# Tag Info

0

Yes, the field outside the conducting surface is 0. If the net charge inside the conducting surface is 0, then the fields outside the surface are zero. It's basically the Faraday Cage effect. And the net charge inside the conducting surface is 0, because that's the definition of a capacitor. You might ask, "Um, well, but what if the net charge is not ...

0

For external points, cylinder behaves like a line of charge. So you just have to put the distance in the formula of electric field of a point charge as (radius + distance from the cylinders surface)

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The rods will get attracted towards the plane due to polarisation of the rods because of the plane.the rodsbwill surely not attract each other.They will repell each other as the induced charges have the same sign(+ or-).

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@Daniel Grissom gives a great answer, but I wanted to drill into the exact question you asked about the direction of flow. There are several points of note here: The comb is not in direct contact with the plates, but is outside it. This means that when the plates have the same charge they are trying to push that charge outwards, if it's energetically ...

2

Ignore the collecting combs and Leiden jars for the moment. As you noted, the Wimhurst disk and 45° shorting bars form a powerful electrostatic pump such that the left side of both disks gets charged one way and the right side of both disks gets charged the other way. This charge feeds back through the shorting bars and builds up until something leaks ...

0

The field of each quarter circle is evaluated separately about its symmetry axis: which is why the limits are taken the way the answer shows.

1

Yes, these thermally generated currents (Johnson noise) generate magnetic fields. This means that even non-magnetic materials generate a very-small magnetic noise if they are conductive. This actually places a limit on very-sensitive magnetic field measurements in shielded environments because the shields are usually conductive. The following Review of ...

4

Yes you are right. You end up having a varying electric field which generates a varying magnetic field which in turn generates an electric field etc... This causes a particular type of radiation called black body radiation.

9

What the picture shows is a corona discharge (see also Wikipedia). It isn't a circuit in the usual sense of the word. It happens because the voltage is so high that it raises the electron energy to above the work function and the electrons just leak off. In effect the coil is charging the air around it. The charge will end up on the furniture, walls, floor, ...

0

The closer the charge gets to the line, the larger the force between the charge and the line... So your teacher is wrong. Now I need to come up with some reason for that... Charge density approaches infinity as a charged wire is made thinner and thinner while keeping the charge same. Electric field on the surface of that wire approaches infinity as the ...

0

It's an induced charge. Since the cloud has a negative charge on its bottom, the tree (which is normally neutral) has extra positive charge in it that is being attracted towards the cloud.

1

As the value of $b$ increases the resistance between the outer and the inner shells will converge to $1/4 \pi \sigma a$. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be $1/4 \pi\sigma a$. We can think of the situation in which there are two shells in the infinite sea of poorly ...

1

Speculative question... highly speculative "answer". If force is independent of distance, then a neutral object could still be polarized (positive charges are displaced one way, negative charges displaced the opposite way, in accordance with the force they feel), but since the neutral object contains the same amount of charge after polarization, there could ...

2

It looks like you're trying to find a vector field $\vec{A}_m$ such that $\vec{E} = \nabla \times \vec{A}_m$. This is only possible in regions of space that are charge-free: the divergence of the curl of a vector field is always zero, so we necessarily have $$\frac{\rho}{\epsilon_0} = \nabla \cdot \vec{E} = \nabla \cdot (\nabla \times \vec{A}_m) = 0.$$ ...

0

It's a bit unfortunate that you called it $J$ because energy is stored in electromagnetic fields and so it moves around but sometimes the net flow of electromagnetic field energy into a region is not completely balanced by an increase in the field energy at that point and one time that happens is when the fields exchange energy with electric charges. And ...

1

The expression for the total potential energy stored in the fields is given by $$\frac{\epsilon_0}{2} \int \left| \mathbf{E}_1 + \mathbf{E}_2 \right|^2 d\tau = \frac{\epsilon_0}{2}\left( \int \left| \mathbf{E}_1 \right|^2 d\tau + \int \left| \mathbf{E}_2 \right|^2 d\tau + 2 \int \mathbf{E}_1 \cdot \mathbf{E}_2 d\tau \right)$$ Notice that the first and ...

0

{1} Here is a link to a paper by Patrick C. Crane on the transmission line disturbances to RF: http://www.faculty.ece.vt.edu/swe/lwa/memo/lwa0168.pdf. I didn't read the paper, but it seems to be an extensive analysis, and mentions at least 3 other references on the subject. On page 3, you'll find a summary of effects on three types of RF: "The phenomena ...

2

The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always. For very ...

2

If you were holding some charge there with some force and always had then an equal charge would distribute throughout the surface of the conductor so that an equal but opposite charge could be right where you are holding your charge. So it is just like the charge was always distributed on the surface. If however you inserted some charge somewhere really ...

0

I think the TiOPc on printer drums may be TiOPc nanoparticles embedded in some kind of organic binder, rather than solid TiOPc. See this: http://patents.justia.com/patent/20140054510 for example for the challenges of dissolving TiOPc in anything. The binder on the other hand should be easy to dissolve; try acetone first, if that doesn't work, perhaps ...

4

Equation $(2)$ is indeed a general solution, but that doesn't mean that all the $A_l$ and $B_l$ have to be nonzero all the time. For a problem in which $r=0$ is part of the domain the $B_l$ coefficients are zero, else the potential diverges at $r=0$ due to the $r^{-(l+1)}$ functions. In the case of the point charge, the point $r=0$ is not part of the ...

