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1

What you are asking is (I think) can one hemisphere exert a force on itself? Yes, of course the charges on one hemisphere repel each other, leading to an outward force which tries to expand the hemisphere. However, as you are considering each hemisphere to be a rigid body, internal forces (eg between adjacent atoms) oppose such an expansion. The same kind ...

1

The macroscopic electric field inside the 'metal' of the conductor is zero in electrostatic conditions. In a hollow cylinder , if a positive charge is place inside the cavity, the field is non zero inside the cavity. Again, the interior of a hollow shell can hold the positive charge there, because of the induced charges on the inner wall of the cavity. ...

1

The dielectric has to be in the space where the electric field is. In an ideal capacitor the fied is only between the two plates. So you can put in an dielectric to increase the capacity of the capacitor. The outside space doesn't matter because there's no relevant field that could be affected by a change of the dielectric constant.

2

If we look at nature from an empirical side, we can deduce the four Maxwell equations from experiments as: $$\nabla\vec{E}=4\pi k \rho$$ $$\nabla \times \vec{E} = - k'' \frac{\partial}{\partial t}\vec{B}$$ $$\nabla\vec{B}=0$$ $$\nabla \times \vec{B} =4 \pi k' \vec{j} + \frac{k'}{k} \frac{\partial}{\partial t}\vec{E}$$ The choice of the value of those ...

1

I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason: In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$) tend to zero. Then we arrive at the result that the ...

2

Gauß law for electrodynamics states that $$\nabla \vec{E}= \frac{\rho}{\varepsilon_0}$$ at any point and no matter by what the electric field $\vec{E}$ is produced. $\vec{E}$ here stands for the total electric field at a point produced by all sources that are present, not just any selected few. So the whole derivation is valid for any surface and any ...

0

I'm rather familiar with Van de Graaff generators, and the answer is no -- the Van de Graaff generator will not work without the belt. If you've ever opened up a Van de Graaff generator (and you should -- this is the whole point of one), you'll notice that a Van de Graaff has a comb and the bottom, and another one at the top. The purpose of the belt is to ...

2

If electrons obeyed classical mechanics, they would rearrange in a new configuration in order to maximize the distance between them. They would not stay still because of thermal motion, as pointed out by CuriousOne, but on average they will still maximize this distance. However, electrons don't obey classical mechanics, but quantum mechanics. The behavior ...

1

Electrons in metals have states which people call Bloch states. If we want to describe these states we should bear in mind that electrons have wave like behavior. Bloch states are a subset of the many infinite wave states which electrons can have in general. They're such that when an electron is in a Bloch state, it somehow fills everywhere in the metal. If ...

1

"Does it have to do with the attraction or repulsion between unlike and like charges?" Yes. The electric field always goes out from positive charges and in towards negative charges. The magnitude of the electric field of the +10e charge is independent of the test charge. $$E=\frac{Q}{4\pi\epsilon r^2}$$ However, the direction ...

8

If you're trying to simulate a 2D solution of the Laplace equation (which is the only unambiguous reading of your post as currently stated; if that's not what you're doing then you should clarify your question with exactly what it is you're doing and how), then your code is wrong. The reason is that your results don't obey the maximum principle: a harmonic ...

1

When you put +ve charge on a conductor, you are really removing the same amount of -ve charge, because only the electrons move. Gauss' Law assumes that charge is infinitely divisible, and can be spread uniformly throughout a volume or over a surface. This is a good approximation when the charge is on the order of $1 \mu C$, corresponding to about $10^{13}$ ...

4

I believe the answer to your question lies in the Gauss theorem itself $$\oint \textbf{E}d\textbf{S} \sim Q$$ and the symmetry of the system, which defines the shape of equipotential surfaces. In case of a point charge there is a rotational symmetry about any axis going through the charge, so the equipotential surfaces are spheres whose area is ...

4

I don't like the "you can't get away" explanation. There is a simple explanation with field lines: In all three cases, the field lines are straight lines from the point charge to infinity. You can easily calculate the density of the field lines for each object. For a point charge, the "number" of field lines through any sphere around the point charge is ...

0

A capacitor is supposed to have infinite dimensions.And Electric field strength of a charged plane sheet of infinite dimensions is constant over infinity i.e. distance does not matter.

0

You can see the system as the whole, and just state that its potential energy is $U(x_1,x_2,...,x_n)=\sum_{i\neq j} \frac{e^2}{4\pi\epsilon_0 |x_i-x_j|}$ The work required to move the configuration of charges to a different one $(x_1^{'},x_2^{'},...,x_n^{'})$ will then be $W=U(x_1^{'},x_2^{'},...,x_n^{'})-U(x_1,x_2,...,x_n)$ If only one charge gets moved,...

