# Tag Info

## New answers tagged electrostatics

0

Your formula for the first energy is incorrect. Instead use: $$W = \frac{1}{2}\sum_{i\neq j}q_{i}V_j(\vec r_{i}) \, .$$ Or even: $$W = \frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{4\pi\epsilon_0|\vec r_i-\vec r_j|} \, .$$ And now you see right away that you are avoiding the energy of the point charges themselves. Because naively it would be zero.

1

The difference is the zero point. When summing over charges, the reference is a state in which this charges are infinitely separated. Those are still distinct, localized charges, just separated from each other. When integrating $E^2$ over all space, the reference state has all charge separated. Even the individual charges from the first method are broken ...

-2

perfect dipoles means forces at center of dipole is zero

1

According to Richard Feynman, the charge is the probability of a particle interacting by the electro magnetic force. More specifically it describes the amplitude of the "probability arrow" of a certain electromagnetic interaction taking place. Much like @Asher has mentioned already, the standard model cannot provide an explanation for why certain particles ...

3

Charge means that the body experiences a force in an electric field. A charge generates an electric field, which generates a force on other charges particles. Two bodies are said to repel if they force each other away and two bodies are said to attract if they force each other closer together. Now, I'm not really answering your question here of "why," I ...

0

a) consider a segment of the cable of length L. Calculate the charge contained on the inner cylindrical core of radius a. Then set the same charge to the outer cylinder of radius b. Calculate sigma from this charge and the area of cylinder with radius b and height L. b) for electrostatic case, the field for r < a must be zero if the inner core is a ...

-1

Is there a small amount of contact information absorbed by both the Electrons at some level. During the repulsion, I'm assuming there is some level of reverberation that gets absorbed by both proton and neutron during the frequency exchange and wave length generated. I'm just guessing here but I would think opposing wavelengths would repel outwards and also ...

1

The first equation assumes the external electric field (caused by the charges on the plates of the capacitor) doesn't change. When a battery is connected, it can fill and discharge the plates as necessary to maintain the voltage. So, the E_0 value increases as you add the dielectric.

1

Note that $\mathbf r(t)$ is the trajectory (a priori unknown) of a charged particle in an external electric field. Now consider the ansatz $\mathbf r(t) = \mathbf r_0(t) - \mathbf a(t) \cos \Omega t$, which is motivated by the solution for a homogeneous electric field $\mathbf E(t) = \frac{m\Omega^2}{q} \mathbf a \cos \Omega t$. Here $\mathbf r_0(t)$ is a ...

0

Without even doing any circuit calculations, you can conclude the voltage between a and b is zero by symmetry. Proof: Assume there's a voltage between the two points. If you close the switch, a current would flow. If you take the mirror image of the circuit, you'd expect the same current to flow, but in the opposite direction. Except the circuit is left ...

0

First, let's assume the left-most terminal is connected to the positive terminal of the battery and the capacitor voltage reference direction is left-most terminal positive. Now, consider a KVL loop clockwise through the top 2C capacitor, the switch, the bottom C capacitor and the battery: $$10 \mathrm V = V_{2C_{top}} + V_{ab} + V_{C_{bot}}$$ So, the ...

0

The potential difference across the top two capacitors must be the same as the difference across the bottom two. I will number the capacitors $C_{11}$ for top left, $C_{12}$ for top right etc. If we assume the charge on each capacitor is the same, then the voltage difference must be zero. But if we can assume that each capacitor may have a different charge ...

2

This the kind of question that can be solved by the method of images. Try placing a fictitious charge on the other side on the conducting plane. You should arrange it in such a way that the electrostatic potential is precisely zero on the surface of the conductor. If your case you put it at equal distance as the first but on the other side. The physical ...

1

I asked a somewhat different, yet similar question.Hope this helps! Why is an $LC$ oscillator lossless, but $C V^2 / 2$ energy is lost to a capacitor connected to an ideal voltage source?

1

I might be erring something basic here, so downvotes are welcomed, but I would love if they include comments to correct this answer, or just erase it. I do not believe the Coulomb law has been tested beyond the order of a few meters. Arguing that light remains unchanged across the universe should be irrelevant. The reason is that the electrostatic and ...

6

There have been lots of experimental attempts to test the validity of Coulomb's $r^{-2}$ law. Many of these are reviewed by Tu & Luo (2004), and is where I am getting the numbers quoted below. Somewhat equivalently, experiments have looked at trying to set an upper limit to the photon mass, which is testing the hypothesis that rather than a $r^{-1}$ ...

0

If you want to research the question more deeply, I would suggest you take a look at the Solar Wind. This is composed of charged particles (mostly protons) emitted by the Sun. The flow and behaviour of the Solar Wind has been studied quite deeply, not least because it affects greatly satellite operations, spaceflight, radio transmission and other important ...

