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14

Charge is a fundamental conserved property of particles. It is, if you like, a measure of how much a particle interacts with electromagnetic fields. A particle with charge can produce and be affected by electromagnetic fields. This is what we mean when we say a particle has charge. Its a simple quantised way to measure the coupling strength of particles with ...


13

Electric field lines are a visualization of the electrical vector field. At each point, the direction (tangent) of the field line is in the direction of the electric field. At each point in space (in the absence of any charge), the electric field has a single direction, whereas crossing field lines would somehow indicate the electric field pointing in two ...


9

If you put your rod in a ultra high vacuum it will stay charged almost forever, but since you probably keep it exposed to air, this is where the electron excess slowly migrates (and the same for the electron defect in the silk). Since the charge exchange requires an hit between an air molecule and a spot of the rod where an electron excess is present, and ...


8

In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where $\mathbf{E}=\mathbf{0}$. Letting the charges be located at $(\cos(2\pi k/N),\sin(2\pi k/N))$ for $k\in\{1,2,...,N\}$, we can generate plots of $|\mathbf{E}|^{-1}$ for various $N$. The zeros of ...


6

Charge is a quantity which arises from Noether's theorem, due to continuuous global symmetries (up to a total derivative) of an action, and as such we have many types of charge, other than electric. For example, consider the Dirac Lagrangian, $$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi$$ which describes fermions. It is invariant by a ...


6

One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...


5

The electric field at any point is the sum of all the fields due to each individual charge in the system. The field has a magnitude and a direction. The field lines are a representation of the magnitude and direction of the field over an illustrated area. The field lines point in the direction of the field. If lines from two sources were to cross, we could ...


4

Electric field lines reveal information about the direction (and the strength) of an electric field within a region of space. If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Every single location in ...


4

You can use $q/\epsilon_0$ to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing. Flux for any closed surface is $\Phi = \oint \vec{E} \cdot d\vec{A}$. There are two ways to calculate this quantity: The hard way, which means evaluating $\vec{E}$ on every part of the surface, and integrating. ...


4

There is indeed a connection. The holomorphy is easily seen in the electrostatic potential. In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified ...


3

A force is said to be conservative if its work along a trajectory to go from a point $A$ to a point $B$ is equal to the difference $U(A)-U(B)$ where $U$ is a function called potential energy. This implies that if $A \equiv B$ then there is no change in potential energy. This fact is independent of the increase or not of the kinetic energy. If a conservative ...


3

The switch really has 2 positions: on and off. However, when you move the switch very slowly, it may leave the closed position slowly. When the switch is just barely open, the field may cause the air to break down and start conducting, to form a spark (as @anna v explained). To rephrase, the reason why sparks happen is because the switch may only be open a ...


3

The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ...


3

The analogy follows with the right definitions. The "flux" of the "vector" $E(z)$ through a contour $\Gamma$ is $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$. I think you may have forgotten the conjugate in the relationship between the "Electric field" and the complex potential $\Omega$: $E(z) = -(\mathrm{d}_z \Omega(z))^*$. So it is the ...


2

John Rennie will probably have more details on the matter, but in general colloids (such as oil dispersed into soap water) are not so much stabilized by a net total charge of the mixture, but rather are stabilized by repulsions from separated charges. For example: This is a cartoon representation of what an oil droplet in soap water looks like; note that ...


2

You need to draw a Gaussian surface. In this case, you can use a rectangular box and look at the electric flux out the top and bottom of the surface of that Gaussian box. The usefulness of Gauss's is that it exploits symmetry in the problem allowing you to treat the electric field as a constant when you take the surface integral of E dot da. Which equals Q ...


2

Air is a bad conductor up to a certain value of the field generated by charges and the distance between them. After that air breaks down and a discharge happens, i.e. sparks. So below this level charges can accumulate by rubbing for example , positive ions left on one surface and negative on the other. When brought close a spark occurs. Why does holding ...


2

Field lines are a visual representation of a mathematical construct, like a graph of a function. The defining properties of this visual representation are The field lines run parallel to the field at every point. The density of the field lines in an area is proportional to the strength of the field. The second property tells you that the field lines can ...


2

You start with a total of $3Q$ on the outer plates - and that charge has nowhere to go. So when you connect the two plates together, they will share the charge equally, and have a net charge of $1.5Q$ on each. Grounding the center plate means that charge can flow freely. So what will happen? First - charge will flow between plate B and ground. The boundary ...


2

You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$: $$ \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ...


2

I presume you are referring to process of nuclear fission of uranium-235, which has the equation: $$^1_0\text{n}+^{235}_{\ \ 92}\text{U}\longrightarrow ^{236}_{\ \ 92}\text{U}$$ However, a subsequent reaction is: $$^{236}_{\ \ 92}\text{U}\longrightarrow^{144}_{\ \ 56}\text{Ba}+^{89}_{36}\text{Kr}+3^{1}_{0}\text{n}$$ The production of neutrons is a feature of ...


2

Remember that, if the net force is zero, velocity is constant (not necessarily zero!). You only need to do push a tiny bit harder for a tiny bit of time to start moving the charge. This extra amount can safely be ignored. Once the charge is moving with some nonzero velocity, equal force is enough.


2

Coulomb's law is indeed a special case between two point charges. to find the force between a point charge and a plate, you would have to integrate the equation over the plate surface to calculate the contributions from all infinitesimal charge elements. It's more practical to figure out what the electric field is, and then use $\vec{F}_e = q\vec{E}$ to ...


1

The whole (pedagogical) point of the slide wire generator is to illustrate that not only do changes in the magnetic field generate current in the loop, changes in the area of the loop - in a constant magnetic field - also generate a current. It's the change in magnetic flux that matters. As long as the wire is moving with some velocity, the magnetic flux ...


1

It does make sense to talk about the number of field lines, but only if you take care to represent the field amplitude as being inversely proportional to the spacing between the lines. With that, the total number of lines crossing a surface is proportional to the flux, etc. Some people, notably textbook authors Chabay and Sherwood, feel that the field line ...


1

No, your reasoning is incorrect, because there's no reason for the forces to cancel in general. Actually, the charge in general will be attracted by the field of the induced opposite charge on the inside surface of the conductor. This is easy to see by use of the fact that $\nabla^2 V=0$ in the region devoid of charges implies that $V$ is a harmonic ...


1

The answer to your question is "no". Put the point charge close to the wall at some spot of the wall. There will a surface charge be collected at this spot of the wall that attracts the point charge. In the following I give an example for which one could even calculate the attracting force analytically. Nevertheless, I keep a bit informal here since the ...


1

The lightest gas that is stable at room temperatue is $H_2$ (two hydrogen atoms bonded to each other). At very high temperature and/or low pressure, the hydrogen molecule dissociates to become atomic hydrogen. At even higher temperature, the hydrogen ionizes. The electron is no longer bound to the proton. This is a plasma state. Electrons and protons ...


1

Yes, the uniqueness theorem guarantees it. If you find a solution in which the potential on the surface of the cube looks just like the potential of a point particle, then the potential outside the cube must be identical to that of a point particle since there is only one unique solution satisfying this this boundary condition. (Assuming no external ...


1

Suppose you have a charged cube as shown below, Say it holds a charge of $Q$ Coulomb. Now, for this cube to behave like a point charge, you need to go very $far$ away from it such that at a particular position, your cube appears like a point charge. In this case, the cube $behaves$ just like a point charge with a charge of $Q$ Coulomb. It is clear ...



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