# Tag Info

1

Agreeing with Bill N 's comment - but since comments tend to vanish on this site, I'm writing this as an answer. It is unusual for the letter $p$ to be used for charge, but common for it to be the dipole moment, $p = qd$. Making that change in the original problem statement, your analysis is correct: a dipole of dipole moment $p$ at an angle $\theta$ to ...

-3

Experiments demonstrate that an electric charge will not pass through a metal ball, but pass about the surface. Faraday suits has been worn by electricians while working on live wires. They touched the suit.

4

What happens with static electricity on a shirt is, if it comes into close proximity to another surface, the difference in electrical potential can become greater than the breakdown voltage of air. Then, current flows through a small electric arc, which is a plasma of air molecules and electrons. Very locally, this has a much higher temperature than the ...

0

The electric field is introduced more as a mathematical object than a physical object. The electromagnetic field is an entity that both stores and transports energy and momentum. Radio and light depend on the existence of this entity. When quantized, the electromagnetic field (four-potential) quanta are physical entities; photons. How the other ...

0

The other way of phrasing your question: if I have a surface of uniform charge density, and the field inside is zero everywhere, does it follow that the surface is a sphere? The answer is "yes". Imagine we have a non spherical surface. We know that it is possible to have zero field inside any conductor regardless of shape. But we also know that the charge ...

2

Witricity uses a rapidly changing magnetic field to induce a current in a receiver (coil). This is based on Faraday's law of induction: $$U = - \frac{d}{dt} \int \vec B \cdot d\vec A$$ From this you see that a changing magnetic field is necessary. The Earth's magnetic field, however, is largely static (compass!) and certainly not fast enough to induce a ...

-1

try to use LTM method to measure ohmic contact between metal and semiconductor?

0

It will work, but not for very long. At best, the solar panel is perhaps 10% efficient, so the system will lose 90% of its energy with each pass. This excludes resistive/heat etc. loses as well.

2

An incandescent light bulb is only about 5% efficient, that is 95% of the power supplied to the bulb is dissipated as heat and not light. Regular PV cells have an efficiency of around 20%, that is 80% of the light hitting them is dissipated as heat instead of producing electricity. Combine these two efficiencies and your lamp + photocells will lose 99% of ...

1

No, it does not imply that the surface is spherical and charged uniformly. Imagine a charged conducting shell of arbitrary shape. (An ellipsoid is a simple example.) Gauss' Law tells us that the charges in the conductor fly to the outside surface of the conductor, and the distribution of charges is such that the E-field inside is zero. But for a ...

0

The photon striking in the region of the n-p junction kicks an electron up to a high "valence band" and that electron then travels to an area of lower potential at the junction between the n-material and one of the sets of conductors. At the same time the hole left behind in the p-material drifts toward the conductor on the other side of the diode and ...

0

How do someone determine the direction of voltage arrows in a circuit? Is there a physical reason to choose a direction or the other? I always use just one simple rule in all talk and drawings about electricity and circuits: Arrows point in the direction a positive charge would move. For current arrows, this explains the direction from a positive to ...

1

(Moving this from my comment to an answer) Yes, the electric field simply penetrates the glass wall and charges (the electrons) placed in that field will feel a force and move. The glass does not really interact with the charges on either side, so you might as well remove it completely (theoretically).

2

An electric current is the flow of electric charge. But electric charge is not an entity, it is a property that must be 'carried' by a charge carrier. An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. However, an electric current is not necessarily an electron current. ...

0

Assuming the outer ring is stationary (an arbitrary assumption just to establish a range of reference), the inner magnet will either move up or down until the bottom of the inner magnet (or top if it moved down, but we'll assumed it moved up) is aligned with the top of the outer magnet. In an ideal experiment, the inner magnet would need to experience ...

0

For a thin wire, with the charges staying inside the wire, the instantaneous change in magnetic flux is equal to the line integral of the Lorentz Force around the instantaneous circuit (which is actually different than the work done as a particle moves about a wire through time). However, the magnetic part of the Lorentz Force is entirely due to the motion ...

1

When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...

0

With a stationary loop of wire and field $\vec B$ varying in time it doesn't seem that you'd be able to exploit the Lorentz force, because it's difficult to introduce the velocity $v$. But another configuration may be more helpful, see picture. A metallic rod (orange) crosses the magnetic lines (light-blue) with a constant velocity $v$ (blue) in the shown ...

0

A battery works by making charge want to move from one terminal to the other (we call this a potential difference, or voltage). The charge is also moving through the inside of the battery itself (the electrolyte). When you hook up a wire to an anode and cathode of two different batteries, the buildup of charge on the anode of the first battery quickly ...