3

Coulombs law as well as Amperes law and similar mathematical formulations of two centuries ago, were incorporated within the strict mathematical format of Maxwell's equations . The apparently disparate laws and phenomena of electricity and magnetism were integrated by James Clerk Maxwell, who published an early form of the equations, which modify ...

1

Remember Lenz's Law: as you change the flux through a coil, an e.m.f. is generated that opposes this change. Therefore, if I have two coils that are a certain distance apart, they will have a certain "shared" flux - flux due to $A$ appearing in coil $B$, for example. Now if we bring $A$ closer to $B$, we change the flux in $B$ due to $A$, and will get a ...

0

For eg if certain amount of clockwise current is sent through a coil nd then a second coil is brought closer....as we bring it closer the repulsion increases and the repulsion is of same magnitude as of the experienced current and hence the current is said to be increased

1

First, the Wikipedia article already says on the derivation of Gauss' law from Coulomb's law: Note that since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more ...

-1

Simply, no! Light takes some time to travel, so the effect of chancing a charge at one point should be retarded further away. If you assume such instantaneous force, it is called quasistatic approximation of Maxwell equations and can for example be used in Mie scattering theory, where a particle is much smaller than the wavelength of light.

0

How [is] electrical energy is transfered [through] a wire In isn't really, the energy comes in front the side and the charge flows through it and the two are different. Let's talk about the energy. Firstly, the electrons fields and magnetic fields themselves have energy. Secondly, when you charged up the battery you changed the electric and magnetic ...

0

only a charged body has electric potential around it. Since the person was positively charged by rubbing on carpet, he has a pistive potential around him. when person rubbed his foot on carpet he lost some of its electrons to carpet giving him positive charge. The definition given in book meant to present a view of how to calculate potential around a charged ...

0

Electrons do move, but much slowly, when you are, for example say lighting a bulb, you give a voltage (Potential) across both the ends of a wire, this Voltage (potential difference) induces a an Electric Field, this travels at the speed of light, the wire just acts as a path for the electric field, the electrons start moving due to this electric field, so ...

0

In circuits it's obvious what a potential difference means, but I agree that it's harder to see what potential means for an isolated obect. To address this we have a standard that the potential of a unit charge is zero at infinity. That is, if you take the unit charge an infinite distance away from the object you're considering, then the potential on the ...

0

This is the way a Faraday cage works: Source Two metal plates isolated from each other will never cancel between them an external static electric field.

0

If the two plates have the same electric potential, there is no electric field inside. It is usually better to think of these problemes in terms of Laplace's equation $$\Delta \phi = 0$$ instead of Gauss's law. A function satisfying Laplace's equation is harmonic and has lots of nice properties, but in particular it is a well-posed problem, meaning any ...

0

$F_1 = F_2 = F_3$. This is essentially the superposition principle. We know that in an atom, for example, in a neutral oxygen atom there are 8 protons and 8 electrons, i.e., 8 positive charged particles and 8 negative charged particles. We know it's nucleus can only carry 8 electrons around it. Now my question is why can't it carry so many electrons, ...

0

A bit of special relativity is required to understand this. Electric and magnetic fields form a 4-vector with their potentials: $$A^{\mu} = (\phi, \vec{A})$$ Where $\phi$ is the scalar electric potential and $\vec{A}$ is the magnetic potential. In the case you described, this would be: $$A^{\mu} = (\phi, 0)$$ Where $\phi$ is constant in time ...

1

In general, no: the field of a charge distribution $\rho$ is not the same as the field of a point charge at some point therein, except for some very particular cases (the one that everyone should know is that any spherical shell of charge has an inner field of 0 but an outer field that looks exactly like all the charge is located at the center point. A ...

1

You are in your reasoning overlooking something. Look at the diagram below: $-q_1,+q_2$ are two point charges at distance $r$. Coulomb's Law dictates that the attractive electrostatic attraction force between them is: $$F=k_e\frac{|q_1q_2|}{r^2}$$ And the electrostatic potential $U(r)$: $$dU(r)=F(r)dr$$ $$U(r)=-k_e\frac{|q_1q_2|}{r}$$ Assume now that ...

1

You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the ...

1

You can't really explain the conductivity in metals without basic quantum mechanics. Metals as made up of lattices of metal atoms, packed at very close distances. The outermost and least tightly bound (to the nucleus) electrons, the valence electrons, occupy atomic orbitals of the least energy. Due to the close vicinity of the atoms and the similarity in ...

0

Actually, the metal is totally filled with positive nuclei and negative electrons, but the outer electrons are quite able to move about from one atom to another, because of the way all the atoms are packed in a lattice. That's why metals conduct heat so well. The electrons are like a gas filling the metal, and they can easily carry momentum. They can't get ...