0

Those are correct. You don't need the second equation, the first equation is constructed from the second, with $\phi = 0$ at infinity. You can use the work done by the electric force: $$W = q_3(V_a - V_b)$$ Where $V_a = 0$ is the potential at the starting point, and $V_b$ at the end point. Note it has higher potential at $A$. the potential at infinity is $0$...

3

The principle of relativity: The laws of physics are the same in all inertial frames of reference. Since the Lorentz force is a valid law of physics, it will not change when we pass from one reference frame to another. First frame, wire is moving. There is no $\mathbf E$ field. Lorentz force $\mathbf F = q\mathbf v\times\mathbf B$. Apparently, you were OK ...

0

Using Gauss theorem we measure electric flux i.e. E.dA and if you use Gauss theorem for this particular geometry of plane parallel plates then you will get the usual result where the distance from the plate of the point where the field is to be measured will not appear. Further, E is proportional to 1/r^2 and the area ds is proportional to r^2 and while ...

1

For an INFINITE parallel plate capacitor, the electric field has the same value everywhere between the 2 plates. An intuitive reason for that is: suppose you have a small test charge +q at a distance $x$ away from the +ve plate and a distance $d-x$ away from the -ve plate. The +ve plate will repel the charge and the -ve plate will attract it. Now if the ...

2

"Water capacitors", where water is the dielectric, are commonly used in very high voltage pulse systems. For example, high-power nitrogen lasers commonly use water capacitors as their energy storage component. When used in these applications, a resin deionizer is used to dramatically reduce the conductivity of water. A great advantage of using water as a ...

0

Because it seems to constitute a complete answer, I am reproducing the comments of Curious (after some editing) as a Community Answer : Wires do have capacitance and charges accumulate. The total capacitance depends on the wire diameter and how far it is from other conductors. A good rule of thumb is to estimate the capacitance of a wire as $10-20 pF/m$....

1

In classical mechanics, all physical interactions are said to happen through collusions. According to Coulomb's law, the force seems to be magically acting from a distance, i.e: a charge at one location influences another charge far away from it without coming into contact with it. For the early scientific age, this seemed to be impossible and there was a ...

-1

$\underline{F}=Q*\underline{E}$ is valid for all situations, but $E=\frac{kq}{r^2}$ is just valid for point charges.

-1

According to me positive charge is present in nucleus so it will attract towards negative charge and -charge is present outward nucleus so attraction is radially outward.

2

1) This means that interaction between charged entities or matter takes place by the mediation of some force. According to standard particle theory, virtual photons are the mediators of electromagnetic interaction. Electric field is a certain region around a charge distribution, where it could influence a force on another charge. This means, the existence of ...

2

1) The electric field is not fundamental to the description of electrodynamics with point charges, one can take the point of view that electric charges simply interact at a distance with a force law proportional to the value of the electric field. 2) This is a bit of a loaded question in that any answer can normally be refuted, but the idea they're trying ...

2

$dl'$ is equivalent to "$d|\mathbf{r}|$", it is essentially a "scalar length measure". The electrodynamics integral you wrote here is a vector-valued integral, so no dotting happens. If you use a linear coordinate system, it may be evaluated as three scalar line integrals, one for each coordinate. Vector valued integrals cannot really be evaluated using a ...

3

The excess charge carriers in the conduction/valence band of Silicon (delocalized so that around silicon atoms there is a slight excess of local charge) are neutralized by the equal opposite charge of the randomly scattered dopants. Thus, the total charge remains zero, and this is actually the only way that an infinite crystal can have a finite ...

1

Doping introduces allowed energy states within the band gap of the material, and these energy states are very close to the energy band that corresponds to the dopant type. For example electron donor impurities create states near the conduction band while electron acceptor impurities create states near the valence band. The gap between these energy ...

0

If charges have to flow between sphere A and sphere B, you need a potential difference between the two. Since your question says that one of them is charaged and one of them is uncharged, there shall exist a potential difference between the two and hence there will be flow of charges but they needn't necessarily equalize. The amount of charge transfered ...

1

By analogy, you are comparing the properities of a solid material, in your first paragraph, to a gas mixture, the atmosphere, in the second. So the path taken would flow through the "weakest" or most conductive sections of air. No matter how small a volume of air you take, the conductivity/ionisation path is very unlikely to be consistent enough to ...

0

Let me tell you the most known phenomenon. You may try this even at home. Take a metallic tumbler and put your mobile phone in it then close the lid so that no part is exposed out of the metallic container. Now with another mobile try calling your first one. You should not be able to connect to the mobile inside the tumbler. This is called shielding. Here ...

0

There seems to be something wrong with the answer. My explanation is intuitive and quite long but worth it Charges outside spherical conductor cannot induce electric field inside a electrically isolated conductor due to electrostatic shielding Electrostatic Shielding-a region enclosed by meat of a conductor will not experience any electric field ...

