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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

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

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 ...


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$\mu$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

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

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

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 ...


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

A simple model that explains the frequency dependency of the resistivity of metals reasonably well is the Drude model (http://en.wikipedia.org/wiki/Drude_model). There we have frequency dependency because the electrons in a plasma are not moving arbitrarily fast, which is consistent with Xurtio's explanation. The cutoff frequencies are usually in the optical ...


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

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

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 ...


2

The electric field means that the potential energy of the polar molecule depends on it's angle relative to the field and is given by: $$ V(\theta) = -p E \cos\theta $$ where $p$ is the dipole, $E$ is the field strength and $\theta$ is the angle the dipole makes to the field lines. Obviously this has a minimum when the dipole is aligned with the field, so ...


2

The total charge density, which is the sum of free charge density and bound charge density is constant all over the sphere, but the free and bound charge densities differ for upper and lower half of the sphere. Your question about symmetry of total charge density can be answered easily, assuming that you know the symmetry of electric field. (your reasoning ...


2

Work done by the electric force is positive and not negative. The change in the potential energy of the electric field $\Delta U$ is equal to the negative of the work done $W$ by the electric force. You have $$\Delta U = -W$$ $$U_1=U_0-W<U_0$$ $$\therefore W>0$$ And as $$W=\int F\cdot dl>0$$, we can say that $F$ acts in the same direction as ...


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

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 ...


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

As Pygmalion points out, the flaw in your reasoning is assuming that the surface of the $K_1$ dielectric is an equipotential, which it need not be. At the triple junction there will be some accumulation of charge and the accompanying electric fields, which will result in a potential difference between the two sides of dielectric 1's surface. Let me explain ...


2

Common microwaves always operate on the principle of dielectric heating; resistive heating is not a factor. In general, microwaves are designed to be non-ionizing, meaning that the individual photons don't have enough energy to knock off electrons. A typical microwave operating at $2.45$ $\text {GHz}$ has photons with an energy $E = hf = ...


2

There are two misconceptions present in your explanation of the problem. $N$ is not number of dipoles, but their volumetric density $Q$ is not total charge, but equivalent charge at boundaries of the dielectric. The idea is that (a) dielectric of the area $A$ and height $L$ polarized homogeneously along its height and (b) two plan-parallel plates of the ...



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