2

A battery is no capacitor, and the actual charge stored in the battery terminals is very low. When you connect the anode of one battery to the cathode of another, that charge is transferred very quickly, and the voltage drops to zero. When you connect anode and cathode of the same battery, a chemical reaction takes place, and charges flow inside the battery ...

0

An electric charge is not necessarily a "physical" property in that it is its own physical entity, rather it is more of a characteristic of a particle. Net charges are created by having either more electrons or less electrons within the atom. This goes into the fundamental behavior of our universe that like charges repel and unlike charges attract. In ...

0

A transformer is essentially two coupled inductors and, thus, the equations are $$v_p = L_1 \frac{di_p}{dt} + M \frac{di_s}{dt}$$ $$v_s = M \frac{di_p}{dt} + L_2 \frac{di_s}{dt}$$ where $$M = k\sqrt{L_1L_2}$$ If the secondary is open, the secondary current is zero (constant) so $$\frac{di_s}{dt} = 0$$ Thus, $$v_p = L_1 \frac{di_p}{dt}$$ In phasor ...

0

I don't have the facilities at hand to draw a circuit diagram, so I'll just describe it. A conventional electromechanical current meter consists of these parts: M1 -- the meter movement itself (which we idealize as having zero internal resistance). R1 -- the innate resistance of the meter movement. This is packaged inside the meter, but we can consider ...

0

Let us assume that the instrument design is the same as in http://en.wikipedia.org/wiki/Ammeter (the words "spring constant / torsional constant", "flux, number of loops" seem to imply that). Then, to get the same deflection for lower current you need more loops, therefore, the resistance of the milliammeter will be higher. It looks like the same ...

0

Answer #1 uses an invalid argument. The conclusion that a milliammeter has lower resistance does not follow from the observed fact that the milliammeter has higher deflection. Consider that some work must be done to deflect the needle. Because of the assumption that the physical constants are the same, the work is equal for equal deflection. Also ...

2

One of the problems with doing experiments on static electricity is that the static charge can leak away while you're trying to do your measurements, and one of the main ways the charge leaks away is due to humidity of the air. I don't know what the humidity is in Kolkata, but I would guess that it's much higher during the monsoon that it is during the ...

1

The rest mass of an electron is 0.511 MeV. When an electron and a positron annihilate their mass turns to energy (two 0.511 MeV photons) so for each annihilation an energy of 1.022 MeV is released. One electron volt is $1.602 \times 10^{-19}$ joules, so in joules the energy released is $1.637 \times 10^{-13}$ J. You ask what happens if $2.3 \times 10^{28}$ ...

2

Resistive Filament - Voltage vs. Current If you referring to measuring the $V(I)$ dependence of the heated filament I would derive an approximate equation as follows. $T: \textrm{Temperature of the filament}$ $P_{dis} \equiv V \cdot I: \textrm{Power dissipated}$ $k: \textrm{Coefficient of temperature vs power}$ $R \equiv V/I : \textrm{Ohmic Resistance}$ ...

1

Would the human start emitting photons and die? EDIT: This answer is an answer to the original question regarding "a billion" positions. The question was subsequently edited to now read "2.3*10^28" positrons. That is not cool. The human would start emitting photons. This is exactly what happens during a PET (Positron Emission Tomography) scan at your ...

1

The apparent brightness of an incandescent bulb is a very strong function of the temperature of the filament, because it behaves approximately like a black body. Thus, much of the power emitted will be in the IR. The black body spectrum for different temperatures can be found, for example at wikipedia: Note this is a visual representation of Planck's law. ...

1

I'm sure the intention of this question was to test your understanding of charging by induction. The assumption of the question is that the wet filament would be a conductor and the dry filament an insulator. The positively charged rod attracts negatively charged electrons. Electrons will move through the wet filament onto the sphere. When the filament ...

0

What a superconductor in a magnetic field does is create (by Lenz's Rule) a current that creates a magnetic field which in turn pushes the external field outside of the volume of the superconductor. Since such a material has no resistance, you have no heat losses that would decrease the internal current. To answer your question, a superconductor does produce ...

1

For a capacitor with closer plate spacing (all else being equal), the electric field in the dielectric between the plates is stronger. Assuming a uniform electric field, the potential difference is given by the product of the spacing and the strength of the electric field: $$\Delta V = E\cdot d$$ So, the potential difference can be the same for both ...

