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1

The filament will be a reasonable approximation to a black body emitter, so it's spectrum will be given by Planck's law: $$ B = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{k\lambda T}} - 1} $$ So just measure the radiance of the light from the filament for a range of wavelengths and do a fit to Planck's law by varying $T$. This will give you an excellent ...


2

You really are asking two questions. First - how do we calculate the temperature: At the typical temperatures of a halogen bulb, the large majority of heat loss is due to thermal radiation (although there is some conductive loss in a halogen bulb as the bulb is not evacuated). Because of this, the most important factor is the "apparent size" of the ...


0

Something like that. In metal shel, charge distribute in the surface. So in this case, when you conect the center sphere with the wire, all the charges goes to the surface of the outer sphere. Then, the charge in the surface of the outer sphere is 2Q-Q=Q.


0

actually, the fire does not conduct electricity, it simple ionises the air around it and the free electrons create a conducting path across the gap


2

Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss.


1

theoretically if its components never wore out then yes. however in practice things do wear out eventually and so no it could not be done in the same way that a perpetual motion machine can work in theory but not in practice.


0

As already mentioned atmospheric humidity can kill experiments on electrostatics. Plus this "defective" hair means, I guess, if the hair isn't clean, ie a bit greasy. In particular long, fine, dry and clean hair will respond electrostatically wonderfully under (almost) any lab conditions. Short, thick, curly hair may respond when clean but not nearly so ...


1

First of all, I'll write the energy with the letter $U$ to not confuse with the electric field $E$. By definition of power, $P=\frac{dU}{dt}$. Now, where is this energy in the equation coming from? $$U=\int \vec{F}\cdot \vec{dl} = q \int \vec{E}\cdot \vec{dl} = qV$$ This is the amount of energy gained by the charge when moving across an electric potential ...


1

First equation: power is defined as energy per unit time. Second equation: if a current flows through a circuit, the power dissipated is the product of voltage and current. This is because the voltage describes the energy each electron is given to traverse the circuit, and the current describes the number of electrons that travel the circuit per unit time. ...


1

Everyone is likely to have a similar pain threshold to electric shocks. If your friend wears leather soled shoes he will conduct static electricity to earth because they offer much less resistance to earth than rubber or plastic soled shoes. Wearing synthetic clothes may cause a similar effect. If he/she has lower resistance than you (by having sweaty ...


2

As you walk across the floor, the soles of your shoes attract electrons from the carpet, and those electrons build up all over your body. The excess electricity wants to find a way off of your body, and when the doorknob provides a path of least resistance, the air between your hand and the doorknob expands, turns into a plasma, heats up, and pops as the ...


2

Since the lamps are in series, the electric current through each is identical; all of the current out of one lamp is in to the other lamp; if there is a flow through one, there is a flow through the other. It cannot be that there is a flow through one and not the other (in the context of this simple model). Thus, if the lamps are identical, their ...


2

The current is the conventional current in the opposite direction to the electrons current but they are the same thing , the current is the same in series connections. If the two lamps are identical they will give the same amount of light.


42

When discharging without a tool, the whole charge exits your body through a small skin surface area, say $0.1 \,mm^2$. When you hold a tool that surface is much bigger; perhaps $100 \,cm^2=10,000 \,mm^2$. That means that the current flowing through neurons in that area is much lower, and perhaps low enough as to not be felt. Pretty much the equivalent of ...


-1

Stainless steel is a relatively good conductor of electricity, as are all metals. The body is also a good conductor, due to water. The skin is a relatively poor conductor of electricity because of dead skin cells. Any break in the skin, i.e. cut, greatly reduces its resistance.


0

In this case,if they are connected one at a time to the battery then, your reasoning is correct. But if both are connected at same time in series, then current will be same and electric field will also be same.


2

Is there a fallacy in this statement? At least two. First, unless one is referring to a perfect (ideal, etc.) conductor, only in the electrostatic case does the electric field inside a conductor vanish. Second, in the case of an ideal conductor, there can be a steady current through without an electric field inside. Recall that an electric field ...


0

You could answer this question easily by a simple "back of the enveloppe" calculation. Assume you have a plate of metal, say iron. Assume you displace all the free electrons from a 1-atom-thickness on one side to the other side. Calculate the charge densities on the surfaces. Calculate the electric field inside the plate. I expect you to find an enormous ...


