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46

Air is normally a bad conductor of electricity, but with enough voltage it can be converted to plasma, which is a good conductor. In a plasma, the electrons constantly bind to and leave atoms. Each time an electron binds to an atom, it emits the energy in light. As a result, the plasma glows the color of a photon with that energy. There are a few different ...


33

The electrons themselves don't move all that fast. The wave energy is the part that moves quickly. Picture it this way. You have 500 meters of pipe, with a small hole at the other end. The pipe is full of water and you increase the pressure at your end. Water will flow out the other end immediately. This is the electrical energy (pressure) and the ...


32

The deexcitation of nitrogen and oxygen, the primary components of air, are that of blue/purple. See http://en.wikipedia.org/wiki/Ionized-air_glow for pictures of nitrogen and oxygen in gas discharge tubes.


15

The answer is that electrical excitation of air molecules is able to produce lots of excited singly ionised nitrogen ions. The electronic structure of singly ionised nitrogen has a number of allowed radiative transitions, where the outer excited valence electrons can rearrange themselves into lower energy configurations. The most prominent turn out to be ...


14

In fact, electron's speed is not so fast that light bulb glows up immediately. It is the electromagnetic field which travels in the circuit at near the speed of light that is resposible for it. After turn on the light, electron only acquires a little speed in addition its thermal speed. The thermal speed of electron can be estimated by $mv^2/2\approx ...


10

The point of the point is to increase the electric field near the point. Small radius curves will have a higher local electric field, eventually creating a localize area where the field is greater than the dielectric strength of the air. This results in what I refer to as "micro-lightning." This microlightning discharges the air (or cloud) before the ...


9

Suppose that you have an negatively charged cloud. Floating over your conductor. Then making your lightning conductor pointy at the edge, facilitates better discharge. Because the electric field set up would be high. ${\sigma}=\frac{q}{4\pi r^2}$, We will take an spherical approximation of the pointed end. It will have a very small radius thus high surface ...


7

A light bulb wouldn't turn off, because no matter what direction the electricity is flowing through it, it is still electricity. It doesn't gain some anti-electricity effect. Here is an analogy with water. The water works flowing forwards and backwards. (Although in this example there is a stop.) If there is still confusion, however, remember that ...


7

If the flare drops energy into the atmosphere, it sets up a changing electric field. The strength of the electric field is measured in $V/m$. This means that the longer the distance, the greater the potential difference. If you have a short wire, the potential between one end and another is not very large. But a long wire can connect points with much ...


6

A negative voltage means that you have hooked your power supply across your device backwards. Purely resistive devices, like resistors and lamp filaments, don't care which way current flows through them, only how much current flows. Such devices will always have current-voltage curves which are "odd" functions, with $I(-V) = -I(V)$, as in your graph. It is ...


6

Lets suppose the amplitude of each wave is $A$ and thus intensity will be $I_0 =A^2$. After superposition amplitude of the resultant wave becomes $2A$. but intensity becomes $I=(2A)^2$ Implies $I=4A^2 =4I_0$


4

The equation: $$\frac{dV}{dI} = R$$ is a definition of $R$. Ohm's law is the statement that $R$ is constant over all voltages and currents (with $I = 0$ when $V = 0$), thereby giving: $$V = IR$$ With this definition, it is all but impossible to say $V = IR$ for any electrical component other than Ohmic resistors. Consider the Shockley equation: $$I = ...


4

Depending on your view, there is electronics with other charge carriers. It is commonplace to have semiconductor devices where the relevant carriers are holes! Furthermore, batteries and electrolysis relies heavily on ions as charge carriers (but hardly count as electronics). I guess genuine electronics with ions will be difficult as charge carrier mobility ...


3

Heat losses (aka $I^2R$ losses) occur because charge has a hard time getting through the conducting element. In a light bulb the filament is purposely made with a higher resistance, $R$, compared to the resistance of the metallic leads on which it is welded to. Heat is energy and power is the rate at which energy is transfered $$P=I^2R$$ and so with a ...


3

... but what happens if you connect it in series? Consider a circuit which has a 3-V (ideal) battery connected in series with a 100-$\Omega$ and a 200-$\Omega$ resistor (series resistors). The voltage across the 100-$\Omega$ resistor will be 1 V, and across the 200-$\Omega$, 2 V. Next, connect a 1 M$\Omega$ resistor (a voltmeter) in parallel with the ...


