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Of course you can define such a quantity, but the question is: does it mean anything physically? Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great ...

5

The force on the conductor must be zero. We will solve the problem in two steps. First, we will write down the external force $d\mathbf{F}$ on each infinitessimal charge $dq$ in terms of the external field $\mathbf{E}_{ext}$ and then we will integrate $d\mathbf{F}$ to get the total force. Note we need only consider the external force (i.e., the force from ...

4

This is true, however the bending is not much until one comes closer to the edge so it is usually neglected or too small to depict. Here's what I get when I simulate the system: This is the same system as a vector plot:

4

What is the electrical potential difference and why we have to talk about a difference and not about the electrical potential itself? Mathematically, the reason is that the force is proportional the gradient of a (not the) potential function. $$\vec F = -\nabla \phi$$ Note that a potential that differs by an additive constant $$\phi' = \phi + C$$ ...

4

Yes, applying an electric field does create a pH gradient and in fact you can observe this simply by adding a suitable indicator to your system. For example see the section Demonstration of pH Gradient Formation in this article.

3

I don't think it is that tough to analyse. If a conductor is present in a uniform electric field then there will be redistribution of charges to counter Electric Field inside the conductor (so that the net field inside the conductor is zero). However in uniform electric field this redistribution of charges will not cause any net force on the conductor. Why? ...

3

(this is a partial answer) One example is the preceding work Barrett, J. W. "The asymmetric monopole and non-newtonian forces." Nature 341.6238 (1989): 131-132. doi:10.1038/341131a0 which is Ref. 13 in the Connes et al. paper. This paper contains one example of asymmetric monopole produced by rotating figure shown about the horizontal axis passing ...

3

Mass and charge are not so similar for the charge having "center of charge". The notion of "center of mass" appears in many applications when number of bodies move. In this situation, the movement can be splitted into movement of center of mass and individual movements of bodies relative to the center of mass. This occurs because of dual role of mass: it ...

2

Your calculation is nice, and the answer seems correct. Another way (less nice) to calculate it is to imagine the charge in the center as a small charged ball (to avoid infinities) and calculate by how much is the field energy lower when it is in the center than when it is far away. Due to symmetry and the Gauss law, the electric field in the former case is ...

2

Assume the contrary,suppose a point exists such that the local charge density is positive,say point A. Now from Gauss' law the total charge on the inner surface is negative.So there must exist a point B at which the local charge density is negative(otherwise the net charge will be positive). Now consider the field line from point A.It will originate from ...

2

I have a guess, although I don't know if it is correct. If you model the globe as a simple insulating circular glass shell with a constant spatial charge density embedded in the glass in the immediate vicinity of the finger and solve Maxwell's equations numerically for the potential, you observe something like this: The rationale for placing a nontrivial ...

2

I first show that we can write $$\Phi_j=\sum_{k=1}^NP_{jk}Q_k\tag{1}$$ for a given configuration of conductors; then it it will be straightforward to deduce $$Q_i = \sum_j k_{ij}V_j$$. To prove $(1)$ we draw on two concepts: principle of superposition, and the uniqueness of the solutions of electrostatics problems. Consider $N$ isolated conductors. ...

2

I think that "because of linearity" should be read as "because of the superposition principle" (which does rely on the linear response of the dielectrics). You do not need to go into the detail of the field distribution: As in your example, set all potentials to 0, except for $V_i$. Denote this situation by $(i)$. The corresponding charges on each ...

2

I'll combine the comments into a guiding answer with some extra details here and there. An empty space containing a single point particle with charge $q$ is defined by two properties: The system's charge distribution $\rho(\vec{r})$ is zero everywhere except at the location $\vec{r}_0$ of the point particle. The integral over the entire space of the ...

1

If the wire is flexible, you could change its bounded area $A$, thus changing the magnetic flux. I'm imagining a "closed" loop where the two ends of the wire meet up. In your case of case of a uniform field that's perpendicular to plane of the wire, $\Phi_B=\pm BA$, depending on your choice for the direction of the corresponding area vector. Then, if you ...

1

Faraday's law of induction can be written as $$\oint\limits_C \vec{E}\text{ }d\vec{l} = - \frac{d}{dt}\int\limits_A \vec{B}\text{ }d\vec{A}$$ where $C$ is some closed curve and $A$ is the area bounded by it. Recalling that the electric field is the gradient of the electric potential (i.e. something like a derivative) and that voltage is nothing but a ...

1

Actually, cell phones do work in Faraday cages these days. What happens is that the conductor in the cage is not ideal, and there is some amount of leakage of electromagnetic radiation to and from the inside of the cage, specially at high frequencies. In order for the cage to be perfectly blocking it would need to have no holes at all (hence it is no longer ...

1

No, glass and indeed all amorphous materials do not exhibit piezoelectricity because piezoelectricity is intimately connected to the crystal structure of the material. Roughly speaking, if the charges within the unit cell are asymetrically distributed then when the crystal is mechanically deformed the positive and negative charges may be displaced by ...

1

The "center-of-charge" is part of a more general concept that is used quite often in physics: Multipole expansion. The general idea of multipole expansion is the following: If you view a charge (or mass) distribution from a large distance, then most of its internal structure is irrelevant to you. Instead, it suffices to do all calculations based on a few, ...

1

The Landau Pole is not a problem for QED because at scales much smaller than it (the Planck scale, which is smaller than the Landau pole by 260 orders of magnitude) the (negative) gravitational self-energy of the particle will more than cancel out its electromagnetic self-energy. So string theory is not necessary in this case, just gravity.

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