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0

Well, the electric field $\vec E$ is different from the force field $\vec F$ a test charge will feel. That difference is exactly the charge of the test particle. That force field is given by the gradient of a function, too $$ q \vec E = \vec F = - \frac{\mathrm d}{\mathrm d r} W$$ where I use the letter $W$ in order not to have confusing notation. The ...


3

I think you are reading a lot into what is a minor distinction. Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what ...


1

What's probably happening here is the following: The fundamental or microscopic fields $\mathbf{E}$ and $\mathbf{B}$ are technically called the electric field strength and the magnetic induction, while $\mathbf{D}$ and $\mathbf{H}$, their macroscopic counterparts, are called the electric displacement and the magnetic field, a quite weird nomenclature, since ...


0

The only thing I can see them going for is the fact that only two of $\epsilon_0$, $\mu_0$ and $c$ are independent, and typically, a modern view will fold $\epsilon_{0}$ into the definition of charge, and declare $c$ to be the fundamental constant used to transform space into time in special relativity, making $\mu_{0}$ a prediction of the theory. I ...


0

There are no sparks/arcs in vacuum. Unless you get field emission, which requires very high charge densities i.e. either high voltages or very small structures on the emitting surface, there has to be an initially non-conducting medium that can be ionized in the field between the electrodes. Once that ionization occurs, there will be a cascade of electrons ...


0

The speed of the electron in a vacuum (or close vacuum) depends on the voltage applied to them. The higher the tension, the quicker they are (speed proportional to the square root of the voltage applied). Dielectric has nothing to do with electrons, but everything to do with light... On top of that, vacuum is a dielectric...


4

In this case, the image method can be used to calculate the potential (and hence the electric field) in the region $z>0$, with a negative charge $-q$ located at $(0,0,-d)$, since the potential would be $V(x,y,0)=0$, in this case. But for points in the region $z<0$, the potential is given by the solution of Lapalace equation $\nabla^2 V=0$, with ...


1

Your method for evaluating the electric field assumes it is appropriate to model it as spatially constant within the wire since you're basically taking a spatial average. You'll have to decide wether or not this is accurate.


-3

we take positive charge as a point charge because positive charge has a high potential and electrons are always move from lower potential to the higher potential.so , we use positive charge as a point charge


1

No, there are several mistakes in your derivation, although you miraculously end up with the right expression. The term $\sin \theta$ comes from taking the horizontal (X) component of the electric field - not from the expression you used for $dq$. The diagram I envisage for your problem is this: So I would say that $$dq = \lambda dy$$ and then $$dE = ...


1

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. So, if there is force acting on a unit charge, then electric field does exist. It is the way by which we can prove the existence of electric field (as per definition demands). I don't ...


0

Yes! Neutrons are electrically neutral, but they have a magnetic moment. You can accelerate a bar magnet with a magnetic field, so you can also accelerate a neutron with a magnetic field. For most beams, the change in energy is pretty negligible, but there's a major exception for ultra-cold neutrons (UCN), which just so happen to be my specialty. UCN have ...


1

The electrostatic field depends only on the total charge distribution. If the charge distribution is known, as it is in your case, then you don't need to worry about the shape or conductivity of the structure supporting the charge. The charge on a conducting solid sphere will, as you say, distribute evenly at the surface. If you by some other method manage ...


4

Using cylindrical coordinates with the origin at the center and the $\phi = 0$ direction 'down' (the OP says the image should be rotated CCW 90 degrees), the electric field appears be have only a radial component with a sign change for $\phi = \frac{-\pi}{2}$ and $\phi = \frac{\pi}{2}$ $$\vec E = E(\rho,\phi)\hat\rho $$ $$E(\rho,\phi) = ...


1

I have came up with this: Charges are the sources of the electric field. So, whatever the point that field lines are "created" or "destroyed", must be a charge. Then, if there are a charge, then must be on the center. Calculating the electric flux: $$ \phi = \iint_S\ \mathbf E\cdot d\mathbf s = \frac{Q}{\epsilon_0} $$ Let's pick a sphere as gaussian ...


