# Tag Info

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I think the main misunderstanding is that the n-layer is neutrally charged overall so it does not attract anything. Only in the depletion region is there an electric field. The electric fields acts to sweep electrons towards the n-layer and holes toward the p-layer. The depletion region partially overlaps both n and p doped layers. The part in the n-side is ...

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This might not answer your question because I don't know what is the "same result" you mentioned. I will use two interpretations of what your "electricity" could mean. A Faraday cage blocks electromagnetic(EM) waves (One kind of "electricity") in both directions. In the case of waves, the surface of the cage is roughly acting ...

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What happens is that the glass rod will regain its lost electrons due to leakage from earth. Conversely the silk will lose its gained electrons due to leakage to earth. The process is slow because these are slow moving charges.

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First of all, saying "... produce heat..." is just not correct. Heat is one of the two forms of energy that can be exchanged between two systems, not produced. If you have an electric heater (with electric I mean it works only if you plug it in), and you consider as system the heater itself, it exchanges electric work and heat with the ...

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Pretty much all the energy consumed by any electric appliance is converted to heat, light or sound... and most of the light and sound ends up as heat before it can escape your home. (Heat pumps don't count. They cheat.) Both of your heaters are therefore equally efficient heat producers, as is your refrigerator, your computer, your TV, etc. The difference ...

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A ceramic heater works by convective heat transfer, i.e., the energy consumed directly goes towards increasing the kinetic energy of air molecules, which are then blown into the surrounding space. The warmth felt due to such a heater is due to molecules in that warm air colliding with your skin, causing random motion of molecules in your skin due to the ...

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In a generator, the free electrons in a rotating coil of wire are pushed along the wire as they move through the field of an external magnet (usually an electromagnet). In an alternator, the electrons are pushed around the loops of a stationary coil by the emf produced by the changing field from a rotating magnet. The alternator (which supplies an AC ...

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Interesting thought aside from loses on reflection as Eric pointed out. But there are a couple other problems, too. Suppose you have a lot of light bouncing around inside. Suppose you open a door so some can get out. How do you allow only a little out at a time at a controlled rate? One problem with batteries is they concentrate a lot of energy in a small ...

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A few things you need to consider: (1) You won't be able to put an arbitrary amount of light into the box -- at some point the pressure of the light inside will be greater than the pressure of the incoming light, and then no more will get in. (2) A far bigger issue: in the real world no reflecting surface is 100% efficient, and the light is going to be ...

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This does not cause any problems with the conservation of energy. The conservation of energy in EM holds locally and is described by Poynting’s theorem. As the theorem says, if the matter gains energy then the fields lose it and vice versa.

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It is because of the general superposition principle in electric circuits with only linear elements (such as resistors): if you have a source (ideal battery) $1$ creating current $I_1$ in some wire and a source (ideal battery) $2$ creating current $I_2$ in the same wire when they are plugged separately, then if you plug them together, they will create ...

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It's not a contradiction! You're just considering two different cases, since it makes a huge different wether you have fixed current or fixed voltage. Now the thing is that if you have fixed voltage, you'd have to reinterpret the current intensity in the second answer as $V=I\cdot R \Rightarrow I=\frac{V}{R}$ and so, as you decreased $R$ by a factor two (by ...

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You just answered your question yourself: if the supply is voltage-limited, the first answer is correct. Under the assumption of constant voltage, what the expression $P = I^2 R$ really means is $$P = I(R)^2 R.$$ That is, current is a function of the resistance. Since $I(R) = V/R$ by Ohm's law, you get back $P = V^2/R$ by substitution. Of course, the ...

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What is meant with that picture is simply that the electric field lines $E_1$ and $E_2$ that the individual positive charged particles $P_1$ and $P_2$ create (representing each electric fieldline from each charged particle of the charged surface) cancel each other out tangentially. This results in an overall electric field $E$ that is orthogonal to the ...

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That's a terrible illustration. Those lines are neither field lines nor vectors. Those lines having arrows at the end don't represent anything. I will give the author the benefit of the doubt and suppose that he or she intended to display vectors anchored at the point of interest. The lines below the dotted line are perhaps to display the length and ...

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The arrows in the diagram are not real electric field lines, but rather the contribution to the electric field given by the infinitesimal line elements PE1 and PE2. These can certainly intersect. Electric field lines are something totally different from what is shown in the figure. For the infinite line coming out of the page it would look like: It is true ...

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They are showing electric field lines due to individual charge segments at different locations along the wire, not the net electric field due to the whole wire.

