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36

Electro-magnetism is a good guess, simply because it's the only force you commonly see that's powerful enough. It's not very useful as an explanation, though, because almost everything you see around you is due to electro-magnetism (e.g. the way the spoon holds together in the first place, or the light that allows you to see the sugar, or the way the water ...


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 ...


17

The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.


17

To add to ACuriousMind's answer on the Liénard-Weichert potentials, you can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula from them for the radiation from a moving charge: $$\vec{E} = ...


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.


15

Not magnetism but static electricity - the other side of the electromagnetic force. Sorry this answer is short, but I think the link will give you all the information you need.


13

First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...


11

You are right - potential is a scalar. But a dipole moment is a vector - it has magnitude and direction. When sodium channels open up, charge flows. Lots of charge moving a little bit causes a change in the dipole moment of the heart. This in turn induces charge to move elsewhere in response (the dielectric properties of tissue cause a propagation of the ...


9

What the picture shows is a corona discharge (see also Wikipedia). It isn't a circuit in the usual sense of the word. It happens because the voltage is so high that it raises the electron energy to above the work function and the electrons just leak off. In effect the coil is charging the air around it. The charge will end up on the furniture, walls, floor, ...


8

If you are talking about point charges then, as explained above, the answer is no. But in the case of non-uniform charge distributions, it is possible for same-charge particles to attract, if they are sufficiently close. As an example, the following two particles are identical, each having a net charge of -1. Plotted below them is their potential energy as ...


8

This may help your, it comes from Rutherford scattering by which they determined that the atom has a hard core. It is positive alphas against positive nucleus, but the math is the same. Determining the closest approach to the nucleus amounts to calculating the minimum distance for the hyperbolic orbit which is produced by the coulomb repulsive force. ...


8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


8

Electrostatic refers to the case where the fields are not time dependent. In that case the Maxwell's equations reduce to: $$\nabla \cdot E =\frac{\rho}{\epsilon_o} \\ \nabla \times E = 0 \implies E=-\nabla \phi \\ \text{then,} \nabla \cdot \nabla \phi = \nabla^2 \phi = -\frac{\rho}{\epsilon_o} $$ The solution to the last equation is: $$ \phi = ...


8

You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner ...


7

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 $r^{-1}$ ...


6

The reason why the method of the images is easily applicable in the case of the sphere or the plane is that it uses the symmetries of the Laplace operator $$\Delta\Phi=\frac{\partial^2\Phi}{\partial x^2}+\frac{\partial^2\Phi}{\partial y^2}+\frac{\partial^2\Phi}{\partial z^2}$$ or in the spherical coordinates $$\Delta\Phi=\frac{1}{r}\frac{\partial^2}{\partial ...


6

The reason is the same as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) until they don't feel a force any more. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


5

So in the first case, when talk about a plain circular ring, I assume you mean an annular ring, with a well defined inner radius and a different well defined out radius. With a positive charge at the center of the annular ring, positive charges will be repelled outward and negative charges attracted inward. Incidentally, not all the positive charge will go ...


5

Gauss law says that the total flux going through a closed surface is equal to the charge inside the surface. You can think of flux as the number of field lines going in (or out) through the surface. In your example there is no charge inside the sphere so the total flux through the surface of the sphere is zero. On one end (the left hand side) the field ...


5

For $H\gg R,L$ And for $L\gg R,H$ you get pretty much the same thing. First off, $(H+L)^2\sim H^2$ and the same goes for $(H-L)^2$. That means that $(H+L)^2+R^2\approx (H+L)^2$. However, $(H+L)\not\approx H$, which means that $\sqrt{R^2+(H+L)^2}\approx H+L$. This makes the first approximation have $\frac{1}{H+L}-\frac{1}{H-L}$ in it. The second ...


5

Now, the constants C1,C2,C3 appearing when we separate variables on Laplace's equation for electrostatic potential has some physical meaning? If they do, what is it? The constants are the related to the square of the spatial (angular) frequency or a spatial growth/decay constant. For an example of spatial frequency, let $$X(x) = A \sin (k_xx) + B ...


5

...why isn't the work done against the net force due to the system considered instead of simply adding up the work done against separate forces caused by individual charges? They're both equivalent, due to the principle of superposition. Basically, the net force is what you get when you add up the separate forces from the individual charges acting on ...


5

Equation $(2)$ is indeed a general solution, but that doesn't mean that all the $A_l$ and $B_l$ have to be nonzero all the time. For a problem in which $r=0$ is part of the domain the $B_l$ coefficients are zero, else the potential diverges at $r=0$ due to the $r^{-(l+1)}$ functions. In the case of the point charge, the point $r=0$ is not part of the ...


5

The problem here is that you've failed to specify a boundary condition. Consider an electrostatics problem where you're given a charge distribution $\rho(\mathbf{r})$ and asked to find the electric field $\mathbf{E}(\mathbf{r})$. The electric field is the solution to the set of differential equations $\nabla \times \mathbf{E} = 0$ and $\nabla \cdot ...


5

It does give "Coulomb's law" with $\frac{1}{r^3}$, it gives it in its proper vectorial form $$ \vec E \propto \frac{\vec r}{r^3}$$ which, when taking the absolute values, yields the form you are probably more familiar with $$ E \propto \frac{1}{r^2}$$ since $\lvert \vec r \rvert = r$.


5

Physics does not answer why questions, except with how from postulates and mathematical models one can describe the data. The how is Coulombs law. Physics is about fitting experimental observations with mathematical models. The answer to the "why attraction" in this case, is, data dictates so. There is no other answer except that Coulomb's law fits the ...


5

Both gravity and electrostatic forces depend on distance ($r$) like $1/r^2$. So changing the separation between 2 atoms changes both forces equally. So whichever force is stronger initially (at any distance) will always be stronger. To determine which is stronger consider the ratio of gravitational to electric force. $$ F_g/F_e = 4\pi \epsilon_0 G ...


5

Yes, electric potential can change discontinuously. All discrete, abrupt changes of anything are idealizations. Real objects have no discontinuities. But our idealized models can have discontinuities, and you used such a model in solving the problem of the charged shell. In a real shell, the charge density would not change abruptly, and the E-field would ...


5

In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...



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