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It is true that the electric field inside a perfect conductor is zero. But consider what is happening on the surface of the conductor. We can only say that an induced field exists, which is opposite to that of the incident one and lives on the surface of the conduit conductor. The incident electromagnetic wave moves the free charges on the conductor which ...


4

The presence of a reflected wave is simply a consequence of Maxwell's equations and the boundary conditions imposed on their solutions. When a wave is incident upon the conducting interface, you are free to try any solution you like for what happens to the electromagnetic fields on either side of the interface. But those fields must be solutions to Maxwell's ...


3

The simplest explanation is the following. For a perfect conductor the incoming light makes the conduction electrons move 180 degrees out of phase with it. These electrons create two waves that are 180 degrees out of phase with the incoming field. One is moving along with the incoming wave and cancels it beyond the surface of the metal. The other moves ...


3

For whoever may struggle with the same problem: $$ \sqrt{1-\frac{4\beta^2}{(1+\beta^2)^2}} = \sqrt{\frac{(1+\beta^2)^2 - 4\beta^2}{(1+\beta^2)^2}} = \sqrt{\frac{(1-\beta^2)^2 }{(1+\beta^2)^2}} = \frac{1-\beta^2 }{1+\beta^2} $$ Substituting back into the expression: $$ \frac{1-\beta^2}{\frac{1-\beta^2 }{1+\beta^2} } -1 = 1+\beta^2 -1 = \beta^2. $$ So $...


3

Imagine you bring a charge close to a conductor, there will be an electric field inside the conductor for a short period of time. But the time period where electric field present in conductor is infinitesimal. This time varies from material to material but it's around $\approx 10^{-16}s$ for metals $$\textbf{J}=\sigma \textbf{E}$$ $$\nabla \cdot \textbf{J} = ...


3

An unpolarized wave can be represented by the sum of two waves with perpendicular polarisations and equal amplitude. The unpolarised incident wave can be considered to be of this nature, with one polarisation in the plane of incidence (p-polarised) and another at right angles to that (s-polarised), and with the electric fields of both being perpendicular to ...


3

An electric field is a vector field, which assigns a vector to each point in space. A vector itself cannot be negative or positive (unless we consider the one-dimensional case where a sign is meant to designate the direction). Arguing about the sign of the electric field vector generally makes no sense. Unfortunately, your question is ambiguous, but I can ...


3

I think positive and negative is irrelevant once you introduce the vector formalism. Instead, contemplate the direction in which the electric field vector point at a point in space? That's a more relevant and important question.


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The equation you cite, which is Faraday's Law, always holds. In fact the four complete Maxwell's equations, which are (in SI units) $$\vec{\nabla}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}\\ \vec{\nabla}\cdot\vec{E}=0\\ \vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}\\ \vec{\nabla}\cdot\vec{B}=\mu_{0}\vec{J}+\epsilon_{0}\mu_{0}\frac{\partial\vec{E}...


3

It is not doable. The reason is that you need to find a geometrical surface on which $\vec E\cdot d\vec S= \vert \vec E\vert dS\cos(\theta)$ so you can “pull out” the constant factor $\vert \vec E\vert$ outside the integral for the flux. There is no such surface for a disk. To use a cylinder you would need to have not a thin disk but an infinitely long one ...


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Maybe this can be a hint for you. The first formula isn't corret. The electric field is: $$\vec{E}(x,y,z)=\frac{1}{4\pi \epsilon}\int_v\frac{\rho(x',y',z')dx'dy'dz'}{r^2}=\frac{1}{4\pi \epsilon}\int_v\frac{\rho(x',y',z')dx'dy'dz'}{[(x-x')^2+(y-y')^2+(z-z')^2]^{\frac{3}{2}}}$$ Your uniform volume charge density between them generates an electric field. Since ...


2

There is no difference between "electromagnetic field" and "electric field and magnetic field". An electric field and a magnetic field are always just two components of an electromagnetic field, witch can be characterized by two vectors $\vec E$, $\vec B$. And electromagnetic field (= electric field + magnetic field) always obey the ...


2

Well this is the first time I hear about this method and I am not sure how it really works and what data you get from it, but I a vague idea that you would want to know the approximate resistivity of the ground below and use that for some kind of analysis. In that case you would want to pass current through the ground by using source above to provide ...


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The arguments regarding the conservation of electric flux in this previous answer of mine apply equally well here, with the minor change that the second angle corresponds to the combined charges here: the flux calculation happens, asymptotically, at infinity, where the electric field looks like that of a point charge with the combined charge, located at the ...


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You can calculate the amount of charge it acquires by $$Q= nq = 100 \times 1.6 \times 10^{-19} C = +1.6 \times 10^{-17}C$$ which is positive since it lost electrons and $q$ is the charge of an electron. The voltage can be obtained from the electric field $$\Delta V=E \Delta x$$ where $V$ is the voltage between two points separated by a distance $\Delta x$ ...


