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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 ...

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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 ...

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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 ...

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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 ...

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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 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 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 ... 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 ... 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 = ...

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1) First of all, screening is a response to an external field, and as such it can never fully counteract the effect, except in the case of a perfect conductor. As $\varepsilon_\text{water}$ is finite, we know we are not in the perfect conductor limit. If an external field is applied, charges will rearrange (via bulk motion of ions, re-orientation of polar ...

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If I remember correctly from Electromagnetism, if you wait for the circuit to reach equilibrium with the capacitor (i.e. current = 0) doesn't the total voltage across the entire circuit have to be zero? So the if $\Delta V_{capacitor} \approx 500V$ and since in a parallel plate capacitor the electric field is relatively constant. So you have: $$\Delta V = EL ... 2 Breakdowns are electron cascades. There are different kinds: 1) Intrinsic breakdown of the material occurs when the electric field is sufficiently strong to ionize an atom of the dielectric (or accelerate a stray electron sufficiently to do the same), with the resultant new free electrons then being accelerated by the field to repeat the process with ... 2 As Pygmalion points out, the flaw in your reasoning is assuming that the surface of the K_1 dielectric is an equipotential, which it need not be. At the triple junction there will be some accumulation of charge and the accompanying electric fields, which will result in a potential difference between the two sides of dielectric 1's surface. Let me explain ... 2 The total charge density, which is the sum of free charge density and bound charge density is constant all over the sphere, but the free and bound charge densities differ for upper and lower half of the sphere. Your question about symmetry of total charge density can be answered easily, assuming that you know the symmetry of electric field. (your reasoning ... 2 A simple model that explains the frequency dependency of the resistivity of metals reasonably well is the Drude model (http://en.wikipedia.org/wiki/Drude_model). There we have frequency dependency because the electrons in a plasma are not moving arbitrarily fast, which is consistent with Xurtio's explanation. The cutoff frequencies are usually in the optical ... 2 A di-electric experiences polarization int he presence of an electric field. The magnitude of polarization will present an effective resistance (more polarization against the field = more apparent resistance). But the polarization takes time (it's not instantaneous). So think about polarization delay vs. the change of the source electric field (i.e. the ... 2 The electric field means that the potential energy of the polar molecule depends on it's angle relative to the field and is given by:$$ 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 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 = -WU_1=U_0-W<U_0\therefore W>0$$And as$$W=\int F\cdot dl>0$$, we can say that F acts in the same direction as ... 2 I imagine that one of my students asks me this very interesting question. This is how I would try to explain it to her/him, without the use of complex mathematics and concepts involving ‘screening currents’ and Higgs mechanisms etc. It is known that in most part matter is empty space (vacuum) between the atoms. Therefore, although photons can be subjected ... 1 First off, the dielectric material on the x-y plane ought to have some thickness, say \Delta, so that the slab occupies the region -\Delta/2 < z < \Delta/2. Now we assume a point charge above the slab on the z axis. This charge causes a polarization P in the dielectric slab. Thus there will be a bound surface charge density of \sigma = P ... 1 But should the fields of other charges outside the block be unaffected?? Nope. The fields outside are affected. It's not always easy to calculate, though. Basically, inside a (linear) dielectric an induced field is formed in the opposite direction which counteracts the electric field, reducing the net electric field. Of course, this new field has to be ... 1 The short answer is yes. Here is an easy way to see it: for simplicity assume that both capacitors have capacitance equal to C and each have voltage V. Then, the charge on each capacitor is equal to CV and the total charge is Q=2CV. Now, a material with dielectric constant \kappa is inserted into one of them so that its new capacitance is ... 1 This is a system of two capacitors in parallel, so the total charge is given by$$ Q_\mathrm{tot} = (C_1 + C_2)V ~.$$When a dielectric is inserted into capacitor 1, it's capacitance changes: C_1 \to \kappa\,C_1. However, the total charge is conserved. Thus$$ Q_\mathrm{tot} = (\kappa\,C_1 + C_2)V' ~. This implies that \begin{align} (C_1 + C_2)V ... 1 I don't see any reason to think anything but the following will happen: Both alcohol and water will evaporate constantly; The microwave will heat the vodka just like any other water based solution; Once the temperature of the vodka reaches 78.2 °C it will boil. On the other hand if you use a tightly corked bottle of gin... this happens! 1 I would use a capacitance bridge. This is by far the most accurate way to measure capacitance. If you're at a university there's probably someone around who has a commercial capacitance bridge. If not, you can easily make one. 1 There are opposing surface charges on the inner faces of both of the electrodes. The is also effective surface charges on the outer faces of the dielectric, due to its being polarized by the applied field. The field lines emanate from and terminate on these surface charges. ## The field up here is small oooooooooooooooooooooooooo -- outer surface of ... 1 It seems like you might be getting confused by the definition of "in the dielectric," as part of the phrase "the field in the dielectric is \epsilon_0 E/\epsilon." This means that the electric field in regions occupied by dielectric material is \epsilon_0 E/\epsilon. If you remove a piece of the dielectric material to make a hole, then that hole is no ... 1 Bound charges are only on the surface of a dielectric if the dielectric is homogeneous and isotropic and free from free charge. Bound charge and polarization vector satisfy the equation\rho= -\nabla\cdot \vec{P},\sigma= \vec{P}\cdot\vec{n}, where $\rho$ is bound charge density and $\sigma$ is bound charge surface density, $\vec{P}$ polarization ...

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