3

Your teacher's explanation makes sense - IFF there is a "missing component" in your circuit diagram. For example, if there is some series resistance (possibly inside the power supply). In that case, the current through $R_1$ will indeed depend on the current through $L$ - since the voltage source plus internal resistance behaves "a little bit" like a ...


3

Resistance is Voltage per Current. $R = V/I$ or $V = RI$ if you like. So if you put two resistors in series, the voltage is that due to the first, plus that due to the second. You understand that perfectly. Now, what is the inverse of Resistance, Current per Voltage? Give it a name, call it Conductance, perhaps. $C = 1/R = I/V$ Then $I = CV$ if you like. ...


3

It perhaps is not as rigorous as you want, but it is simple and intuitive: $$ R = \frac{\rho L}{A} $$ Where $\rho$ is the resistivity, $L$ is the length, $A$ is the cross-section area. When you plug resistances in series, you "are" increasing $L$, and thus $R$ increases. If you put in parallel, you "are" increasing $A$, and thus $R$ decreases.


3

You appear to be using energy to move all the "stuff" out of a hole, and then filling in the hole again with "special magnetic stuff" to extract the same energy as electricity, with inevitable losses... Perhaps you could practice by pumping water from Lake Ontario up into Lake Erie, and then selling the electricity you get when the water flows through the ...


3

The resistance of a lamp filament is not constant. As the current increases and the filament heats up the resistance increases. That's why the statement you quote is phrased that way. It means that the resistance is 4 Ohms when the current is low enough that the heating and resistance change are negligable. More verbosely it could be phrased: The limit ...


2

Vacuum tubes can conduct hundreds of amps of electricity quite readily. The effect depends on heating the negative terminal so that electrons can leave the metal surface which otherwise keeps them in the surface owing to a phenomenon called work function.


2

A negative charge attracts a positive charge and repels an also negative charge. If you have many (negative) electrons in one end of a wire and non in the other end, then these electrons will be pushed away from each other towards the end from which they are not pushed away. This principle is the key - the electron-dense end is called the negative end, the ...


2

It is absolutely ridiculous. The deepest drilling on earth, the Kola Superdeep Borehole, is $12.262\,\mathrm{km}$ deep, the earths diameter is about $13000\,\mathrm{km}$, that is the deepest hole on earth is one thousandth of earth's diameter! Furthermore, the inner of the earth is hot, your coils would melt, your magnets would not be magnetic (as they have ...


2

Ohm's law assumes the temperature remains constant. An Ohmic conductor is one in which the current flowing through it is proportional to the voltage applied across it. A non-ohmic conductor is one in which the voltage and current are not linear. A) The resistance of most conductors increases as the temperature increases, however being ohmic and not ohmic ...


2

Your analysis doesn't apply to what you are applying it to. $P=V^2/r$ is the power dissipated in the internal resistance of the battery. It is the power dissipated by the internal resistance if the battery is shorted. What you want is "the power supplied by the battery", which is the power dissipated in the external resistor: $P = V^2/R$, where care ...


2

Your misconception seems to be that a conductor connected to the ground must have zero charge. The actually effect of grounding a conductor is to put it at zero potential, not charge. The ground is assumed to have an effectively unlimited supply of charges/capacity to accept charges in order to make this condition true. After step 2 and before step 3, the ...


2

What is your goal - to stop bullets or to generate electricity? The key to stopping bullets is not so much to absorb the energy (although that matters too) but to absorb the momentum. You may know that $$p = F\Delta t$$ In other words, given a certain momentum $mv=p$, you need to apply a force F for a time $\Delta t$ in order to slow it down. The ...


2

You say the lamp is plugged into a AC outlet, but then talk of a "wall switch". Apparently you mean that this switch controls the power to the outlet, and that a switch on the lamp is kept on, or that the lamp has no switch. If so, you should clarify this as a switched AC outlet, since most aren't. In the case of a switched AC outlet, the switch will be ...


2

Depending on the location of the switch, the answer will change. A properly wired lamp would have no signal on the live (phase) wire, and therefore there would be no field. However, if you interrupt the neutral wire (or the switch is in the lamp, not the wall) then you will have a varying AC field because the voltage on the wire changes (and thus a small ...


2

Your first two equations are effectively the same once you consider that $F = q E$. You should use the second one if you're concerned about the electric field in general, but the first one if you want to know what actually happens to a particle (i.e., what force a particle feels). When in doubt, you can usually do alright by working out first the field, and ...



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