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You are absolutely correct, the electric field does fall off with distance from the battery. However, this is only true during the transient state (the state of the field when the battery is first connected). In fact not only are the magnitudes inconsistent, but so is the direction of the field. The field doesn't always point in the direction of the wire. ...

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The electric field of a collection of static or stationary electric charge(s) would indeed diminish as the inverse square of the distance from the charge(s) outside of the conductor, but not inside a conductor in which an electric current is flowing. Examples of stationary electric charges would be a metal sphere atop a Van de Graff generator, or charges ...

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Quoting from the article http://www.edn.com/design/consumer/4423155/Loudspeaker-operation--The-superiority-of-current-drive-over-voltage-drive of Esa Merilainen "The most remarkable thing here regarding loudspeakers is that the voltage between the ends of the wire does not appear anywhere in these equations. That is, the speaker driver in the end obeys only ...

1

Well, if you look at a practical circuit which produces that graph, there must be a sudden drastic change in the circuit at that instant of time to cause the capacitor to abruptly shift mode from charging to discharging - quite possibly a switch/switches were turned on/off effectively putting the capacitor in a different circuit. If you are asking the ...

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In figure (a), the voltage is continuous but the time derivative is not; the capacitor current would discontinuously change sign from positive to negative. In figure (b) however, the voltage is discontinuous. It is typically said that the voltage across an ideal capacitor is continuous since, for the current to exist, the time derivative of the voltage ...

