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59

Since cables carry electricity moving at the speed of light, why aren't computer networks much faster? Perhaps I can address your confusion with a rhetorical question: Since air carries sound moving at the speed of sound, why can't I talk to you much faster? The speed of sound is much slower than light, but at 340 m/s in air, it's still pretty damn ...


21

As you've probably guessed the speed of light isn't the limitation. Photons in a vacuum travel at the speed of light ($c_o$). Photons in anything else travel slower, like in your cable ($0.64c_o$). The amount the speed is reduced by depends on the material by the permittivity. Information itself is slower still. One photon doesn't carry much ...


11

A transmission line is made of a pair of conductors which have some resistance, inductance, capacitance, and leakage conductance. We can take all of these per unit length: The wave equation for signals in this line, in the limit of a lossless cable with $R=0$, $G=0$, is $$ \frac{\partial^2 V(x)}{\partial x^2} + \omega^2 LC \cdot V(x) = 0 $$ You have to be ...


4

"Surely this is a bottleneck" - No, it's really not. Any real-life network connection is not speed-limited by the propagation speed of the signal in the cable, but by the processing delays in the various routers, switches, and network interface processing at each end.


4

In principle, sure. That's what microphones are, as ACuriousMind points out. But if you want to power anything substantial, an important issue to overcome is the relatively small amount of energy contained in sound waves. According to this website, the front rows of a rock concert have a sound intensity of $10^{-1}~\text{W m}^{-2}$. So even if you had a ...


3

How sure are you that electricity travels at the speed of light? Although electricity propagation moves at the speed of an E/M wave, and not electrons, its speed depends on the dielectric constant of the material. Only in a vacuum, I think, would it travel at the speed of light.


3

Why only 64% What does propagation speed mean? I know there are other variables effecting the latency and perceived speed of computer network connections, but surely this is a bottle neck. Speed of signal propagation is distance the signal (packet) travels in one second. It is usually lower than $c$ because EM waves that carry the information travel in ...


2

Two reasons: 1) The speed of light in a "medium" is (almost*) always slower than the speed of light in a vacuum. 2) Electricity propagating in a wire is subject to inductive and capacitive effects which slow it's progress. And even if wires were infinitely fast, integrated circuits are not. Again, inductive (a little) and capacitive (a lot) effects limit ...


2

Assuming that the bulb is powered by some variation of alternating current, and that there is a static magnetic field present in the environment, the vibration could be caused by the interaction of the fixed external magnetic field, and the alternating magnetic field caused by the AC flow through the filament. Bulb filaments are often in the form of a ...


2

This really depends on the design of the bulb in question. High-power or hard use bulbs have filaments under some tension and strong supports. Other (mainly decorative) bulbs have very wispy support wires and straight, loose filaments. In this case, the wire can be very sensitive to mechanical vibration through the floor. The bulb and lamp ( a bulb will be ...


2

Is the shaking immediate upon turning the bulb on or after a while? This could mean heating causes Is this the only bulb doing this or is it same with other (of same type)? This could mean bulb towards end of its life or bulbs of this type do that due to construction of filaments Is this the only bulb doing this or is it same with any other? This could mean ...


2

Assuming the ground to be at $0 V$, there is a potential difference of $220V$ between the person's right hand and the ground. The only way to prevent an electric current to flow from his right hand to the ground is if the pathways of current between those two are insulated, blocking the path of current. Holding an insulator in the left hand does not prevent ...


2

Yes - electrons flow from the negative to the positive, so in the opposite direction to the conventional direction of the electric field (which points from positive to negative). So if the E field points outwards, the electrons flow from the outer to the inner cylinder. The direction does not affect the answer (the calculation of the flow) though - at least ...


2

You have a rather precise power meter in your home, which is a "gift" of the electrical power company. Turn off every other load that is connected to that power meter and do your measurement. Alternatively, you can invest $20 in an electronic power meter that is available online and in many stores.


2

The easiest way would be to use an energy monitor device (the most popular one seems to be the Kill-A-Watt but there are others). They simply plug into the wall and then you plug your appliance into it. It displays instantaneous voltage, current, power, power factor, etc. and can keep total energy over time. Another option would be to buy an electric ...


