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-2

The better way of looking at it would be from the macroscopic point of view.In a equipotential surface a particle that was under acceleration due to external force now no longer needs this external force to be applied. Hence the value of force become 0, resulting in zero work done...


0

you can either evalute the integral numerically or search for a good coordinate transformation to evaluate it "by hand". This transformation is the one to spherical coordinates, though. Also you could try this: \begin{align} Q & = 2 \pi A \int_{-\infty}^{\infty}\int_0^{\infty} r \cdot H(R^2-r^2-z^2)~dr dz \\ & = 2 \pi A ...


0

You continue as follows \begin{align*} Q &= 2\pi A\int_0 ^{\infty} \rho d\rho \left(\int_{-\infty}^{+\infty} dz H(R^2 - \rho^2 - z^2)\right) \\ &= 2\pi A \int_0 ^R \rho d\rho \left(\int_{-\sqrt{R^2 - \rho^2}}^{+\sqrt{R^2 - \rho^2}} dz \right) \\ & = 2\pi A \left(\frac{2}{3}R^3\right) \end{align*} Where we used \begin{align*} H\left(R^2 - \rho^2 ...


2

My guess; you are mixing up quadripoles and quadrupoles. Quadripoles are two-port networks used in electric circuit analysis. The original German word is "Vierpol Theorie", which means Four-pol because of 4 Poles. https://en.wikipedia.org/wiki/Two-port_network Quadrupoles are related to multipole expansion used in electromagnetic, atomic orbital,.. theory. ...


1

Without equations: The ideal dipole is made up of two oppositely charged particles infinitely close to each other. So we can immediately deduce that if the electric field does not change along the direction of the dipole, it exerts no force (because the force it exerts at the positive particle will identically cancel that at the negative one). The only way ...


0

When you are inside a cylinder that is infinitely long and has uniform charge density, you can "look" in a particular direction. Following a narrow cone, you will arrive at a bit of the surface $dA$ that is distance $r$ away. This gives rise to an electric field along the $\vec{r}$ direction, with a magnitude given by the amount of charge $dQ = \sigma dA$ ...


1

Your brain/mind might be processing the information in the wrong way. As field is uniform, force remains constant and acceleration remains constant. You see, the acceleration remains constant, but velocity doesn't remain constant, as the time to which your proton is accelerated increases, your proton's velocity also increases. So, your proton will be ...


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NO the force does not change with distance as E=F/Q F=QE electric field is constant and charge is also constant then force is also constant.


0

First, when you found your volume it looks like you missed a sign. In my last integration over $\phi$, I had $$ \frac{4\pi}{3}\int^{\phi}_{0}d\phi R^3\sin\phi-\frac{d^3\sin\phi}{8\cos^3\phi}\Rightarrow$$ $$ \frac{4\pi}{3}\left[ -R^3\cos\phi - \frac{d^3}{16\cos^2\phi}\right]_0^{\cos\phi=\frac{d}{2R}}.$$ If you switch the sign on the second term, it looks ...


0

You cable consists of (at least) one conductor, usually a metal wire. In a conductor, electrons are free to move. An electrostatic field, which you must have encountered in your studies, moves electrons. So if on one end of the wire we create a suitable field, electrons along the entire conductor will be slightly pulled towards or pushed away from the end. ...


0

For a distribution of charges in space, the electrical potential, $\phi$ at a location in space can be calculated as the change in potential energy $\Delta U$ of a system of charges caused by adding a small ($\lim q\rightarrow 0$) charge $q$ to the system in that location, divided by the charge, i.e., $$\phi= {\tiny \lim q\rightarrow 0} \frac{\Delta U}{q}. ...


0

Yes you can have negative electrostatic potentials. Consider a system of 4 charges $q$ and four $-q$ organized in a cube of lengths $d$ such that no charges of the same sign are adjacent to each other. This system has potential energy $$U=-\frac{q^2}{\pi\epsilon_0 d}\left(3+\frac{1}{\sqrt{3}}-\frac{3}{\sqrt{2}}\right)\le 0$$ This is a model of a crystal of ...


4

Potentials are defined up to an arbitrary constant, so there is no particular meaning associated to a negative potential. It is the difference of potential that really matters, since the arbitrary constant is then washed away.


0

Compare With Known Positive/Negative Charge Find some material which you know takes up or gives away electrons. Bring it close to the object in question. If they are of like charge, they will come together, otherwise they will repel. The Lorentz Force Take advantage of the Lorentz Force and move the object in question near a strong magnet. Use the Lorentz ...


0

The earliest definition seems to be: http://en.wikipedia.org/wiki/Electric_charge In 1839, Michael Faraday showed that the apparent division between static electricity, current electricity, and bioelectricity was incorrect, and all were a consequence of the behavior of a single kind of electricity appearing in opposite polarities. It is arbitrary ...