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If you have just given the voltage signal with $$\def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V$$ then the current at $t=2\rm s$ is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ... 0 There never is any confusion or contradiction because, unlike charge, the current does not have to be continuous, that is i(t-0) = Cv'(t-0) and i(t+0)=Cv'(t+0). Since time's flow is unidirectional, from  -  to  + , the time derivative of the voltage is well defined to mean f'(t-0) and f'(t+0). 0 (As far as I know, 300kV is the exclusive domain of HVDC so I will not mention AC here.) Ionization potential is the energy difference between a free electron at rest outside an atom, and a particular electron energy state inside the atom. If you zap a bound electron with enough energy, it can assume a new, unbound wavefunction and travel away. But, ... 0 The electron energy in the line is 300kV * the electron charge q. Whether or not electrons are emitted depends on the gradient of the voltage, i.e the electric field, at the surface of the wire, which determines the force on the electron. See the wikipedia article on field electron emission. A well-designed high-voltage line keeps the peak electric ... 0 It depends. Is the electric current flowing briefly in a circuit that contains a large capacitor? None of the other answers seem to have contemplated this possibility. If it is a steady current through a resistor, definitely the answer is no. 0 The speed of sound is only about 0.1% - 0.6% faster in dry air than in humid air at the same pressure and temperature, but humid air is also less dense than dry air, so my best guess is that the vibration from the wires couples less strongly to air that has moisture in it than it does to denser dry air. 0 A single resistor wired in parallel with voltage source like a battery may give confusing reasons for a novice, so a few other things need to be taken into account. Any battery also has an internal resistance (it is like a resistor in series with the voltage source), and this resistance also varies with the state of charge of the battery, which is to say, ... 3 Jerk_dadt is correct. Electric current is the flow of free electrons in the conductor. At any instant, the number of electrons leaving the wire is always equal to the number of electrons flowing from the battery into it. Hence, the net charge on the wire is zero. If you say the current carrying conductor is charged, it will violate the Kirchoff junction ... 2 No, not necessarily. Current is the just the movement of electrons already in the wire (that is neutral). The electric field (or voltage) applied causes them to have a net movement in one particular direction (i.e. opposite direction of conventional current). So copper has 29 protons and 29 electrons per atom. A copper wire would have a net zero charge. It ... 2 \def\vE{{\vec{E}}} \def\vS{{\vec{S}}} \def\vA{{\vec{A}}} \def\rot{\operatorname{rot}} \def\grad{\operatorname{grad}} \def\div{\operatorname{div}} \def\ph{{\varphi}} \def\vn{{\vec{n}}} \def\vr{{\vec{r}}} The charge floating through the wire causes a current density \vec{S} in the wire (not only on the surface). This causes an electrical ... 2 Yes, alternating current will radiate electromagnetic waves. For example, in telecommunication, the transmitter itself generates a radio frequency alternating current, which is applied to the antenna. When excited by this alternating current, the antenna radiates radio waves. 1 An inductor stores energy in a magnetic field. After current has been flowing in the inductor for a period of time, it has built up a magnetic field around the wire making up the inductor. In that state the inductor offers no opposition to current flow. If it were then disconnected from it's energy source (battery perhaps) then the the magnetic field will ... 1 is it possible (with current technology, not "is it theoretically possible?") is it possible (with current technology, not "is it theoretically possible?") Well we will start off with "is it theoretically possible?" since if it is not even theoretically possible any further discussion is a waste of time. The maximum possible amount of work that can be ... 1 It depends on how bright the LED needs to be. Using body heat is a bad idea as the temperature gradient is small (effectively zero if it's covered by clothing). You could use vibrations to power your device for example. However I doubt it'd meet the power requirement. 0 1) With a constant and DC power source eventually the solenoid will become fully 'charged'. At that point its 'resistance' term vanishes because it no longer produces an emf against the battery. At this point, the \frac{di}{dt} term will be zero, because the current isn't changing. 2) When you cut power, the magnetic flux is no longer maintained by the ... 0 Inside the solenoid there's no (component of the) magnetic field that isn't perpendicular to the loops, so treating a solenoid as if it has closed loops is valid approximation. Perhaps you can understand it better if you approach the solenoid as a single wire and see what the emf does from that point of view. 0 I suspect this is an approximation that works well. Each piece of the coils is almost exactly aligned with the induced electric field \vec{E} at that position. (To be more accurate, some are exactly aligned, but most aren't. Try imagining a coil wrapped around a cylinder.) In this way, the EMF \oint \vec{E}\cdot d\vec{l} for the closed loop is nearly ... 0 Do you mean a coil in which conductors are made of ferromagnetic material instead? If that's the case, you would probably not be better off: the field which is created is concentrated around the conductor (mostly at core of the solenoid) and in another direction, so it would not benefit from the increased magnetic permeability (which is what ferromagnetic ... 1 I think you have some misconceptions about voltage. You mention "net voltage" but voltage is always a difference in electric potential. In a circuit, that means you never talk about a "net voltage" or a voltage at a certain point. Voltage is always meant to be read as "the voltage between two points" or "the voltage at A with respect to B." It is never ... 1 The definition of the word 'electricity' is broad. I like Wikipedia's disposition, which was clearly written by someone who was science-minded, and worded carefully to avoid creating more confusion: Electricity is the set of physical phenomena associated with the presence and flow of electric charge. Without subdividing the concept, we can't answer ... 0 The net voltage around the entire circuit is zero. However, the voltage across the resistor is equal in magnitude and opposite in sign to the voltage inside the battery. You could view this conceptually as electrons flowing from the minus to the plus terminal (through the resistor) while the internal mechanism of the battery produces more electrons at one ... 3 You are right in that a magnetic field is build up, which generates a electric field opposing the given potential. But the consequence is not an oscillation of current, but only a damping of the increase of the current. Therefore, if you have a Heaviside step function for the voltage, it'll result in an "exponential" increase of your current (I(t) = I_0 ... 0 Electricity does have mass, yes. Indeed, one of Einstein's 1905 papers, "On the Electrodynamics of Moving Bodies" specifically demonstrates this. A moving magnet becomes more massive due to its increase of energy, and this additional inertia causes its electric field to increase in strength as well. Hence E = mc^2. If you wished, with sufficiently accurate ... 0 It is showing the direction of current flow reversed through the meter with the reversal of the leads. -1 Any ordinary LED (a light emitting device) would probably do this, by sourcing an opposing voltage when exposed to light (and the effect / voltage would also be bigger if the source of the light contained the wavelength that would normally be emitted by the device). 0 A Short answer is here, Firstly the capacitor gets charged.Though it takes infinite time to reach the battery potential the current is reduced to considerably low values. So obviously bulb won't glow. Note:As capacitor gains more and more charge its potential increases finally reaching the value same as the battery.So potential across bulb tends to zero. ... 1 in welding a plasma is created, it's a mix of ions, electrons and atoms. alltogether they are a neutral mix. once you get plasma you get a ton of UV coming out of it, very dangerous to eyes not only on the direct contact, but also via reflection from other objects. in your case, I still don't know what exactly is the device you are creating. It sounds like ... 2 All materials emit thermal radiation (such as light). The hotter the material, the more the radiation is shifted to high frequencies (shorter wavelengths). The radiation comes from oscillating electrons (regardless of whether there is an electric current). Welding reaches temperatures high enough to cause significant emission of UV light. Oxyacetylene and ... 0 In order to induce an EMF in a current loop, there must be a changing magnetic field. HOWEVER You are correct in surmising that there would be a force on a moving charge in the presence of a magnetic field. This is known as the HALL EFFECT. If the ring were replaced by a current carrying wire running the length of the path of the ring, charges would most ... 2 A general strategy for these questions is to start at the battery and trace the current through the circuit. So, starting from the batter, we can see that the entire current passes through R_1. After that, the current hits a split (at the top of the circuit in your drawing), where part of it goes to the left through R_2 and the other part of it goes to ... 2 The parallel of R2 and R3 is in series with R1. The current that flows through R1 gets divided between R2 and R3. 1 The force on each electron is qv\times B, which is toward the left. This clearly doesn't induce them to flow in a circle. (Remember that if electrons flow to the left, current flows to the right.) The entire ring is not deflected to the right because there is an equal force on the nuclei toward the right. 0 From a microscopic point of view you can image metal (conductors) in a lot of different ways. The easiest model is the Drude model in which atoms are fixed in the space and everyone have one or two (in a metal) free electrons. When you apply an external electric field this particles move as a consequence of Coulomb force. It's important to say that electric ... 1 Imagine two coils, coil 1 and coil 2 with (self) inductance L_1 and L_2. If the two coils are not coupled, we have:v_1 = L_1 \frac{di_1}{dt}v_2 = L_2 \frac{di_2}{dt}$$Now, if the coils are coupled, we have:$$v_1 = L_1 \frac{di_1}{dt} + M\frac{di_2}{dt}v_2 = M\frac{di_1}{dt} + L_2 \frac{di_2}{dt}$$where M, the mutual inductance is ... 0 When a coil connected to an AC generator creates an EMF in another nearby coil (mutual inductance), is self inductance simultaneously occurring in both coils? As pointed by Satwik Pasani and others, yes there will be self inductance occurring in both the coils. As you are using AC generator (word "generator" is misnomer: it generates nothing, it ... 0 Yes, sometimes it is negligible, like in the case of two coupled loop. But it would be important for two coils where each has large number of turns. 0 In the simplest case, and some unmentioned assumptions, the minimum charging time would be given by the formula already mentioned:$$ T = \frac{V \times Ah}{W} this gives 12x150/25 = 72 hours. However, this case makes at least two major assumptions, the battery is fully discharged and there are no losses of any kind! In the real world, the battery ...

