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The water molecule is neutral on overall basis, i.e: the water molecule as a whole has no net charge. The water molecule is not linear rather it has a bent shape with two hydrogens on the same side. This happens because of the lone-pair-bond-pair repulsions. The oxygen has is a more electronegative element than hydrogen, i.e: oxygen has high electron-...


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The converse is not a "theorem" in that you can't prove that it is true, even assuming Maxwell's equations as axioms. It is simply a "hunch" that $|\vec{S}|$ represents the power intensity and $U=\frac{1}{2}\,\epsilon\,|\vec{E}|^2+\frac{1}{2}\,\mu\,|\vec{H}|^2$ the energy density. What you can prove (and what you already understand) from Maxwell's equations ...


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For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is ...


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Imagine you are looking through a telescope that zooms in and only lets you see a very small region of the plane. Let's think about what happens to the electric field from the area you are looking at as you move the plane away from you, while you keep the direction of your telescope fixed. If you double your distance from the plane two things happen, and ...


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The detailed answer by NowIGetToLearnWhatAHeadIs (and Andrea) is the correct one and I will leave my answer as it is so it can be viewed by other users who might have thought of this themselves in order to understand the fundamental mistake they make. This answer Implies that this goes for any kind of force field with a source of similar geometry, while the ...


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A flame has many charged particles (ions and electrons) within it. THose with the appropriate sign are attracted to a charged object and neutralise it. There is a standard demonstration to show that there are charges of both signs within a flame. A candle is placed between two vertical conducting plates which have a potential difference of about 1000 V ...


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It might help to think of the energy of the electric field as a kind of potential energy. If you bring two like charges together, that takes energy. They're trying to repel each other. Thus, once they're brought together, they have a large amount of potential energy. One way to account for that energy is to say each particle has some potential energy. ...


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The result is almost correct, the divergence has the term $$ -\frac{ b\exp(-br)}{r^2} $$ which looks simple but it is not the behavior of the Yukawa potential which only has $1/r$, not $1/r^2$, in 3+1 dimensions. Near $r=0$, the most singular term with the $\exp(-br)\sim 1$ behaves as $E_r=1/r^2$ which is the same as in the Coulomb potential $V\sim -1/r$ ...


2

A dipole may be represented by a pair of nearby charges of the opposite sign (but the same absolute values), $+Q$ and $-Q$ at a distance $\vec r$ from each other. At the distance $R$ from the dipole (and I will only consider the case when we measure the field on the axis that includes the dipole itself), one adds $E$ going like $+1/R^2$ from $+Q$, and $-1/(...


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Looks like, in general, $\frac{\partial E_x}{\partial y}\neq\frac{\partial E_y}{\partial x}$, so $\overrightarrow{E}$ cannot be a gradient of any decent function. Therefore, I don't think the problem has an unambiguous answer.


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Here is another attempt at an answer, from a slightly different angle. It is less sophisticated and may add insight. You say that the capacitors are isolated before they are connected. Literally, that is only possible if they are in different universes. In the real world, they are connected through an infinitesimal capacitance - the capacitance between the ...


2

Start by noting that the electrical potential is an energy per unit charge. In an electric field $E$ the field produces a force on a charge $Q$ of: $$ F=EQ $$ so if we move the charge a distance $dr$ the work done is just force times distance or: $$ W=EQ\,dr $$ The work done per unit charge is $E\,dr$, and this is what we mean by the change in the ...


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With static electricity, the electrons cannot move because the material used is an insulator. Hence there is no current. If the material were conductive, then a current would flow, and there would be no accumulation of charge. Electromagnetic fields will induce a voltage in a conductor, so there will be a current as well. You also need to remember that ...


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This is a more general answer. Atoms and molecules are quantum mechanical entities. This means that the "shape" of atoms depends on the solution of quantum mechanical equations, which give probabilities for locating in space the electrons that are bound to the nucleus of the atom with the electric potential provided by the protons of the nucleus. The ...


2

Your statement is correct. The charge distribution is such that the hollow cavity of the conductor has a equal amount of negative charge induced on its inner part. This distribution is such that field due the cavity (including the charge inside the cavity) cancels out everywhere outside the cavity. So looking it the other way around external sources do not ...