-4

Carbon is the only gas, when solidified, that can conduct electricity. This is of the form Graphite, when it only has 3 bonds between the atoms. This leaves electrons to flow through the network structure.

0

As I understand your question, you have a set of 5 'good' sources in series. You also have 4 'good' sources and 1 source that starts out 'good' but turns into an infinite resistor, which are in series. You connect the two sets in parallel, and then measure the open-circuit voltage between the two ends. The short answer is that nothing would happen to the ...

3

Your answer : The bowlers and fielders rub the ball to make the ball smooth and shiny on one side and leave it rough on the other side. To generate reverse swing. There is a lot of science behind swinging the ball. The ball gets reverse swing when it is quite old and you see not only the bowlers but the fielders as well rubbing the ball before every ...

0

As you say: "If there is any stray charge outside of a certain conductor, couldn't there be some on an internal surface of a huge, hollow conductor?" Certainly! I think your confusion arises from a misunderstanding of what "outside" means in this context. Outside does not mean "the furthest surface from the center of mass of an object." It means "the ...

1

The electron gun produces electrons by heating the cathode; this shakes out electrons from the metal ("boils them off"), and as soon as they're out, they are repelled and accelerate away to do your bidding. This is called Thermionic Emission. When an electron is emitted, another one comes in from the cable connecting the cathode to the power source. The ...

1

We do! (Or at least, we should, to be complete.) Imagine the basic arrangement of a bar magnet, north up, moving upward toward a solenoid. Faraday's Law has a very important negative sign for this question, which dictates that the new current generated by this increasing flux is going to be clockwise from the top (use the right-hand rule, and flip it for ...

0

Ampere's law doesn't say that in this example you would have a time varying magnetic field (which is what you would need in order to generate another electric field), but rather just some sort of magnetic field which behaves according to the equation: $$\nabla \times \textbf{B} = \mu_0 \textbf{J} + \mu_0 \epsilon_0 \frac{\partial \textbf{E}}{\partial t}$$ ...

6

First, there is nothing wrong with our charge polarity conventions. They are predict electrical phenomena just as accurately as the opposite convention would have done. Did we have a problem if since the begin of their discovery we called them positive particles and negative to protons? We could predict the behavior of electrical phenomena equally ...

1

There are a few key bits of physics and math to understand here, but one does need to be very careful to avoid producing a circular argument. The key, concept I think is that of a linear circuit element, for which the output is precisely proportional to the input. The precise meaning of 'input' and 'output' depend on the precise device, but this does ...

0

Superposition is essentially a mathematical concept. Inspection of a physical phenomenon and the choice of the mathematical model to represent it, define the mathematical relations that the phenomenon is suggested to be subject to, and superposition may be one of them. For example, when we say that a capcitor follows the mathematical rule $I = ... 0 Well this can be explained by the work function of materials. Due to rubbing, heat is generated which supplies energy for removal of electrons. As the work function of the glass rod is smaller than the silk cloth, it easily loses electrons to the silk cloth which then releases energy (electron gain enthalpy) and thus ensures conservation of energy. 3 To simplify, let's think of a prismatic volume, as if a planar figure (in$xy$plane, area$A$) had been extruded in$z$direction. Let's assume that conductivity ($\sigma$) is dependent on$x$and$y$, but not on$z$. And let$L$be the prism length in$z$direction. We want to compute total resistance between both ($xy$) parallel faces. If we take a small ... -1 insulators because they make a heat which turns into electricity by rubbing together. 0 Olly, the first part of your thinking is correct, as the atoms receive more energy, the electrons do collide more energetically, but they also move "away" from the atom's center. The further they are from the center, the easier it is for an electric field to "move" them. This means that for the same effort (voltage), more electrons are moved (larger ... 2 Exactly what is this "circuitous route"? Does the thing I touch also have to be touching the carpet? Though I'm not a native English speaker I am pretty sure that a circuitous route is a path that combines you shoes and the carpet as were they a part of a circuit. The thing you touch has to be connected to the carpet (by touching the carpet itself or ... 1 The formula that you quote from the online is correct if the curve is a circle, i.e. the radius is constant and one calculates the magnetic field in the middle of the circle. Your formula uses the same assumption because you also took radius equal to$R$, i.e. $$r_s=Rcos\phi \ \hat{i} + Rsin\phi \ \hat{j},$$ but there are some mistakes: 1)$\text d \vec ...

2

What you describe essentially happens in superconductors. There "perpetual electricity" exists in the form of super currents. In normal inductors, though, the resistance of the conductor steadily dissipates the current leading to field collapse.

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