0

Imagine identical charges moving through a conductor. Said conductor has a cross-sectional area $A$. The volume of an element of length $\Delta x$ of the conductor is simply $A \Delta x$. If $n$ represents the number of mobile charge carriers per unit volume, the mobile charge $\Delta Q$ is given $\Delta Q = (nA \Delta x) q$, or the number of carriers times ...

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Suppose you consider the popular analogy of a water pipe circuit with a pump supplying power and a turbine or something similar for the water to do work on. The water flows in a loop, so the net work should be zero as per your argument, but nevertheless work is done - the turbine turns. The solution is that the water does work on the turbine but has work ...

1

The electrical energy you pay for doesn't just go into the appliances; some amount of it ends up as heat in the electrical path from the meter to the loads. Also, heat from a load such as a heating element can travel along the electrical conductors away from the application (the water tank) - remember: most good electrical conductors are also good thermal ...

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Yes, it's true that hard water reduces energy efficiency of boiler. Explanation: The scale buildup has an insulating effect that reduces the element’s ability to heat the surrounding water. The boiler’s thermostat continues to call for heat, so the element heats longer and more often. This causes wasted energy and high electricity bills and ultimately ...

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It should be associated with the work function, which is the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface, and different materials have different work functions. Consider a very simple case, that a spherical electrical object exists in vacuum. Considering ...

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Many times this kind of problems becomes very simple just by redrawing the circuit in a more standard way. Have a look at this: Now I think you will have no problem to solve it!

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