0

Take this as the definition of the electric flux going through some surface S in some electric field E $$ \Phi_E = \int_S {\bf E} \cdot d { \bf a} $$ This says that to find the flux on the surface we simply add up the electric field at each point on the surface. On the other hand this is Gauss's law: $$ \int_S {\bf E} \cdot d { \bf a} = ...


2

How is it possible to accelerate a neutron? Neutrons have a dipole moment, so they may be 'accelerated' insofar as they will turn in a magnetic field – that is their primary interaction with the electromagnetic field. It is possible to accelerate a charged particle in an electric field, how is it possible to accelerate a neutron? Neutrons also ...


1

The neutron can be attached to a proton via the Strong Force by colliding a high-energy proton with the neutron, and then the proton-neutron atom can be accelerated with a regular electric field. Gravity can also accelerate a neutron.


14

Although a neutron is electrically neutral, it has a non-zero magnetic dipole moment. It interacts with a magnetic field to give a potential $$ U = \vec{\mu} \cdot \vec{B} $$ A gradient of magnetic field strength will give a force $$ \vec{F} = \nabla|\vec{\mu} \cdot \vec{B} | $$ It's not possible to produce large, sustained field gradients, nor is it ...


27

Basically, the answer is no, it's not possible. When we produce neutrons for research purposes, we have to produce them using nuclear reactions. They come out of the nuclear reactions with energies that are determined by the reaction, are not otherwise under our control, and that are on the MeV energy scale of nuclear physics. Examples of a neutron source ...


1

The answer-as requested by HDE. So my foolish misconstruct of the electric field is that since it accelerates charged particles more or less depending on the distance from the source, it should create less motion in electrons a further distance from the source. I'm not sure why I thought resistance had much to do with it-on further analysis resistance ...


4

This table should answer your question which is about intermolecular interactions. The largest distance that the electromagnetic forces can be effective for neutral atoms and molecules is in fractions of a nanometer. When in classical distances then the electric field coming from neutral atoms is effectively zero. Nano structures though, if you make your ...


0

I think as we know E = V/d, and the field is same, so for field remains constant between the plates of the capacitor, while increasing the distance the potential also increases. In the same manner as that of distance so that the ratio of V and D is same always.


1

I agree with Chris, the imaginary part here is just the mathematical way of showing a rotation. In this case, it happens to be that the second ket is 90 degrees out of phase with the first. You can think of the imaginary axis as a direction orthogonal to any real axis, thus it is similar to saying the imaginary part is at some angle with respect to the ...


1

The $i$ in front of the $|01\rangle$ tells you that the first order mode is $90^\circ$ out of phase with the fundamental mode. The absolute phase generally has no meaning so you could just as well have put $-i$ in front of the fundamental mode. The phase difference between the two has physical meaning. When you take the projection onto the real axis one ...


1

Wikipedia> Electric charge: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric flux: In electromagnetism, electric flux is the rate of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a ...


0

Resolution of force, or any vector in general, into its components along a particular choice of co-ordinate vectors $\{e_i\}$ is not specific to any particular assumption about the medium/space. It is convenient to employ an ortho-normal set of co-ordinate vectors (e.g. ${\hat i} \cdot {\hat j} = 0$ in the Cartesian case), but even if the vectors aren't ...


1

The equation of the electric field of a sphere shows that $$E = \frac{Q}{4\pi\epsilon_0 r^2}$$ for values of $r\ge R$. Solving for $E(x) = \frac{E(R)}{2}$ you find $x = R\sqrt{2}$ - in other words, you need to move a distance $R(\sqrt{2}-1)$ from the surface for the field to be halved (minus 1 because you start at the surface...)


0

The assumption is that the electric flux lines are going to go through the conductor parallel to the conductor. Under those conditions, the electric field within the conductor is going to have a constant magnitude, and point parallel to the conductor. I assume you're familiar with $$E=-\nabla \phi\ \ ,$$ where $\phi$ is the electric potential. We then ...


2

The path a free positive test charge would follow if acted upon no other force but the force due to the field itself. This is wrong. (Did you actually have a book that said, this? If so, what was the book? This would be a serious error.) A charge in free space will have an acceleration parallel to the field, but the acceleration is not typically in the ...



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