# Tag Info

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 ...

0

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.

12

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 ...

25

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 ...

0

Recall that the force between two point charges is given by $$F=k\frac{q_1q_2}{r^2}$$ If the net electric force on S is zero, then the sum of the forces on S from R and T will be 0. Use this to express d2 in terms of d1. For 110., it is important to realise that when two identical charged conductors come into contact with one another, their total charge ...

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 ...

1

The Lorentz force on a charge in an electromagnetic field is $$F=q(E+v \times B) \ \ .$$ For an electron between the cylinders, $q$ is negative, and $E$ is defined as pointing outward, so the electron will experience a force radially inward. But due to the unfortunate sign convention used for currents, electrons flowing inward means that the conventional ...

2

Yes - electrons flow from the negative to the positive, so in the opposite direction to the conventional direction of the electric field (which points from positive to negative). So if the E field points outwards, the electrons flow from the outer to the inner cylinder. The direction does not affect the answer (the calculation of the flow) though - at least ...

0

gee whizz, its like Maxwell and Faraday never existed! Remember, a current carrying wire gives rise to a concentric magnetic field. This will be accompanied with a radial Electric field.

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The total amount of charge on the two plates is 0, i.e. $q_1=-q_2$, in the case of a plate capacitor, the field is twice that of a field produced by an infinite charged plate, so $E=\frac{q}{A\epsilon}$ where q is the charge on the plate, A is the area of the plate, and $\epsilon$ the relative dielectric constant of the material between the plates.

1

The field from a current in a wire is purely magnetic for a static current. When the current varies with time, there will be radiation. What you are missing is that a radio antenna doesn't operate with DC. You can perhaps understand it like this: the magnetic field from a current loop depends on its magnetic moment, which is current times area. When the ...

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