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## New answers tagged electric-fields

0

When you choose an arbitrary gaussian the flux is still given by the charge inside the surface. However it would be useless in calculating the electric field unless you are able to write $$\int\vec E\cdot d\vec A=\int EdA=E\int dA.$$ That is, Gauss law is useful when the surfaces elements are always parallel, antiparallel or perpendicular to the electric ...

0

You would calculate the flux $\Phi$ as $$\Phi=\int \vec{E}\cdot d\vec{A},$$ just as you would for a sphere or a cylinder. Gauss's law is true for any closed surface. Just take your electric field and take the dot product with the differential area $d\vec{A}$ of the surface and integrate this over the entire surface. The tricky ...

2

The RHS side of Gauss's Law, that is the charge enclosed should remain the same is indeed true. The apparent confusion if any, should be in the LHS of the equation, the integral of the 'dot product' of field and area vectors. Consider the diagram, Now, when we take the dot product of the field vector with the area vector in the initial case, the field and ...

0

The electric field inside the diode will be opposite to that of the electric field of the source. Thus will cause a barrier potential of 0.3 or 0.7 volts opposite to the source when its forward biased. That is why the the diode start conducting when forward biased when source is at more than 0.3 or 0.7 volts. Less than 0.3 or 0.7 volts it will conduct very ...

0

Electric fields obey the superposition principle. So let's call the two fields $\mathbf{E}_{1}\left( \omega_{1} \right)$ and $\mathbf{E}_{2}\left( \omega_{2} \right)$. Then the total field, $\mathbf{E}_{t}$, is given by: $$\mathbf{E}_{t} = \mathbf{E}_{1} + \mathbf{E}_{2} \tag{1}$$ Note that the field intensity or energy density is given by: $$W = ... 1 The answer is that it depends on the field on the outside (boundary conditions) and the dielectric constant of the insulator. For a imaginary insulating sphere of vacuum in a vacuum, it should be obvious that the sphere does not affect the electric field at all. Inside a dielectric, the field will be weaker than on the outside. For a dielectric sphere in a ... 0 2) Does it include energy of electron due to electric field too? We do not know. If the electron was composed of charged elementary parts, its apparent inertial mass would be greater than sum of inertial masses of the parts. This is because such parts will act on each other with electromagnetic forces and in accelerated motion, sum of these internal ... 0 In the situation you depicted, the electric field is different from 0 in the hollow region enclosed by the conductor and equal to 0 inside the volume of the conductor (at equilibrium). Why is \vec E = 0 inside the volume of the conductor at equilibrium? We will proceed by reductio ad absurdum. Since we are at equilibrium by hypothesis, there can ... 1 I have found your question and the diagram a little difficult to interpret. I have redrawn you diagram to show a charge of +Q on the outer shell and a charge of -Q at the centre together with two conducting shells shaded grey. What else the electric field inside the conductors is zero. If there was an electric field then the mobile charge carrier ... 1 There is the concept of the self energy of the electron that can be described by its electric field. The model, which goes back to classical physics, can be described as a spherically distributed charges (that equal the charge of the electron) at infinity and the work required to bring the charges to the radius of the electron. The calculation would show the ... 0 To begin with, I can't think of any reason why an increase in the mass of an electrically charged object should affect the magnitude of the electric field it produces. You can easily check that E=kq/r^2 doesn't depend on m. Moving charges do produce magnetic fields, but this is a different matter (and the magnitude of the magnetic field has nothing to do ... 2 Too long for a comment: Consider the Maxwell equations:$$\nabla \cdot {\bf E}=\rho/\epsilon_0 \qquad \nabla \cdot {\bf B}=0\nabla\times {\bf E}=-\dfrac{\partial {\bf B}}{\partial t} \qquad\nabla\times {\bf B}=\mu_0 {\bf J}+\dfrac{1}{c^2}\dfrac{\partial {\bf E}}{\partial t}$$According to Heras (when commenting on a paper by Griffiths and Heald) the ... 0 The capacitance of a capacitor depends on the dielectric constant (relative permittivity) between the plates. The maximum charge that can be stored on a capacitor depends on the maximum potential difference across the plates of the capacitor which in turn depends on the maximum electric field (breakdown potential gradient - dielectric strength) which the ... 1 Equipotential lines are like contour lines on a map which connect land which is at the same height above sea level. Finding the electric field is like finding the slope on the map. The slope (the steepest line of ascent or descent) is always perpendicular to the contour lines, and it is steepest where the contour lines are closest together (provided that ... 1 Kirchoff's laws are the electric equivalents of more fundamental laws. Kirchoff's Law of Current: Is a formulation of the Charge Conservation Law, in a node there is no net charge, so from the continuity equation the net current going through the node is 0. Kirchoff's Law of voltage: Corresponds to the energy conservation, altough you need current's law to ... 0 The Short answer is no. Kirchhoff's Law is derived by the assumption that the electric field is conservative, while the electric field induced by the change of an magnetic field is not conservative. The Long answer is, that you have to clarify what voltages you are looking on. Kirchhoffs Law is usualy used to make a statement about the voltage drops in a ... 1 This is a homework-like question, so I will not provide a full answer. Here are a couple of good things to think about on your way to the answer: What is special about the velocities (or momenta or kinetic energies) of the particles at the instant of minimum separation? What quantities are conserved throughout the interaction? 4 You are confused due to the vectorial form of the equation, so you should write it by components (I will use cartesian coordinates), and I will use \partial_x = \dfrac{d}{dx} for comfort, that being said, your equation can be written as:$$ F=-\int{(dr)}{(\vec{\nabla} \cdot \vec{P}) \vec{E} }=-\int{(dr)}{(\partial_x P_x +\partial_y P_y + \partial_z P_z ) ...