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The idea that in a dipole, the further you remove something, the stronger the interaction becomes is equally counter-intuitive to me. The dipole moment does not define an 'interaction'. You can just interpret it to be a useful mathematical quantity - nothing more, nothing less. For example, if I removed two charges to infinity from each other, their ...


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You can always see potential as an 'energy difference' between two points. For example in some physical problem you choose a certain zero point of energy and you take that as a reference point for the other energies in the problem. You can define that the potential energy of a mass is zero when it is at a certain height and then calculate the potential ...


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Field is not any abstract fairy-tale 'mathematical fiction' concept; it has momentum and energy density; it exchanges momentum and energy with charged entities. Energy conservation is a local process which evidently implies electromagnetic field between two interacting charges must mediate the energy and momentum exchange between the charges and hence ...


2

The phenomenon you are talking about is called dielectric absorption. The way it works is this: Let's say you've just discharged a capacitor. An ideal capacitor would remain at zero volts after this. However, in real life, the capacitor will develop a small voltage from time-delayed dipole discharging (also known as dielectric relaxation). Dielectric ...


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You are not completely mistaken. What you have to do to get the desired formula is a Taylor expansion of the term which contains $R/x$ and then consider the limit when $x \to \infty$. Also be careful with notation, $\sigma=Q/ (\pi R^2)$. As Andrea Di Biagio mentions in his comment, $1/(1-u) \approx 1 - u$ when $u$ is small. In your situation $u=R^2/x^2$. ...


1

An electron gun is used to shoot electrons at the ink which then gives the ink droplets a negative charge, varying based on where the ink needs to go. Then, the charged ink droplet passes between two metal plates, which deflect the ink to its appropriate location on the paper. The ink does not acquire charge from the metal plates, but from the electron gun. ...


1

An isolated charge (like, an electron) is produced by ionizing an atom, using energy to pull a single electron free of the atom and pulling that charge far from the opposite-charge ion. So, it DOES take energy to isolate the charge. The isolated charge has an E field around it, but the original uncharged atom had none. Similarly, when you apply a ...


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Indeed, there must be dipole moments of such magnitude occurring (if you just base it on the definition)! But usually, there is no extensive electric field to support the continuous rotation of the dipoles, because the net electric field in space is zero. Though there will occur some abrupt torque (again, based on definition) of intense magnitude due to the ...


1

I have found this clear answer to your question: I mean Raziman T.V.'s explanation. Enjoy it! Here's the answer: Feynman explains this well in his lectures (Chapter 13 : Work and Potential energy (A)). The explanation is for the gravitational field of a plane, but the same argument works for the electric field of a charged sheet as well. The ...


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The hollow neutral conducting sphere will experience an induced charge separation if there is a point charge in the cavity. If the charge is negative, then negative charge in the neutral conductor is repelled and will move far away. Positive charge is attracted and moves closer. So the inside surface will be positively charged and the outside surface ...


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Your reasoning is faulty. It is true that in electro-STATICS (ie when there is no movement of electric charges) the electric field inside a conductor is zero. But this is not the case for electro-DYNAMICS (ie when charges are in motion), eg when there is an electrical current flowing. It is also true that, for alternating currents, the current becomes ...


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Often the easiest way to do such problems is using potential rather than the electric field. And always, it is best to use the appropriate coordinates. In this case, cylindrical coordinates with the rod at $\rho=0$ and extending from $z=-L/2$ to $z=+L/2$ are natural. So what you do is say that for an element of rod with a tiny length $d\ell$ and charge $\...


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Charges at rest move when a force is applied on them and this is due to Newton's laws. Now to apply a force, we need a field, like electric/gravitational field. Each field acts upon certain measurable properties of a system, like gravitational on mass, electric on charge etc. Now potential is just a fancy name of height in electromagnetism. I hope you're ...


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If you found the total x and y component of the total field just use: Magnitude E_{tot}: $$E_{tot}=sqrt(E^2_{tot-x}+E^2_{tot-y})$$ The direction in angle: $$\theta = arctan(\frac{E_{tot-y}}{E_{tot-x}})$$ The magnitude of force would just be $F=qE$ Note: you should be careful in angles as it may sometimes not the angle measured from +x as the norm, for ...


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a) Add the x and y components separately, then calculate the magnitude E (the resultant field) and the angle it makes with the x axis. b) The force is qE, in the same direction as E.



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