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The electric charge difference between the earth and the atmosphere grows with altitude, at around 88 DC volts per meter. This electric potential may be shorted out when a thermonuclear explosion releases radiation which ionizes the atmosphere. About 5% of a nuclear explosion's energy is in the form of ionizing radiation. A study of lightning flashes ...


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Your statement implies that the Electric field $\boldsymbol E$ will be parallel, during the whole motion, to the instant velocity $\boldsymbol u(t)$, i.e: $$\boldsymbol E = \boldsymbol E_\parallel + \boldsymbol E_\perp= \boldsymbol E_\parallel.$$ If you have a curved trajectory there must be a component of the force which is perpendicular to the velocity in ...


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Here is an analogy. Let's consider dropping a ball on the earth's surface. We all know that Earth's gravitational field is a vector towards its centre. Only if the ball is dropped from rest, or was a given a downward shove, it's path will be straight(at least till it hits the ground). On the other hand, if you lob the ball horizontally, it's path is curved. ...


0

When the charge has gained momentum in a particular direction, it will experience a force and acceleration whose direction is that of the field lines, yes, but the velocity will not be in this direction. Compare with circular motion - the velocity is tangential to the circle, but, the acceleration is perpendicular to it.


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In particular, I want to know if the fact that accelerated charges radiate (Larmor's formula) can be derived from the Hamilton's equation of the system. If the Hamilton's equation include the electric and magnetic fields as dynamical, then yes, it should be do-able... However, if you are just including the electrostatic interaction between the ...


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hahaha Actually, EMF is always applicable, even in a blackhole. All stars have perfect symmetry, to a degree, gravity is always trying to crush the star and thru fusion SNF / the EMF electrons repel this force outward causing this balance. Now you can have GRB's and the formation of blackholes if the star is large enough or if not Neutron stars or even ...


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For the first question: there may be several species of particle in a theory, so charge conservation can be upheld without creating a particle and its antiparticle together. For example, in beta decay, an electron is `created', but no positron, and charge conservation is upheld by a neutron being exchanged for a proton. This partially answers question two ...


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No. Charges are contained within the wire or conductor, but those are in random motion, battery only aligns the charges using electromotive force, in a direction, in order to produce current. Thanks :)


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The charged sphere induces surface charge density on the conductor. This is necessary because field lines are always perpendicular to the surface of the conductor. This induced surface charge density modifies electric potential in the region. It will no longer be $V(d)=Q/4\pi\epsilon_0.d$. If you know the potential at the conductor(which I think is necessary ...


2

My question is do the byproducts(elementary particles such as higgs boson) emit photons as they are produced to be filmed on tape by the sensitive detectors? If you're imagining that the particles are emitting photons as the leave the collision center and that the tracks of the particles on the computer screen were recorded by collecting these ...


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If you move q/2, both end up with charge q/2 and the force will scale with $\frac{q^2}{4}$. If you move 2Q, the product of the charges will be $-2Q^2$ the magnitude of which is obviously bigger (although in the first place the force will be repulsive and in the second case it will be attractive).


1

Suposse that -Q=ne, where e is the electron charge and n a natural number. So, you can "add" positive charge to an object removing n electrons. Now imagine you have a object with charge 2Q, if you add 3n electrons, the final charge would be -2ne+3n3=(-2+3)ne=ne=-Q. Remember that electrons has negative charge.


1

The charge parity operator $\newcommand{\C}{\mathcal C}\C$ takes each particle to its antiparticle, which has opposite charge. As such, its eigenstates are those states that remain the same if you change all particles for their antiparticles. This can happen, for example, in a state that contains two particles which are each other's antiparticle, such as ...


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Here is an explanation of electrical flow that might illuminate this matter: http://amasci.com/amateur/elecdir.html


1

That there are two distinct types of electric charge is a metaphysical fact. But nature is indifferent to what we choose to label these charges; up / down, left / right, positive / negative, black / white, etc. Electrons will still flow to the plate in a CRT regardless of how we choose to label the polarity of the charge on the electron and plate. ...


1

Benjamin Franklin proposed electric fluid theory and considered electric current to be flow of a charged fluid. He meant to use positive to denote a surplus of the fluid, negative as a deficit of it. No one knows how he came up with the choice, but it became the convention and as a result lead also to the labeling of charge. I know of no fact that could ...