4

I think the answer is clearer if you consider the equipotential as shown in the diagram below. Forgive their straight line nature as they were easier to draw that way. Given that $\vec E$ must be perpendicular to an equipotential surface then in your computation of potential difference $\displaystyle V_{AB}=-\int_B^A\vec{E\,}\cdot\mathrm d\vec{r\,}$ the ...

0

Since the charged conductor is present it will create an electric field and corresponding potential but to measure it experimentally you will need another known charge at one position, say A and then move it or allow it to move ( that depends on whether the charged body and our test charge are of same sign or the opposite sign) to a position B and then ...

0

You have to have probe wires in order to bring any charge into the voltmeter so it can be measured. Electrons will flow in a wire because they are attracted or repelled by other nearby charges. For example, if the charge is positive and you place the probe near the charge, electrons in the probe wire will move toward the charge, causing the other end of the ...

1

I think your confusion is coming from the fact that you are actually using theta for two different things here. Let's use phi for the angle between the velocity of the rod and the magnetic field, and use theta as it is depicted in the diagram. Then your expression should be written as $$qvBsin\phi$$ This comes from the fact that $qvBsin\phi$ is derived from ...

2

A very obvious answer is just to heat the air enough (flame). You can also accelerate electrons with electric field and inject them into air (plasma needles and a lot of commercial and lab plasma setups). Microwaves can be used to create and sustain ionization (think of microwave plasma experiments). The same goes for other wavelengths that resonate with ...

0

In this situation, I just want to quote what Richard P Feynman once said in an interview. "If you hold two like poles of a magnet together, they repel apart, which means there is some force existing in between them avoiding them to have a contact. That's an experimental truth. But if you ask me why there is a force in between them that do not want them to ...

3

But I don't understand the mechanism of the force creation But the concept of electric charge and electric field is, by definition, the mechanism of the force creation - that humans have invented to model that which has been observed. Never forget that the observed is the metaphysically given. It is up to us, as beings possessing a rational faculty, ...

1

In addition to the answer by dmckee and to answer the question how high in energy you could get a photon it might be worth thinking about 'Gamma Ray Astronomy' where the highest energy photons are detected. The record highest photon energy observed is apparently currently 80 TeV, which corresponsds to a wavelength of $1.5 \times10^{-20}m$ wavelength (if I ...

4

Higher energy gamma and longer wavelength radio? Keep in mind that the different 'kinds' are merely human labeling conventions for a spectrum that is continuous in the mathematical sense. There is no feature of "radio" that distinguishes it objectively from microwaves. We just pick a boundary on the basis of some technological limitations that apply when ...

1

The electric field due to the outer cylinder has no contribution inside. One way to view it using Gauss's law, the other way is that if you took a slice from that cylinder, and considered a point inside it other than the center, you'll find a point producing electric field in the near side of the point (small charge, small distance) and a corresponding arc ...

2

The electric field between the conductors is due to both sets of charges. however when finding a value for the electric field using Gauss's law it is only the charges inside the surface which are of interest and it is easier to choose the charge on the centre conductor and the red Gaussian surface $S_+$ which would be a cylider. You could find the ...

