# Tag Info

28

In general the answer is "yes it is possible" - but in your case the answer is "that is not a Faraday cage". Radio waves are (partially) reflected by any discontinuity in dielectric constant of the medium they propagate through. The ones that propagate (through walls etc) will also experience attenuation. A faraday cage is a continuous conducting structure ...

16

Just to add to what Floris has said. It is frequent (in the UK) that institutional settings would have toughened glass in windows, particularly in bathrooms, gyms etc. that would have the form of a wire mesh (of order 1cm grid) embedded in the glass. That would do a particularly good job of blocking phone signals that would otherwise penetrate the glass.

6

There have been lots of experimental attempts to test the validity of Coulomb's $r^{-2}$ law. Many of these are reviewed by Tu & Luo (2004), and is where I am getting the numbers quoted below. Somewhat equivalently, experiments have looked at trying to set an upper limit to the photon mass, which is testing the hypothesis that rather than a $r^{-1}$ ...

4

By capacitor charge is meant the absolute value of the charge on each capacitor plate: $\mid Q \mid$. If the battery generates the potential difference $V$ and you connect the capacitor to the battery through a conducting wire, as shown in your picture, once the equilibrium is reached each plate of the capacitor will have a charge $Q = CV$, where $C$ is ...

4

The key to understand the issue is that between the upper and lower "corner" of the circuit the voltage is always zero, therefore no current will flow across $\mathrm C_5$ and $\mathrm C_6$. In "corners" I mean the points common to $\mathrm C_3$-$\mathrm C_4$ and $\mathrm C_1$-$\mathrm C_2$. These pairs of capacitors are effectively voltage dividers. ...

3

Coulombs law as well as Amperes law and similar mathematical formulations of two centuries ago, were incorporated within the strict mathematical format of Maxwell's equations . The apparently disparate laws and phenomena of electricity and magnetism were integrated by James Clerk Maxwell, who published an early form of the equations, which modify ...

3

Charge means that the body experiences a force in an electric field. A charge generates an electric field, which generates a force on other charges particles. Two bodies are said to repel if they force each other away and two bodies are said to attract if they force each other closer together. Now, I'm not really answering your question here of "why," I ...

2

This the kind of question that can be solved by the method of images. Try placing a fictitious charge on the other side on the conducting plane. You should arrange it in such a way that the electrostatic potential is precisely zero on the surface of the conductor. If your case you put it at equal distance as the first but on the other side. The physical ...

2

$E_1 = E_2$ . since $E$ is independent of dielectric as long as potential b/w plates is constant. $$E= = -\frac{dV}{dr}$$ So, it is independent of dielectric b/w it. So, correct statement would be $$E_1d = E_2d$$ $$Ed = Ed$$

1

Remember, the dipole is a vector. So, its not simply $p = qd$. For a general charge distribution $\rho(\mathbf r')$, you need the multipole expansion of potential in spherical coordinates, for powers of $1/r$. Meaning, take the potential, make expansion of $1/r$ powers. The $1/r$ term is the monopole. The $1/r^2$ term is the dipole. The $1/r^3$ term is the ...

1

The heart only acts as a big dipole, when it is electrically active. ECG measures the potential difference between different places on the skin. These potential differences are created when different parts of the heart muscle are in dfferent stages of their action potential. For example, when the septum and the subendocardial myocytes are depolarized, but ...

1

You seems to assume both capacitors has the same plate separation $d$. So, lets assume that. Assume there is no dielectric material. Therefore, nicely $Ed = Ed$ in both capacitors. Which is nice. :). Now, I think I understand your confusion. Have an isolated capacitor with electric field inside plates of $E$. Insert dielectric $K$. Under this case, the ...

1

The difference is that batteries chemically "pump" electrons from one side to the other. There is a small amount of charge separation in a battery even when it is not connected to a circuit. This charge creates an electric field that opposed the chemical action of the battery to prevent further charge separation. This makes the battery act somewhat like a ...

1

When the point charge is not at the center of the sphere, the electric field lines will not intersect the sphere at right angles. Consequently, there is an initial component of electric field along the surface of a conductor. We know this results in a force on the charge carriers inside the conductor, and these charge carriers will re-arrange until the ...