2

An informal definition of insulator is that electrons are bounded enough so that they cannot flow all over the material, they remain next to the atom where they belong. In a conductor electrons can move freely and flow to different parts of the solid, detached from the original atoms. Why that happens depends on many factors, but it does not mean that you ...


2

Conductivity is not just about how tightly bound electrons are, but equally about how easy it is for them to travel. Example: a bunch of islands in a shark-infested sea. You cannot swim from one island to the next although it is close. At low tide you can walk across no problem. The first example is an insulator, the second is a conductor. Rubbing (google ...


2

If atoms have well define energy levels and the differences correspond to the frequencies of light that can be absorbed, This is correct how is it that opaque objects absorb all or most visible light frequencies get absorbed and you basically don't have any visible light coming out on the other end? The crux of the matter is the word "objects". ...


2

Your requirement that the measurement be made with equipment available in a kitchen is a severe constraint as I can't think of any way of measuring the electrical power supplied. If it's impossible to measure the electrical power in then the only other approach is to measure the thermal power out - i.e. measure the heat produced by the appliance. Given that ...


1

You can only measure a part of EU compliance. By using your utility meter and measuring the difference between the power consumed over say 10 minutes with the appliance on and off (with everything else in the house as off or steady as possible), you can measure consumed power. However, that's just one part of EU compliance. The other is power factor. ...


1

For a capacitor, the voltage across must be continuous since the current through since $$i_C = C \frac{dv_C}{dt}$$ Since the current through is proportional to the time derivative of the voltage across, the $v_C(t)$ must be differentiable, i.e., there can be no discontinuous change. There is no such limitation on the capacitor current, the direction ...


1

If you have a complete circuit, every piece of metal will gain and lose the same number of electrons and will not have a net charge. If you connect two plates, one to each end of a battery, the battery will take charges from the plate connected to the positive terminal and send charges to the plate connected to the negative terminal until the voltage ...


1

In layman's terms.Elect is inverse mag.Hold powerful magnets on opposite sides of a thin glass window.Report findings.Next q.v.fiber optics.Photons being electro messenger particles. The longer and thinner the glass rod,the greater charge on surface possible. A magnetic field does not transmit through fiber optic thread. Good thing. Glass (silica) is a ...


1

The Lorentz force on a charge in an electromagnetic field is $$F=q(E+v \times B) \ \ .$$ For an electron between the cylinders, $q$ is negative, and $E$ is defined as pointing outward, so the electron will experience a force radially inward. But due to the unfortunate sign convention used for currents, electrons flowing inward means that the conventional ...


1

As long as your product $|q_{1}||q_{2}|$ remains the same as in the case where you had the equally charged spheres, then yes, you will get the same value for the angle (provided the masses are equal). This is because the electrostatic force acts equally on both charged spheres.


1

I've found an explanation. As Duncan said, as the sum of the charges stays equal the greatest product of two integers with a certain sum is half of the sum when (for example 5+5=10 and 5x5=25 is the greatest product. NOTE: Here the sum is always even as we take two equal charges 'q' initially). Hence all other series yields a force lesser than the ideal ...


1

Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is E=ρr/(2ε) Now, to calculate the potential difference between the surface and axis of the cylinder, deltaV=-∫ρr/(2ε)dr from 0 to R ...


1

You are increasing the number of electrons in the wire, but only by a very small amount. There's a somewhat clichéd but still excellent analogy for electrical circuits called the hydraulic analogy. In the hydraulic analogy the power supply is a pump, and the pressure is the voltage. The water represents the electrons, so the pressure generated by the pump ...


1

It's quite common for ac-powered electrical devices which are designed without moving parts to "hum," usually with an audible frequency of 60 Hz, 120 Hz, 180 Hz, or other 60 Hz harmonics. (In other parts of the world, where the power distribution system frequency is 50 Hz, you'd obviously get harmonics of that frequency instead.) For instance, when I'm ...


1

There is no way one can answer this question without further requirements. In practice, engineering considerations will put wireless charging of vehicles probably somewhere between the 20kHz and 150kHz range (unless the regulator permits a higher frequency). Why 20kHz? Because all wireless charging solutions will involve the generation of some amount of ...



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