2

Potential refers to a particular point - or set of points which are "equipotential". So you can talk about the potential of one of the capacitor plates (because each is an equipotential surface) but not the potential of the capacitor (because when charged the $2$ plates are at different potentials). When talking about a capacitor, potential usually means ...

1

Is there an electric field around the poles of the battery before the circuit is attached, and is there still, after the circuit is connected? Yes. However adding the connecting wires is likely to change the distribution of the field. And why is the field equally strong everywhere in the wire, no matter the shape? Usually we design our circuits ...

0

Electric potential energy is stored in your system of a large mass charged body $A$ and a small mass charged body $B$. When an external force is applied to body $B$ that external force will do work on the system (body $A$ and body $B$) and change the electric potential energy of the system and possibly change the kinetic energy of body $B$. However it is ...

0

Perhaps the confusion comes from the often belief that if a force field is only a function of position then the force is conservative, and that non-conservative forces are those that include friction dependence on speed, and other variations. But this is incorrect, as alfred centauri commented this field is not conservative. But this is not just because ...

-1

Apparently you are not considering the work done by the electric field, and you are confusing yourself with the "energy storage" concept. After traveling a $\Delta x$ distance into the electric field $\vec{E}$ (avoid using $dx$ because $d$ usually stands for infinitesimal changes), the particle B with charge $q$ will have suffered change in its "horizontal" ...

0

Even though we call it as a plane sheet of charge it is not really a plane sheet. there is an elemental volume with the limit the volume is nearly zero. So we have a few layers of atoms in a sheet. In the case of conductors charges can reside only on the surface (consider that you roll the sheet into a cylinder; there can't be any electric field or charge ...

2

On a local level, when you measure the electric field in matter (conductor or otherwise) there is a rapid variation both in time and space. Semiclassically, the electric field get's very large as you approach the nucleus, then drops off, the electrons are moving around the nucleus at very large speeds. So, yes, you are correct when you say there will be ...

0

The direction of the force acting on a charge q with velocity ${\bf v}$ moving in a magnetic field ${\bf B}$ is given as ${\bf F} = q {\bf v} \times {\bf B}$. In your question you have ${\bf v} = v {\bf i}$ and the electric field is (eventually) pointing in the direction ${\bf j}$. This means the direction of the magnetic field must be in the ${\bf k}$ ...

0

As another answer pointed out, your formula is for electric field around an isolated point charge. It doesn't apply to the case of parallel plate capacitor. Normally we use Gauss's Law to find the electric field between the plates of the capacitor. We know that the field between the plates will be uniform from the differential form of Gauss's Law ...

0

You reference the equation giving the electric field near a finite point charge. There is no finite point charge in a capacitor (unless we count a single electron, but I think you'll find a single electron won't produce a very large field measurement on a human scale...). The charge is distributed uniformly, and as you (you're a test charge) get closer to ...

-1

there are lots of questions and explanations, why the field of an infinite plain is homogeneous and does not depend on distance. That is the approximation involved: that the plates are big. So outside the plates the fields add to zero, and inside it's double. No, charge is not brought from one plate to the other. If you have an alternating current, it will ...

2

The principle of relativity says that there is no experiment that can determine absolute motion. So all observers, regardless of relative motion, need to agree on the outcome of any experiment. Because to the relativity of observers' measuring devices, they may not numerically agree on the measurements. By applying the laws of relativity they will be able ...

2

A few comments before doing the calculation: In the CM frame, there is only an attractive force, while in the given frame, there is both an attractive and a repulsive force. This is no more mysterious than the fact that a vertical object in my frame can look tilted to somebody with rotated axes. Going from the CM frame to your frame mixes the electric ...

4

Lorentz Transformations Suppose we call the lab frame the K-frame and a frame moving at velocity, $\mathbf{v}$, relative to the K-frame called the K'-frame. Then we can express the electromagnetic fields in the K'-frame in terms of the K-frame fields as:  \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \boldsymbol{\beta} \times \mathbf{B} ...

2

The observer moving with the CM will measure that the force of repulsion the electrons is given by $F=\frac{e^2}{4 \pi \epsilon_0 d^2}$ ($d$ is their separation), he can only make measurements in his reference frame (that is moving with speed $v$), and will not be able to be determine this speed.

3

This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ...

3

Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ...

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