1

For Capacitors, the charge stored in it is directly proportional to the potential difference across it. Hence, the charge stored in the capacitor is given by the relation $$Q=CV$$ where C is a constant known as capacitance which is an inherent property of the capacitor. In a parallel combination, the charge through each capacitor has the same entry and exit ...

1

Why are potential differences equal across two capacitors in series, but charge on each capacitor is not? This is based on a false premise. There is no rule that says that "potential differences are equal across two capacitors in series". In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor ...

1

There is not really a difference between the two lines of reasoning. In each case you are taking a path integral along a chosen path and calculating all of the contributions of $\vec{E}\cdot d\vec{l}$. You could choose an arbitrary path, and this would still be the case, but the integral would be more complicated. An easy way to think about this is to use ...

1

"I find lots of solutions on the internet that say you can replace the cavity with a negative density, why?" Because they use a trick to calculate the potential easier. They assume that the empty hole is neutral, but composed of a positive charge density equal to that of the sphere plus a negative charge density of the same amount. In this way you can ...

1

You can't really explain the conductivity in metals without basic quantum mechanics. Metals as made up of lattices of metal atoms, packed at very close distances. The outermost and least tightly bound (to the nucleus) electrons, the valence electrons, occupy atomic orbitals of the least energy. Due to the close vicinity of the atoms and the similarity in ...

1

You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the ...

1

You are in your reasoning overlooking something. Look at the diagram below: $-q_1,+q_2$ are two point charges at distance $r$. Coulomb's Law dictates that the attractive electrostatic attraction force between them is: $$F=k_e\frac{|q_1q_2|}{r^2}$$ And the electrostatic potential $U(r)$: $$dU(r)=F(r)dr$$ $$U(r)=-k_e\frac{|q_1q_2|}{r}$$ Assume now that ...

1

In general, no: the field of a charge distribution $\rho$ is not the same as the field of a point charge at some point therein, except for some very particular cases (the one that everyone should know is that any spherical shell of charge has an inner field of 0 but an outer field that looks exactly like all the charge is located at the center point. A ...

1

First, the Wikipedia article already says on the derivation of Gauss' law from Coulomb's law: Note that since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more ...

1

Remember Lenz's Law: as you change the flux through a coil, an e.m.f. is generated that opposes this change. Therefore, if I have two coils that are a certain distance apart, they will have a certain "shared" flux - flux due to $A$ appearing in coil $B$, for example. Now if we bring $A$ closer to $B$, we change the flux in $B$ due to $A$, and will get a ...

1

Note that $\mathbf r(t)$ is the trajectory (a priori unknown) of a charged particle in an external electric field. Now consider the ansatz $\mathbf r(t) = \mathbf r_0(t) - \mathbf a(t) \cos \Omega t$, which is motivated by the solution for a homogeneous electric field $\mathbf E(t) = \frac{m\Omega^2}{q} \mathbf a \cos \Omega t$. Here $\mathbf r_0(t)$ is a ...

1

The first equation assumes the external electric field (caused by the charges on the plates of the capacitor) doesn't change. When a battery is connected, it can fill and discharge the plates as necessary to maintain the voltage. So, the E_0 value increases as you add the dielectric.

1

I might be erring something basic here, so downvotes are welcomed, but I would love if they include comments to correct this answer, or just erase it. I do not believe the Coulomb law has been tested beyond the order of a few meters. Arguing that light remains unchanged across the universe should be irrelevant. The reason is that the electrostatic and ...

1

I asked a somewhat different, yet similar question.Hope this helps! Why is an $LC$ oscillator lossless, but $C V^2 / 2$ energy is lost to a capacitor connected to an ideal voltage source?

1

According to Richard Feynman, the charge is the probability of a particle interacting by the electro magnetic force. More specifically it describes the amplitude of the "probability arrow" of a certain electromagnetic interaction taking place. Much like @Asher has mentioned already, the standard model cannot provide an explanation for why certain particles ...

1

The difference is the zero point. When summing over charges, the reference is a state in which this charges are infinitely separated. Those are still distinct, localized charges, just separated from each other. When integrating $E^2$ over all space, the reference state has all charge separated. Even the individual charges from the first method are broken ...

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