0

The electric force is the negative derivative of the potential. The negative derivative of 2y+3 is -2. Which means that a constant force would accelerate a positive charge in the y's decreasing direction. Because in the example an electron is given which is a negative charge it will accelerate to the direction where y is increasing. The answer requires more ...


1

The conductivity of plasmas is very high, though not infinite. In a metal wire the applied voltage accelerates the conduction electrons, but the electrons collide with and scatter off the atoms that make up the crystal lattice of the metal. This transfers energy from the electrons to the metal and the metal heats up as a result. The energy lost as heat is ...


1

The rating 2600mAh (or 2.6Ah) means the battery will produce 2600mA for one hour, or 1000mA for 2.6 hours of indeed 1mA for 2600 hours. The rating is the current multiplied by the time the battery can produce that current. In practice the rating depends on the current. The figure of 2600mAh will have been obtained for whatever current is optimal for that ...


0

You need to know what current your tracking device consumes to find out how long your battery will last. You also need some extra information to find out how long it would take to charge your battery. So the information you provide is not enough to answer these questions. It is also possible that your battery is indeed faulty.


1

Static comes from the same root as stasis, meaning stop, immovable, To create static electricity, you have to rub two different materials. At the moment you rub them, the electrons already moved Note the word "create", creation is not static, and yes there are transient fields and currents during creation of a static field. The static describes the ...


3

If you are seeing 4.5 volts across three batteries and 3.3 volts across two others, then the join you think you made is not a join. There is a common thing in electronics called a "cold solder joint". It usually happens when your solder is not quite hot enough when it touches the metallic surface - just the kind of thing that happens when you solder to a ...


1

Your calculations are fine, and the fact that they come out close to what you find on the market (good for you for checking) supports that. The difference in charge/volume ratio may come from the fact that the quoted energy density is for just a cell, while the commercial batteries have cases, charging electronics, etc. increasing the volume and decreasing ...


1

I didn't really understand the "W.h/L" unit, I expected it to mean W.h-1.L-1 Think about the meaning of units. You are trying to find the volumetric energy density, that is you want $$ \frac{\text{energy contained}}{\text{volume occupied}} \,.$$ Now Watts ($\mathrm{W}$) are a measure of power which is $\text{energy}/\text{time}$, so $\mathrm{W \cdot ...


-1


1

In response to your comment, there is a way to find the voltage without finding equivalent resistance. You can write the potentials at each junction point. Taking the points on the lower line to be zero potential, the potential gain on going up the last branch is $1$V. Then going left, add another $1$V. Then, current through the last branch is 1A and the ...


0

Not immediately, it takes a while. Even with just a simple bit of wire, the electrons flowing in are not the electrons flowing out on the other end. You have to understand there are a lot of metal atoms in the wire itself. When you push electrons into the wire, this displaces the electrons which are already there. This is just the same as with a filled water ...


0

The straightforward answer to your question is no, they are not the same electrons. To understand why requires us to give some deeper consideration to what is going on chemically within the two cells. I am going to give my explanation using the simplest chemistry I can imagine, which is a Zinc-Copper battery. The Zinc-Copper cell is a rod of zinc and a ...


3

How can electrons travel in these beams if they repel? First of all, the picture you posted looks like lightning which is basically arcing, i.e., ionization of gas to create a conductive path. This is not what I would typically consider an "electron beam". To answer you question: Creating and maintaining the integrity of an electron beam is not easy. ...


0

If you take a simple circuit of battery-switch-inductor, and close the switch, at the moment of closing the inductor sees a voltage but no current, because the current has to build up over time, as energy is put into the magnetic field. If instead of a battery you have an AC source and no switch, voltage and current across the inductor are out of phase, so ...


0

Is it possible for an inductor to have a voltage across it without any current passing through it ? Yes, if electromotive force due to core counteracts the applied voltage by the rest of the network. For harmonic driving voltage the current oscillates harmonically so this occurs only at one instant.