0

We remove an overall constant for simplicity. Let us use cylindrical coordinates $(\rho,\phi,z)$, where $$\tag{1} x ~=~\rho \cos\phi, \qquad y ~=~\rho \sin\phi .$$ Also assume the standard metric $$\tag{2} ds^2~=~\mathrm{d}x\odot \mathrm{d}x +\mathrm{d}y\odot \mathrm{d}y +\mathrm{d}z\odot \mathrm{d}z ~=~\mathrm{d}\rho\odot \mathrm{d}\rho ... 0 So, I think I got it thanks to the tips you two gave me. As you mentioned, my single electric fields weren't incorrect, but I have to put them into the same coordinate system. The electric field for the wire running along the x-axis should look something like this then: ... 0 I doubt that your expression is correct. Your original equation is of the form \begin{equation*} \partial _{\mathbf{x}}\cdot \mathbf{E(x})=\rho (\mathbf{x}) \end{equation*} where \rho (\mathbf{x}) vanishes away from the x_{3}-axis. You can write \begin{equation*} \mathbf{E(x})=\mathbf{E}_{1}\mathbf{(x})+\mathbf{E}_{2}\mathbf{(x})=\partial ... 0 The principle in general is called superposition in physics or linearity in mathematics. It is very useful when you want to study a system for that system to be approximately linear. What linearity is, in a more general context. Here "linear" is the property of a function (or an operator, or whatever) to distribute over addition. So for example if you have ... 1 "I find lots of solutions on the internet that say you can replace the cavity with a negative density, why?" Because they use a trick to calculate the potential easier. They assume that the empty hole is neutral, but composed of a positive charge density equal to that of the sphere plus a negative charge density of the same amount. In this way you can ... 0 When dealing with infinitely long line charges (basically a cylindrical geometry) calculating the potential relative to infinity becomes a problem. You have to establish a reference (ground/earth) at a finite location. So, your result of an infinite potential difference is not incorrect, although it is confusing the first time you see it. This site ... 0 I think we should start with the local form of Gauss's law \nabla.\vec{E}=\rho Now$$ \int \nabla.\vec{E}\,dv=\int \rho\,dv$$Using Gauss's divergence theorem we have$$\int\vec{E}.\vec{ds}=q$$I assume \epsilon_{0} to be 1 but you can always put that back into this. I think this way of looking at it does not assume any coordinate dependence.. Ofcourse ... 0 There's a problem in this equation:$$ d\vec{l}=dy \cdot -\vec{j} $$here dy needs a minus sign. It is easier to see this writing the path. The path of integration is parameterized by:$$ \vec{l}=\vec{P}+(\vec{N}-\vec{P})\frac{y-y_P}{y_N-y_P} = \vec{P}+\hat{j}(y-7a) \\ y_p \le y \le y_N $$Therefore$$ d\vec{l}=\hat{j}dy $$and$$ \int_P^N E_y \cdot ...

0

An external electrical field leads to rearrangement of the charges, and this cancels the field inside. Electric fields (applied externally) create forces on electrons in the conductor, creating a current, which will further result in charge rearrangement. The current will cease when the charges rearrange and the applied field inside is canceled.

1

There is not really a difference between the two lines of reasoning. In each case you are taking a path integral along a chosen path and calculating all of the contributions of $\vec{E}\cdot d\vec{l}$. You could choose an arbitrary path, and this would still be the case, but the integral would be more complicated. An easy way to think about this is to use ...

1

Why are potential differences equal across two capacitors in series, but charge on each capacitor is not? This is based on a false premise. There is no rule that says that "potential differences are equal across two capacitors in series". In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor ...

1

For Capacitors, the charge stored in it is directly proportional to the potential difference across it. Hence, the charge stored in the capacitor is given by the relation $$Q=CV$$ where C is a constant known as capacitance which is an inherent property of the capacitor. In a parallel combination, the charge through each capacitor has the same entry and exit ...

0

as the plane is non conducting, the lines that you draw for the electric field should not stop on the plane. therefore a line that starts from the charge, then goes towards the plane, should at some point be deviated and at very large distances be almost parallel to the plane. if you would put a negative charge for the image charge, then remove the plane, ...

0

The dipole moment of a continuous charge distribution is given by $$\mathbf{p} = \int\mathrm{d}^3\mathbf{r} \; \mathbf{r} \rho(\mathbf{r}),$$ (the moment is taken with respect to the point $\mathbf{r} = 0$). For a displaced charge distribution $\rho'(\mathbf{r}) = \rho(\mathbf{r} - \mathbf{b})$, you can use a change of integration variables to show that ...

1

When the point charge is not at the center of the sphere, the electric field lines will not intersect the sphere at right angles. Consequently, there is an initial component of electric field along the surface of a conductor. We know this results in a force on the charge carriers inside the conductor, and these charge carriers will re-arrange until the ...

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