0

Ohm's law is $$J=\sigma E$$ Substituting Ohm's law in $\nabla . E=\frac{\rho}{\epsilon}$ gives $$\nabla .J=\frac{\sigma \rho}{\epsilon}$$ The continuity equation (charge conservation) is $$\nabla . J=- \frac{\partial\rho}{\partial t}$$ Equating the above two equations gives \frac{\partial\rho}{\partial t}=-\frac{\sigma \...

3

The boundary conditions by themselves can't tell you anything about a conductor. The boundary conditions can't even tell which side of the surface has the conductor! One way to model a conductor is as an Ohmic conductor where there is a constant $\sigma$ (different than the surface charge density listed in your boundary conditions) and then you assert the ...

2

You need to use Ohms law: $J = \sigma E$ which has to be added to Maxwell's equations as a bulk observation, as explained by this answer. You can then conclude that the electric field is zero in a conductor for: perfect conductor where $\rho = 1/\sigma = 0$ and $J$ is finite static case where $J = 0$ and $\sigma$ is finite

1

No. The battery is already neutral, and remains neutral during operation. (-ve charge leaving one terminal has to be replaced at the other terminal.) If some -ve charge flowed to Earth the battery would become +ve, attracting electrons back to it.

1

Technically, you can neutralize the electrostatic potential of the entire battery this way. However, batteries do not primarily work by electrostatics. They work by creating a potential difference between the two terminals which encourages electrons to flow out of one (the negative side) and into the other (the positive side). This encouragement is ...

6

Initially, when first glass rods were systematically being rubbed, the "charging" phenomena was observed. The electric charges were hypothesized to be positive and negative, and the pioneer (Franklin? forgot the name...) pretty much arbitrarily decided to call one positive and the other negative. Further experiments helped him deduce that two like charges ...

16

The Hall Effect shows that negative charge is moving. In the Hall effect, one passes a current through a wide strip of metal exposed to a perpendicular magnetic field. If positive charges moved, we'd expect the positive charges to be travelling in the same direction as $\vec{I}$, and the magnetic force $q\vec{v}\times\vec{B}$ would be to the right. Thus, ...

11

Physics's don't know that only negatively-charged particles move. We can create ion currents on demand in many environments. We do know that the current flowing in a metal wire is negatively charged particles in motion. As for how to determine that, you do a Hall effect measurement. The measurement works by subjecting a current in a relatively wide bar to ...

0

The potential will remain constant and the the electric field at any point between the shells will be zero.

0

The electric force is just a vector, not a vector field. The Coulomb force formula you wrote depends on "some" position vectors $\vec r_1,\vec r_2$, much like fields $\Phi(\vec r)$ depend on the position vector $\vec r$, but it's a different kind of a dependence. First, a field must depend on a single position in space, $\vec r$, while the Coulomb force ...

1

It is not a gauge boson for a static field. Nothing is. It is the EM gauge boson, and the quanta of that field, representing an excitation. You can also say that it carries the EM force as it is exchanged between particles (though the word force already makes the statement more a classical analogy than a description of fact). More exactly any interaction ...

4

Technically "potential difference" is the difference in electrical potential, i.e. $\Delta V$, not the difference in electrical potential energy, $\Delta U$. Potential difference ($\Delta V$) is also called voltage, in certain contexts. However, many people and sources are sloppy about their terminology, and they will say just "potential" when they really ...

1

The mathematical definition that John Rennie gave explains it well, but I'd like to give an intuitive answer to your question. Imagine a rubber sheet horizontally stretched. The height of the rubber sheet at a point is equivalent to the electric potential at that point. Now, since as of now there is no disturbance at all, the height (potential) throughout ...

2

Among competing hypotheses, the one with the fewest assumptions should be selected. Some electrified objects repel, some attract. This can be explained by two kinds of charge. Nothing that cannot be explained by two charges can be explained by adding a third kind of charge. So we continue to describe electricity as occurring in two kinds.

1

The symmetry arises because of the boundary conditions, which are independent of $z$ and $r$. Let us place one conducting plate at $\theta=0$ with potential $\varphi=0$, and another at $\theta_0$ at potential $\varphi=V$. We now want to find $\varphi(r, \theta, z)$ in the gap. You are right that while $\varphi(z) = \varphi(-z)$ at arbitrary $(r,\theta,z)$, ...

0

1) Maybe this could provide you with some insight. This is from Griffiths Electrodynamics 3rd Edition. I'm not sure why the solutions you've found require the cosθ′ but this one doesn't include it. 2) When using spherical coordinates and trying to outline a full sphere, theta outlines a half-circle (existing in the xz plane), phi then rotates and ...

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