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


3

Take a capacitor and put it across a battery. There will be a transient current as the electrons go towards the anode . This happens very fast and the current is small. If you short the capacitor with a wire, the battery will empty all its charge on the short, which, depending on the battery can really be damaging. Your body accumulates some charge which ...


8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


2

The whole electrical power grid is connected to ground. I don't know the details of other regions, but if you are in North America, the two current carrying conductors in a residential electrical outlet are called "hot" and "neutral". The "neutral" conductor is connected to the Earth at many places. If your bare feet touch wet Earth, and your hand touches ...


1

If you are not in a complete electrical circuit, any electric shock caused by touching a charged object or wire is brief. These "static shocks" are slightly painful, but they are (rarely) dangerous or fatal. I'm sure you've experienced a minor static shock. By wearing insulating footwear, you break a complete circuit and forbid a flow of electricity from ...


0

I don't think this is the case since grounding only defines the potential of the plate to be 0. There's more. An ideal ground can supply / absorb an arbitrary amount of charge while remaining at zero potential. "A grounded conductor is a special type of equipotential: infinite amounts of electrical charge (± Q) can flow from / to ground to / from ...


0

I would say- in ideal static case - conductor has no charge gradient inside (and total charge is 0, as you say). The electric field of both systems, grounded and not-grounded, will be the same, assuming that the plate is in both cases on zero potential (Q=0).


0

It depends where and how you ground your conducting plate. If it is grounded at infinity, where the potential of your charge is zero, it should not change anything over an already (net) uncharged plate. However, if you tie any bit of the plate that would otherwise be at a nonzero potential to zero, you will cause the plate to become charged as a consequence, ...


0

The field lines either has to be attracted if chargers are opposite or repulsed if chargers are same. Hence can never cross


0

All the above answers are correct,although none gives you an answer as to WHY you should NOT use a sphere and none addresses your "infinite gaussian surface" problem and you seem to be a bit confused on how things actually work(you have to understand how things work before you delve into the mathematics part). So,if you use a sphere,then your integral of ...


0

I want to respond to your edited question. Firstly, if your annular ring is a conductor, you don't ground a side, you ground the whole thing (because it is an equipotential surface), and then yes an amount of charge equal to the charge in the center flows away. For any conductor, there is a charge distribution $\rho_1$ associated with it being charged but ...


2

If you had a spherical piece of paper, any point on the paper would be surrounded by paper on two dimensions. You could cut out a little circle with that point in the center. If you had a normal sheet of paper, most of the paper would be like that, but there'd be a boundary where the points only have paper on one side and you could only cut out a semicircle. ...


1

The best rubber material to use is pure gum rubber. Rubber from an inner tube (tire) contains carbon and balloons may have other colorants which are conductive and will not work well. A wider belt will carry more charge. The 'combs' or charge pick-offs at the base and collecting sphere should not touch the belt but come very close - about 1/16". The tube ...


4

The product of the permittivity and permeability is encoded into the geometry of spacetime because the product $\varepsilon_0\mu_0 = 1/c^2$ and the speed of light is special. So the value of the product is telling us about the geometry of spacetime. The relative values of $\varepsilon_0$ and $\mu_0$ tell us about the relative strengths of the electric and ...


0

Technically the resistance of a wire is never 0. Typical wires of copper have electrical conductivity of $10^{7} S/m$ at room temperature. So there is indeed a very slight potential drop but highly negligible compared to the one that would be experience at a proper resistor. As for the problem of electrostatic, indeed from a single charged particle the ...


1

Start from the assumption that space is isotropic (independent of direction). Under that assumption, if you rotate your imagined empty-universe-with-two-charges by any angle around the line joining the two charges, then you will wind up with exactly the same physical system you started with. In particular, the charge configuration is unchanged. Now, if the ...


1

Since the sphere has no electrical properties, I am taking it to just be a mathematical surface. In this case, we are just computing a special case of the electric flux through a disk of radius $R$ from a charge centered on its axis a distance $\ell$ away which is $$ \Phi = \int_0^R E(r) \cos\theta\,dS = \int_0^R dr \frac{Q}{\ell^2 + r^2} (2\pi ...


1

The solid angle approach works, it's just when using that formula you need the full charge. The distance from the charge to the edge of the disc formed by your hemisphere is just $R\sqrt{2}$. So draw a sphere of radius $R\sqrt{2}$ around your point charge. The total flux through this sphere will be the full $\frac{Q}{\epsilon_0}$. Because the electric field ...