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The beta decay is more precisely written as $$n \to p^{+} + e^{-} + \bar\nu_e$$ which means a neutron from the nucleus decays into a proton, an electron, and an electron anti-neutrino. In this equation the electric charge is conserved, as well as the lepton number. Hence, you definitely have to change your equation. Usually, $^{22}Ne$ is considered to be ...

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For simplicity assume that magnetic field is uniform in a downward direction and increasing in magnitude at a constant rate. I will explain in a minute why I have made this assumption from the information that you have provided. Without saying anything about the nature of the loop other than it is of constant area $A$, the rate of change of magnetic flux is $... 0 Electrons can flow around superconducting wires forever. They use these in MRI machines as magnets. No energy is generated. 0 Let me try again. The question says: "A square loop, (cut out of a thick sheet of material), with each side of resistance R", is dropped from within a uniform magnetic field. Then the changing flux, (or the motion of the top through the field) generates an electromotive force, ε, which causes a current flow: ε = I(4R). In this case, the loop is ... 0 Insulators can be charged and can exchange charge (electrons) when they come into contact with another insulator. Keep in mind that a conductor is just better at conducting a flow of electricity whereas an insulator is a very bad conductor of electricity. In this case, we are however talking about static electricity, that is the buildup of excess charge on ... 0 To get a spark in air (under standard conditions) you need a potential of approximately 3kV per millimeter of distance. If you see a spark just before your finger touches a doorknob, you can use that to roughly estimate what the voltage difference was. Electric charge flows (very briefly) between your finger and the doorknob, until the potential difference ... 0 The total flux out of the cylinder must be zero. You can use Gauss's law (reduced with solid angles) to find the flux in and out through spherical caps at each end of the cylinder. The difference must go out through the sides. 1 For an Ohmic resistor, the second equation applies. Similarly for a parallel plate capacitor.$V$and$I$or$V$and$Q$will have a set relationship; that cannot be varied independently for the given circuit element. However, your first equations apply more generally for any system. i.e. if you take something (doesn't need to be Ohmic) and apply a voltage ... 0 Ohm's law states that if the external resistance goes up, the current goes down. The voltage is constant if you ignore the voltage source's internal resistance. If you account for the small internal resistance the voltage of the source goes a little up. 1 Temperature increases and energy flows to the neighborhood. While the power dissipated in the resistance is greater than this heat flow rate, the temperature keeps rising. Once an equilibium is reached, the temperature remains constant. It is the case of an incandescent lamp. The resistive electric power equals the energy per unit of time emitted as EM waves.... 1 The flux$\Phi$in Gauss' law is the net flux through the Gaussian surface. It is perfectly possible for the field to be$0$at one point but the net flux over the surface be non-zero. In order to infer$\vec E$from Gauss's law, it is essential that$\vec E\cdot d\vec S$be constant on the surface. This is clearly not the case here, i.e. the$\vec E$is ... 1 Gauss's law says that $$\int_\mathcal{S}\mathbf{E}\cdot d\mathbf{S}=\frac{Q_\text{enc}}{\epsilon_0}$$ which says that flux through the closed surface equals$1/\epsilon_0$times of$Q_\text{enc}$. Now it is possible that the electric field at any point on the surface is zero. That doesn't make the electric flux to be zero as it's sum through all the points ... 2 The resistance of the human skin is given in the literature as between ~1 kOhms (wet skin) and ~1 MOhms (dry skin). I couldn't quickly find values for the body interior, but given that the body inside is "wet", it cannot be higher than 1 kOhms. In the scenario where the electrician has dry skin and closes circuit by putting his hands on either line ... 2 Based on the image below, the only thing your wrist wire accomplishes is putting the skin of the wrist in contact with the wrist wire of the left hand at the same potential as the skin of the wrist in contact with wrist wire of the right hand. But it does not necessarily put the skin of each hand touching the hazardous wires at the same potential. All it ... 0 The vault measure how much we can allocate of potential energy to the charges in mouvement inside a potential differencial ,why? because more good is the differential more potential energy is disponible to the movement of charges , when a lot of charges, say électrons negatively charged, accumulate in a certain point of space they are modifying the ... 2 “Voltage” and “concentration of charge” are two names for the same thing — or at least, two ways of talking about the same phenomenon. Physicists love to talk about energy (it’s even in your quote already). When we say that there is an “electrical potential difference” between two points, what we mean is that an electric charge would have to gain or lose ... 