic strength, which is 1E-7 for pure, pH-neutral water. That gives a screening length of 1$\mu$m. So at DC, there will be an electric field in the bottom ...

3

Your professor is right. Capacitors K2 and K3 are not parallel and then in series with capacitor K1, because the vertical line that is separating K1 on left and K2 and K3 on right is not an equipotential line. That is, potentials on the left side of K2 and on the left side of K3 are not the same! You actually have upper half of K1 and K2 in series and ...

3

Not always. All of your Gaussian surface should be in a linear dielectric with constant electric permittivity $\epsilon$ to be able to use gauss law and derive that formula. With this conditions it's true most of the times. Here you can use again the gauss law: $\vec D = {Q_a \over 4 \pi r^2} \hat r$ But we know that for linear dielectrics: $\vec D = ... 3 Breakdowns are electron cascades. There are different kinds: 1) Intrinsic breakdown of the material occurs when the electric field is sufficiently strong to ionize an atom of the dielectric (or accelerate a stray electron sufficiently to do the same), with the resultant new free electrons then being accelerated by the field to repeat the process with ... 3 There are (at least) two ways to get at the Brewster angle. One is to consider little electric dipoles that are set oscillating by the incident light - as you mention and which I won't expand upon. Where I work, this is how we teach it in basic optics. Then in electromagnetism we adopt the other approach which is to use the Fresnel equations (Fresnel ... 3 You're not doing anything wrong. To make the most of your equation,$\rho_\text{free}=\epsilon_r(\rho_\text{free}+\rho_\text{bound})$, it is best to rearrange it as $$\rho_\text{bound}=\frac{1-\epsilon_r}{\epsilon_r}\rho_\text{free}= -\frac{\chi_\text e}{1+\chi_\text e}\rho_\text{free}.$$ This equation expresses the fact that a free charge in a dielectric ... 3 A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is$K$and the thickness of the dielectric is$t$, then for calculating the force it behaves as if the thickness was$t\sqrt{K}$. To see this let's take the example we know about where the dielectric fills the space between the charges: In (a) the thickness of ... 3 You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction$n$is given by: $$n=\sqrt{\frac{\mu}{\mu_{0}}\frac{\epsilon}{\epsilon_{0}}} = \sqrt{\mu_{r}\epsilon_{r}}.$$ But Jackson notes that for most optical frequencies (and non-meta-material media),$\frac{\mu}{\mu_{0}}=\mu_{r}\approx 1$, which is why ... 3 Everywhere inside of the dielectric, the following (Gauss's Law inside of meadia) equation holds $$\nabla\cdot\mathbf D = \rho_\mathrm{free}, \qquad \mathbf D = \epsilon\mathbf E$$ Inside of the dielectric, there is no free charge, so we have the equation $$\nabla\cdot(\epsilon\mathbf E) = 0, \qquad 0<z<a$$ Now, we recall the definition of the ... 2 There isn't really a good physical explanation - this simply arises from the conventions we choose to represent our electromagnetic fields. The electric constant$\epsilon_0$was defined as the constant needed to make Gauss's law for electricity and Coulomb's law work for whatever units of length, charge and force you want to choose. When we add a medium, ... 2 Well first of all the major points here are potential difference across both capacitors will remain same after insertion of dielectric in$C1$, and the total charge of the system will remain constant. Now in$C1$after insertion of dielectric, calculate change of capacitance, assume that it has new potential$V1$and charge$Q1$, assume potential and charge ... 2 In electromagnetism, absolute permittivity is the measure of the resistance that is encountered when forming an electric field in a medium. In other words, permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium. Yes, metals have infinite permittivity as they completely negate the electric field inside their ... 2 From skimming a few articles and patents on e-ink driver technology, my impression is that the primary reason is that each microcapsule acts as a capacitor. Once voltage is applied, the particles move to one electrode or the other and remain there because there is no drain path for the charge. The 'gooiness' of the fluid helps, as evidenced by the typical ... 2 If the plates are disconnected, the charge has nowhere to go. Rather U will have to change. What happens is the charged capacitor does work on the dielectric (pulling it in), resulting in a change in the energy stored in the capacitor. 2 There are no sparks/arcs in vacuum. Unless you get field emission, which requires very high charge densities i.e. either high voltages or very small structures on the emitting surface, there has to be an initially non-conducting medium that can be ionized in the field between the electrodes. Once that ionization occurs, there will be a cascade of electrons ... 2 The problem here isn't a simple algebra error, but rather an issue with the physics. A medium which at rest is isotropic no longer behaves as an isotropic medium when it is moving relativistically. Instead, it behaves as a nonreciprocal bianisotropic material. In particular, the phase velocity of light at a particular frequency in a medium is no longer ... 2 I imagine that one of my students asks me this very interesting question. This is how I would try to explain it to her/him, without the use of complex mathematics and concepts involving ‘screening currents’ and Higgs mechanisms etc. It is known that in most part matter is empty space (vacuum) between the atoms. Therefore, although photons can be subjected ... 2 A di-electric experiences polarization int he presence of an electric field. The magnitude of polarization will present an effective resistance (more polarization against the field = more apparent resistance). But the polarization takes time (it's not instantaneous). So think about polarization delay vs. the change of the source electric field (i.e. the ... 2 A big carrier concentration can be a bad thing. The advantage of semiconductors is that they change their properties under external actions (generally, the carrier concentration$\Rightarrow\$ conductivity). If you already have a lot of carriers than it is difficult to make a considerable change. For example, narrow-gap semiconductors are used in ...

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