0

A point charge is given at a single, specific location in space. Think of it as an idealized source of charge. At the point where r = (0, 0, 0), electric field is zero and V(r) = $\infty$. As the value of r approaches 0 from any direction, V increases towards infinity but it is never discontinuous. For any other surface, the concept is well-explained by ...


0

Consider a point charge $q$ at $(0,0,0)$, the potential at $\bf{r}$ is given by $V(\bf{r})$ $= \frac{q}{4\pi\epsilon_0r}$. If you consider a path through $(0,0,0)$, you encounter a discontinuity in electric field and the direction of field changes. Is the potential for the field of point charge discontinuous at the location of the point charge? ...


0

There are also electric fields outside of a real capacitor as well, any capacitor with finite-sized plates. The energy in a capacitor is stored in the electric field, and since some of the electric field is outside the plates, some of the energy is also already outside the plates already. Imagine a bunch of surfaces everywhere in space that are orthogonal ...


2

In a parallel plate capacitor there's accumulation of electrons on one side and lack of them on the other. Since one plate is in front of another, the fields on each are equal in magnitude and opposite and therefore the field lines are straight (away from the boundaries) and cancel. In a battery the field is chemically produced inside the structure of the ...


0

As far as I understand the question. The question is when EMF source is attached and detached. EMF Source Attached. Consider the EMF source to be a battery in a circuit.A battery maintains a potential difference between its ends and hence provides an electrostatic force for the charges in the circuit. EMF Source Unattached. When the EMF source (battery) ...


0

The capacitor plates are made of some kind of conductor having non-zero thickness. The charges on the plate will reside on the surface of the conductor that faces the opposite plate. We say that these are surface charges. These are the only free charges in the system. We model them as a "sheet of charge". So the same description actually applies to both ...


0

Just answering the second question (since the first was well covered already): Grounding works the same that it always does - it means there is no longer a constraint that the net charge on the ring is zero. Charges will distribute to minimize energy - if there is a negative charge in the center of the ring, negative charges will try to get as far away as ...


2

The total energy can be expressed as the sum of the pairwise energys excluding the self-interaction energy. $$W=\frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{r_{ij}}=\frac{1}{2}\sum_iq_i\sum_{j\neq i}\frac{q_j}{r_{ij}}=\frac{1}{2}\sum_iq_iV_i,$$ where $r_{ij}$ is the distance between two point charge. The factor $\frac{1}{2}$ comes from the fact that the summation ...


0

The Poynting formula for electrostatic energy in volume $V$ $$ E = \int_V \frac{1}{2}\epsilon_0 E^2 dV $$ can be derived from the Coulomb law only for cases where the field acting on the particles is defined everywhere. However, point particle has infinite charge density at the point it is present and the field is not defined at that point. So the ...


-2

The direction is wrt the applied field....dipole opposes the field hence p has direction from negative to positive...


1

I believe the problem with your proof is that packing an area with spheres only takes up about 75% of a volume, and this problem doesn't disappear when you shrink the spheres down to infinitesimal sizes. No matter how small you shrink the spheres, those same packing properties should hold. Your proof relies on saying that the integral over a large surface ...


0

Think of charges as a bunch of $+1$s (protons) and $-1$s (electrons). It doesn't matter how many $0$s (uncharged particles) we have because $0 + 0 + 0 + 0 +... +0 = 0$. Now in most cases it's pretty difficult to change the number of $+1$s a body has, so we can see if it has more $-1$s than $+1$s to know if it's negatively charged or not. Let's call the ...


0

You get the electric shock in exactly the same way you get it on the ground: walking around, or (as @Glen suggested) by rubbing the blankets you accumulate electric charge on your body. This charge is transferred to the rest of the system (the plane or the earth) when you touch the knob. Glen's link explains how the charge forms in the first place. Your ...


-1

the positively charged particles attracted the negatively charged particles so when dust particles fly near the t.v. screen.the screen induces the charge in the dust particles pulling negatively charged particles closer to it.


1

If you connect two lossless capacitors in parallel, charge will flow back and forth. The loop formed by them will store energy in the form of magnetic energy (due to the current flowing) and you will end up with a resonant circuit. The frequency of resonance will be determined by the series capacitance and the inductance of the loop. So there will not be a ...


5

So in the first case, when talk about a plain circular ring, I assume you mean an annular ring, with a well defined inner radius and a different well defined out radius. With a positive charge at the center of the annular ring, positive charges will be repelled outward and negative charges attracted inward. Incidentally, not all the positive charge will go ...


1

It sounds like you may have seen a Blue Jet. I don't know if you can see these with the naked eye at ground level but the article linked seems to imply that you may be able to. Assuming a spherical earth of 3,960mi radius and 24,900mi circumference, you can work out how far something needs to be along the surface of the earth before you can't see it over ...



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