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As Zeldredge said the name is arbitrary and does not matter electron could have been positive and protron negative just the name.


2

Now, from Coulomb's Law we can find a vector for the electric field due to this electron at all points in this space. When you read about Coulomb's law, you can see that it describes the force between two charged particles. When you set them down such that they have no initial relative motion, they move together or apart in a linear fashion. But note that ...


2

The electron is an elementary particle in the underlying building blocks of matter organized in the elementary particles table of the standard model of particle physics. Elementary particles are point particles. The standard model is a precis of a very large number of measurements (data) fitted by mathematical models of theoretical physics. A point ...


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In case the question concerned the case $r \rightarrow 0$, you would reach the situation where the charge (represented by a charged particle like electron, proton, positron) approaches the Coulomb field of the other particle and they would have a tendency to create a kind of a planetary system - but - quantum effects start to play a role here and those two ...


3

If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to ...


1

What is wrong with my reasoning? Opposite charges attract because one of the charges has a negative sign. The force on the negatively charged particle is thus $$\vec F_- = \frac{kQ(-Q)}{r^2}\hat r = -Q\,\frac{kQ}{r^2}\hat r = -Q\,\vec E_+ $$ The force on the negatively charged particle is opposite the direction of the field from the positively ...


0

The electric field lines show the direction of the electric force acting on a unit positive charge at a particular point in space. So, therefore, the force acting on a negative charge, as is in your question, will act in the opposite direction shown by the electric field lines. I believe that the fact that field lines are defined in terms of a unit positive ...


1

Suppose charge is not conserved when we have a change in velocity (i.e. when we move from static to moving or vice versa). Then we wouldn't expect the change in the Coulombs attraction to be solely dependent on $r$ $$F = k_e \frac{q_1q_2}{r^2}$$ because as one particle attracts another, the values in the charges $q_1,q_2$ would change and so $F$ would ...


1

Take for example, the case when the charge Q at the centre is negative, so the force is attractive. If it gets displaced even by a small value $\delta r$ towards either, the separation with this particular vertex would be $(\frac{l}{\sqrt 2}-\delta r)$, while it is slightly larger for the others. Due to the inverse dependence of the force on this ...


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Consider the square to be in the x-y plane. Then see what happens when you displace the central charge in the z-direction. Of course, the answer will depend upon whether Q is positive or negative and possibly also on the direction of displacement.


2

a discrete charge distribution is approximated by a continious charge distribution if the ratio $$ \frac{\bar{d}}{D} << 1 $$ for $D$ the smallest diameter of the distribution and $\bar{d}$ the mean distance of discrete charges in any subvolume of the distribution (e.g this ratio has to be small in any subvolume of the distribution).


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Even though a continuous charge distribution does not exist in nature, it is a very useful concept in situations where there are so many very small charge carriers that you don't see them individually. Think of it like water. If you are not using sophisticated tools, you can't see the molecules. It looks continuous. Hydrodynamics is the study of the water ...


0

$N=CV/m$, hence your solution turns into $\frac{C}{m}$, which doesn't look right (electric field in V/m = force in N / charge in C) I suppose, that answers your question. You can use Wolfram Alpha for unit conversion, for example in your case. And here you'll see how force relates to voltage: ...


1

1) Every "bare charge" contains an infinite part. The reason why we introduce the infinite part is precisely so that the divergences in loop integrals cancel. Physically you never measure bare charges: you always measured a suitably "dressed" charge that is typically scale dependent and is subject to screening effects, etc. 2) The finite part is always ...


1

You will have to use Kirchoff's law to get the answer. How can some of positive numbers be zero? No, the sum of all charges will be zero while that of positive plate will be finite. Now the net sum would be zero since charge is conserved on the system having all the right plates. Use Kirchoff's Loop Law and Kirchoff's current law to find charges and ...


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. ...


2

The problem is about the energy needed to assemble the charges, assuming that they already exist. So, imagine that the charges are separated from each other by a very large distance. Then your third expression is effectively zero. As you bring the charges together, the first two expressions don't change at all, and the third expression does. So the work ...


1

There is no need for taking the mod of the charge. The voltage has the same sign as the charge. So if you start out with +20 µC on one capacitor (they give the + sign for a reason - so that's the side where we will put the positive charge) and +60 µC on the other (from $Q=CV$) then it follows that the total of redistributed charge is 80 µC as you correctly ...



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