2

To get a significant charge on your transfer sphere, the sphere would need to be brought to a high potential relative to the “ground”. But that would require the other end of your line of batteries to be “grounded”. The batteries have pumped charge from the ground and put it on the sphere. This will not change the potential difference between the two ends ...


2

You don't need the speed of light: $$ k_x=\frac{\omega}{c}\sqrt{\frac{\epsilon_d \epsilon_m}{\epsilon_d+\epsilon_m}}=\frac{2\pi}{\lambda}\sqrt{\frac{\epsilon_d \epsilon_m}{\epsilon_d+\epsilon_m}}, $$ since $$ \omega=2\pi\nu \,\,\,\text{and}\,\,\,c=\lambda\nu. $$ Or you could simply say: $$ k_x=k_{\rm Vak}\cdot n=\frac{2\pi}{\lambda}\cdot\sqrt{\frac{\...


2

My first answer gives a more "holistic" approach, just arguing on the basis of having valid solutions to Maxwell's equations. However, here is another approach that looks in detail at the co-existing fields and current densities, using the microscopic version of Ohm's law. This demonstrates that the reflected wave is indeed produced by currents ...


2

Isn't Gauss' Law for electric fields $=\frac{q_{enc}}{\epsilon_0}$derived based on the assumption that the $r^2$ (one from $d\vec{S}_{spherical}$ and the other from $\vec{E}$) cancel, otherwise the flux would depend on the radius of the sphere and we would need to know the relation between $k$ and $q_{enc}$? It is easiest to see that $\Phi_E = \oint \vec{E} ...


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As it is pointed out in the comment that $1/r^2$ is for a point charge. And a charge distribution can make of any kind of field like in this case that goes like $r^5$. As an example, consider potential due to a change distribution $\rho(x',y',z')$. If you recall the multipole expansion that looks like $$\phi_A=\frac{1}{4\pi\epsilon_0}\left[\frac{K_0}{r}+\...


1

$\vec E $ is identical to $E_x \hat i +E_y \hat j+ E_z \hat k$. Right? So, ${\rm d}\vec{E}={\rm d} E_x \hat i +{\rm d}E_y \hat j+ {\rm d} E_z \hat k$. If I say now that your vector integral is actually sum of three different scalar integrals, would you agree? $$\int{\rm d}\vec E = \int{\rm d} E_x \hat i +\int{\rm d}E_y \hat j+ \int{\rm d} E_z \hat k$$ Now ...


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Yes, the formula $E=\frac{1}{2}\epsilon_0 E^2$ is valid for electric field energy density in vacuum (or other medium such as air that interacts only very weakly with electric field) whether the electric field is purely electrostatic or general (including induced electric field or field of EM waves), but it is necessary that point charges or line charges are ...


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(a) On a macroscopic scale (that is considering only electric fields averaged over distances spanning many atoms) the electric field in a conductor in electrostatic conditions is zero everywhere. This must be the case because if there were a resultant electric field it would exert forces on the mobile charge carriers and they would be moving, that is we ...


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The electric field is given by the gradient of the potential, E = -$\nabla$$\Phi$, and is a vector. If you calculate the gradient as in situation 2, its component in that direction is 0 so the component of the electric field in that direction is 0, as you know. That doesn't mean the electric field itself is 0. It has a component in the radial direction which ...


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If the charge inside a Gaussian surface is zero, then every field line which enters the surface must leave at some other point. That does not require a zero field inside.


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I recommend that you try to forget the contents of your paragraph that starts 'However...'. The dielectrics that you put between the plates of a capacitor should all be insulators, whether they are easy or hard to polarise. [Polarisation is a separation of charges within molecules, not movement of charge through the dielectric.] And electrons don't jump ...


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My understanding leads to one definition saying that less polarizable substances have higher permittivity and the other definition saying the opposite. Where is my mistake? I'm having trouble following your logic, but the greater the electrical permittivity, the lower the electric field strength, and the greater the capacitance. The greater the capacitance ...


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Since the problem specifies large plates, Gauss's law can be used in the central regions. With a positive charge density the field would start at zero and point out from the center. Putting Gaussian surfaces at + and – x: 2EA = 2ρAx/$ε_o$. Suggesting that the non-conductor may be polarized would conflict with the given condition of uniform charge density.


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This is really a group theoretical question. The best references for group theory in solid state are (in my opinion) "The Mathematical Theory of Symmetry in Solids: Representation Theory for Point Groups and Space Groups" by Bradley and Cracknell and "Group Theory: Application to the Physics of Condensed Matter" by Dresselhaus. The latter ...


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It actually does not matter as long as you keep track of the signs. It was probably defined this way in order to give a similar formulation for torque $$\vec{\tau} = \vec{r}\times\vec{F}$$ dipole moment is defined as $\vec{p} = q\vec{d}$ and torque on a dipole $$\vec{\tau} = \vec{p}\times\vec{E}$$ If it was defined otherwise you had to write it down like $\...


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