1

In computer science, there's a semi-joking saying that $\log n < 50$ for all $n$. Of course this isn't true – as you say the logarithm is unbounded. But what is true is that it grows so slowly that, if you can only put in quantities like memory size, mass of some material, time etc., then the logarithm is “effectively bounded”, because ...


1

Presumably it is the linear temperature coefficient of resistivity (or of resistance as they come down to the same thing).


4

As others have mentioned, for all intents and purposes, yes it reaches %99 charge after 5 tau. However, as the current gets smaller and smaller as we reach full charge, technically it will never become 'fully' charged, even in practice. The current will continue to get smaller and smaller, until it is unmeasurable and therefore negligible.


3

To fully charge a capacitor to 5 Volts, say, you could connect it to a 10 Volts source until it is half charged, then connect it to your 5 V source. This is of courcse a ridiculous method, since you could hardly hit the moment of correct charge so precisely; any micorvolt error would start an exponential curve as in your original setup. That being said, ...


14

This gives me a feeling that a capacitor never gets charged fully. Am I right? Why not? In the context of ideal circuit theory, it is true that the current through the capacitor asymptotically approaches zero and thus, the capacitor asymptotically approaches full charge. But this is of no practical interest since this is just an elementary ...


4

A battery in a circuit normally sources electrons at the negative terminal and sinks electrons at the positive terminal. The chemical reaction at the positive terminal consumes them. The total number of electrons in the battery does not change. Similarly, a diode accepts electrons at the negative terminal and sources electrons at the positive terminal, ...


0

You are roughly correct. However, you must be careful because the surface you would choose for finding the magnetic field from $\mathbf{J} \cdot \mathrm{d}\mathbf{S}$ is NOT the same surface you would use to find the electric field. The concept you want to solve this problem is self-inductance. Defining the magnetic flux $\Phi=\int\mathbf{B} \cdot ...


0

The relationship between electric field and electric potential is just that the electric field is (minus) the gradient of the potential. Thus in the case of a uniform field extending from a uniformly charged plate (let's call it along the z-axis, with the late in the x,y plane) $$ E_z = - \frac{dV}{dz}$$. This of course means that $$ V(z) = - \int E_z\ dz = ...


-1

Hint: The 10 micro farad capacitor can be removed ignored since it gets stuck in a wheatstone like network.


0

Take the simplest case of a uniform magnetic field and a circular wire perpendicular to the magnetic field. You know that a changing magnetic field induces an electric field. Because the magnetic field is uniform and the wire perpendicular, the electric field has the same magnitude on all points of the wire. You can find the total emf from the change in ...


0

The key to this question is the position of the voltmeter and the flux it contains. Consider the diagram below: Let us say we are finding the voltage between A and D using the volt meter $M_1$. We can do this using one of two loops: One going straight from A to D. The one going from A to B to C then to D. The first one will give an answer of ...


2

In brief: No. But to be safe you need a better appreciation than the wording of your question suggests. Numbers can be arbitrary Touching a point labelled 0V might kill you. Touching a point labelled 1000V might not - if no part of you is in contact with anything that isn't also also labelled 1000V, using a common reference point. From the perspective of ...


2

Electricity will only flow if there is a difference of voltage. If the earth, the object, and your body are all at the same voltage (lets call it zero volts) then there is no difference in voltage and hence no electrical current.


2

You are touching something that has OV and the Earth which again has 0V. No potential difference there.So why should you have a current through you?There is no electric field to drive charges through you.


1

Electric fields produces a difference of potential on two points with different distances of the field source. Magnetic fields induces current on a closed loop if the loop is not on parallel in relation of the lines of field and the magnitude of the field does have to change (you have to have a flux). If you have a magnetic field interfering on your ...


2

The paper strips were ironed to make them flat and easy to stick to the CRT screen. Old CRT color screens used (IIRC) around 25kV to accelerate the electrons to excite the phosphors on the screen itself. This resulted in the buildup of a static charge on the inside of the screen, and a corresponding charge on the outer, with the glass acting as a dielectric. ...


1

Incandescent light in filaments is a result of Joule heating: $$ \text{Heating Rate}\propto \text{Current}^2 \times\text{Resistance} $$ It is not the presence of electrons alone, but the rate at which they are being pushed through the material, and how resistive that material, that matters when it comes to generating heat (and therefore light). When you ...



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