1 He may be referring to the fact that information can be stored as accumulation of charge, e.g. on a capacitor's plates. In this case, a digital '1' is represented by a positive voltage across the capacitor, which gets positively charged by injecting current on its "top" plate (generally speaking, the one where the voltage is read), i.e. by removing ... 2 Yes, resistance is$R=V/I$, even if$V/I$is not constant. You would just have something like$R=R(I)=V(I)/I$, where$R(I)$and$V(I)$are functions of$I$(or you could do a similar expression with functions of$V$). To be nitpicky,$V=IR$is not Ohm's law. Ohm's law is that$R=V/I$is constant. So by using$R=V/I$you are just using the definition of ... 1 Well semiconductors behave quite different than conductors, the I-V characteristics aren't linear like conductors and in fact, they even differ when they are forward biased and reverse biased. So to avoid all these, instead of resistance, we define dynamic Resistance which is given as $$R_{dynamic}=\frac{\Delta V}{\Delta I}$$ -1 Yes,$W$and$VA$are the same. Similarly, you could measure pressure in$Pa$or$Nm^{-2}$.$Hz$and$Bq$are both$s^{-1}$but one is usual for frequency and the other for the rate of radioactive decays. The usage is just conventional. If you wanted to totally pure and consistent then don't use any derived units, just base ones. So,$kg m^2 s^{-3}$... 3 Complex Power is composed of a Real component (dissipated power in Watts) and an Imaginary component (Reactive stored energy in Volt-Amperes Reactive (VARs)). The total is combined as a vector quantity (Volt-Amperes = VA). A common quantity to be concerned with is Power Factor (PF). PF is the cosine of the relative phase angle between the AC Voltage ... 1 Residential electricity meters measure real power in watts not reactive power in Volt-Amps. The former is the average value of the instantaneous voltage times current. The latter is the rms voltage times the rms current. If the sinusoidal voltage and current are exactly in phase then the two measures are identical. If the sinusoidal voltage and current ... 2 You pay the electricity company for both the "pure" (resistive, real) watt components AND the reactive (capacitive or inductive, imaginary) watt components in the power that enters your building through the meter box. This means it is very important to trim or null the inductive component out by adding the right amount of capacitance, or vice versa,... 0 This is an important point which is commonly overlooked. The crux is that the voltage drop on each capacitor is not necessarily the batteries voltage. Let's work out the ideas step by step. Firstly, to find the charge on capacitor system let us take equivalent capacitance. When we replace the two with an equivalent one, all voltage will drop on that, hence ... 0 The motion of a single charged particle all by itself does not constitute a current. The notion of a current implies that there is a flow of charge bearing particles, whether this be electrons or protons. Notably, the notion of a current arose phenomenologically well before the atomic notion of electron was postulated and discovered. 2 The most "authoritative" answer on this is to look at Maxwell's equations, because they are at the heart of EM theory. And the answer there is yes, it does count as "current" for their purposes, in that it contributes to the$\mathbf{J}$term. The$\mathbf{J}$part is defined as $$\mathbf{J} = \rho_q \cdot \mathbf{v}$$ where$\rho_q$is ... 0 For a particle with charge$q,$velocity$v$, and position$x'$you can define the current: $$I(x)={q}{v}\, \delta( x-x')$$ where$\delta$is the Dirac-delta. Note that this is not a stationary current, so we can't apply formulas from magnetostatics. 0 I've been trying to understand this as well. This is what I've been able to figure out so far. I welcome comments and corrections. A charged particle in motion is electrically equivalent to an electric current even if you can't calculate the current in the conventional sense of measuring Coulombs per second. We can say this because a current made up of many ... 3 A) ....Since C1 and C2 are two different capacitors, why gain or loss of same amount of charge would cause them to have the same magnitude of potential? They have the same amount of charge but the magnitudes of the voltages (potential differences) of the capacitors cannot be the same if the capacitances are not the same. The relationship between voltage, ... 1 That depends a lot on what type of bulb we are talking about (LED, incandescent, halogen, fluorescent, etc). For simplicity we (erroneously) assume that the bulb behaves like a linear resistor. But because one of the bulbs has fused, would the new resistance depend on the other bulbs? The new load resistance of all bulbs combined would be higher. Would ... 1 The new equivalent resistance would be change, not the individual resistance. Suppose there bulb in parallel with resistance$R_1$,$R_2$and$R_3$. $$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$$ If one of them, says the third one, get fused, the equivalent resistance would change to $$\frac{1}{R'_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$$ So the ... 1 The sign we label on both plates of a capacitor represents the sign of charge accumulated on that plate, which is not an indicator of the sign of electric potential. In fact, the electric potential can take any values (the potential across the battery can be from$V_0$to$0$, or from$0$to$-V_0$, or even from$100V_0$to$99V_0\$ if you like) as long as ...

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