# 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.

5

First make a parallel plate capacitor with plates of area A and spacing d. Fill the space between the plates with the dielectric whose complex permittivity $\epsilon(\omega)$ you wish to measure. The formula for this capacitance is a complex function of frequency because the permittivity is a complex function of frequency. $$... 4 The energy is used to polarize the dielectric, i.e.: Moving charges inside the dielectric. 4 Use a setup that looks like this: The level of the water in the fine tube changes with the average density of the ice/water mixture so as the ice mets it will go down and as water freezes to ice it will go up. 3 About the autoionization of water ... Wikipedia (http://en.wikipedia.org/wiki/Debye_length) gives a formula for water$$\text{debye length in nm} = \frac{0.304}{\sqrt{I\text{ in molar}}}$$where I is ionic strength, which is 1E-7 for pure, pH-neutral water. That gives a screening length of 1\mum. So at DC, there will be an electric field in the bottom ... 3 Your professor is right. Capacitors K2 and K3 are not parallel and then in series with capacitor K1, because the vertical line that is separating K1 on left and K2 and K3 on right is not an equipotential line. That is, potentials on the left side of K2 and on the left side of K3 are not the same! You actually have upper half of K1 and K2 in series and ... 3 Breakdowns are electron cascades. There are different kinds: 1) Intrinsic breakdown of the material occurs when the electric field is sufficiently strong to ionize an atom of the dielectric (or accelerate a stray electron sufficiently to do the same), with the resultant new free electrons then being accelerated by the field to repeat the process with ... 3 Not always. All of your Gaussian surface should be in a linear dielectric with constant electric permittivity \epsilon to be able to use gauss law and derive that formula. With this conditions it's true most of the times. Here you can use again the gauss law: \vec D = {Q_a \over 4 \pi r^2} \hat r But we know that for linear dielectrics: \vec D = ... 3 There isn't really a good physical explanation - this simply arises from the conventions we choose to represent our electromagnetic fields. The electric constant \epsilon_0 was defined as the constant needed to make Gauss's law for electricity and Coulomb's law work for whatever units of length, charge and force you want to choose. When we add a medium, ... 3 Everywhere inside of the dielectric, the following (Gauss's Law inside of meadia) equation holds$$ \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 $$Now, we recall the definition of the ... 3 There are (at least) two ways to get at the Brewster angle. One is to consider little electric dipoles that are set oscillating by the incident light - as you mention and which I won't expand upon. Where I work, this is how we teach it in basic optics. Then in electromagnetism we adopt the other approach which is to use the Fresnel equations (Fresnel ... 3 You're not doing anything wrong. To make the most of your equation, \rho_\text{free}=\epsilon_r(\rho_\text{free}+\rho_\text{bound}), it is best to rearrange it as$$\rho_\text{bound}=\frac{1-\epsilon_r}{\epsilon_r}\rho_\text{free}= -\frac{\chi_\text e}{1+\chi_\text e}\rho_\text{free}.$$This equation expresses the fact that a free charge in a dielectric ... 3 It's not altogether clear what you're asking, but I'm guessing you're doubting the "standard" set:$$\nabla\cdot\vec{D} = \rho\nabla\cdot\vec{B} = 0\nabla \times \vec{E} = -\partial_t\vec{B}\nabla \times \vec{H} = \vec{J} + \partial_t\vec{D}$$These are the set you use in linear, isotropic inhomogeneous mediums. So, for example, from the ... 3 From plasma physics perspective, those branches are called streamers. What happens here is that the pointed conductive object creates a high electric field because of it is pointy geometry. Of course it is connected to external power supply so it is biased at a certain voltage. The high electric field at the edge of the nail causes the loosely confined ... 3 Ohm's Law is an idealization. No material follows it exactly, even for small applied voltages. But if your voltage is small enough, the deviation from Ohm's Law will be so small as to be unmeasurable ... but it's still there. What we usually mean in Ohm's Law is that the behavior is linear for the purpose of whatever the application is. Once we have ... 3 A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is K and the thickness of the dielectric is t, then for calculating the force it behaves as if the thickness was t\sqrt{K}. To see this let's take the example we know about where the dielectric fills the space between the charges: In (a) the thickness of ... 3 You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction n is given by:$$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

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

So, this is an old post that I came across when I had a similar question. Here's a paper where they dissolve different amounts of ions in the water and found that the ability for the microwave oven to heat the water actually reduces as more ions are introduced. So from this study, you could conclude that ion drag is not source of heating in the water.

3

The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...

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

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

I'll assume that you meant that the disk has a uniform surface charge across the entire disk, the situation is stable, and there are no externally imposed electric fields. If this is true, then the charges must be held in place, because otherwise the charges at the center of the disk will be redistributed towards the periphery, violating the purported ...

3

No, this is essentially never true. In a linear dielectric, $D=\varepsilon E$, so $D=0$ can only be true if $E=0$. In other words, the statement is saying that any linear dielectric always perfectly screens a uniform external electric field. That's not true. Conductors perfectly screen electric fields in their interior, but dielectrics don't (unless the ...

2

Since many metals form a thin oxide layer of relatively high resistivity, they can be used to affect the dielectric properties of a capacitor made from that material. For example, aluminium, tantalum and even copper all form oxide layers with high resistivity (compared to the respective metals) and high relative permittivity (dielectric constant). Since ...

2

Both of these technologies rely on radio frequency waves, which are blocked by dense matter eg, a hill, a building.

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