2

If you solve for just inside a sphere you might need to throw out your B terms. If you solve for just outside a sphere you might need to throw out your A terms. But if you are solving for the region between two spherical shells you might need both. That's why it is general, because in general you need them.

0

The Poisson equation inside the (homogeneous) semiconductor is $\Delta \phi = - \frac{\rho}{\epsilon_0 \epsilon_r}$ whereas outside it, the relavite permittivity $\epsilon_r$ is different, e.g., if the material is sitting in vacuum $\Delta \phi = - \frac{\rho}{\epsilon_0}$ The solution you propose does not fulfill both equations simultaneously. So, the ...

0

The red lines represent equipotential lines, which have the additional property of having an electric field of zero. The electric field at a point on the red line can be defined as $$\mathbf E_{red}=\sum_{i=1}^{4}\mathbf E_i$$ $$\mathbf E_i=\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\cdot{\mathbf{\hat{r}_{0i}}}$$ Let's take the case where all charges ...

2

Lets to this step by step and take care of the signs! Let $q_1=3$µC, $q_2=5$µC and $q_3=-8$µC. The formula for the force, acting on particle one due two the presence of particle two, is given by $$\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{{| \vec{r}_{21}|}^2} {\hat{r}}_{21},$$ where $\hat{r}_{21}$ is the unit vector pointing from charge two to ...

0

The problems solved by the method of images are, at their heart, boundary value problem involving the divergence of a scalar field, and those problem have uniqueness theorems that says there is only one configuration of fields in the volume that generates a particular set of boundary conditions. So if you find any method of generating a field that meets the ...

0

In dV=-E.dr, E is electrostatic electric field and the negative sign implies that the electric potential V decreases with increasing electric field E. This is the significance of this negative sign. In E is not electrostatic, but induced electric field. This field is induced due to time-varying magnetic fields or motional electromotive force causing a ...

4

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system. I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on ...

1

$q$ and $q'$ together produce a potential that is zero on the surface of the sphere. If you find the surface charge density determined by the radial change in that potential you get a charge distribution whose total over the whole surface is $q'.$ If you take that surface charge and $q$ then they together make a combined field that is exactly zero ...

1

Imagine two situations. Situation one, you have a point charge of charge $q'$ at the image charge location. Situation two, you have some charge distributed on the surface of the conductor. In both cases you have the exact same electric field on the surface right outside the conductor. So you have the same electric flux there. But electric flux is ...

1

Your description is not very complete, but I guess what happens is exactly what you expected: to get significant repulsion (to counteract the atmospheric pressure), you need very high charge, which will be necessarily limited due to air discharge (maybe that is why you observed sparks). I don't see how replacing an aluminum shell with a plastic bag ...

1

Think of potential as like potential energy. If the mobile charges are electrons and they are mostly at rest they will mostly move towards lower potential energy which is higher potential. So think of it as like a hill with electrons free to roll down hill (higher potential) until they get to a surface of the conductor at which point they are not free to ...

1

After all, you can't say:" since you are studying electrostatics, there must be equipotential region on the surface no matter what happens; that's it" Actually, it is almost that simple - otherwise there is a contradiction (1) Assume the electrostatic case (the electric field is constant with time and there is no electric current, i.e., no electric ...

-1

Although what follows is about an electric field in space, rather than the surface of a conductor, the ideas may help: An equipotential surface is perpendicular to the lines of an electric field, and equidistant from the source of the field, so that the electric potential at any point on the surface is the same as at any other point on the surface. In ...

0

You're confusing "equipotential" with "zero potential", and I think also "electric field" with "electric potential". The electric field at point A is the force that would be experienced by a unit charge at A. The electric potential at A is the potential energy of a unit charge at A, referenced to some arbitrary zero potential point B; it's equal to the work ...

1

Static electricity is not like regular electricity in that it does not involve closing a complete circuit; it just needs a large difference in voltage potential between one object and another. When you shuffle your feed on a nylon carpet and touch your finger to a doorknob, you are not closing a circuit. Instead, you are building up a large negative charge ...

1

I think an insulator does not completely stop charge transfer, if viewing it to act in the same way as maybe ie. a thermos cup, which does significantly increase the cooling time of a hot coffee inside, but does not fully prevent heat from escaping, hence allowing the hot coffee/material inside to cool.

0

Just to add, nonlocal dielectric response also leads to the permittivity being dependent on the wavevector. Nonlocality is thus tightly bound to the notion of spatial dispersion. This has profound implications on the light propagation is nonlocal media. The dispersion curves can be bent upwards or downwards with the wavenumber. Therefore, it may occur that ...

1

You've run into a classic approximation problem! Usually, you can approximate $X + \epsilon$ as just $X$ if $\epsilon$ is very small compared to $X$. And usually, you can do this to every variable in a problem, and the answer will come out fine. It only doesn't work when the answer itself is also very small; for example, let's say here the field from the ...

0

Consider infinitesimal charges distributed around the ring. At a point on the axis the radial (perpendicular to the axis) components of their fields cancel by symmetry. The field in the axial direction is the sum of the axial components of their fields. Thus, the field (due to the ring) on the axis is equal to the axial component of the field of a single ...

0

I found a paper that discusses this nonlocality in the context of intermolecular interactions. Not the same context as the article you linked, but it may help with understanding the concept. "The nonlocal dielectric function of a molecule determines the effective potential at a certain point due to an applied external potential at a different point, within ...

0

It means polarization $\mathbf P$ at point $\mathbf x$ is an integral involving electric field $\mathbf E$ at points $\mathbf x'$ different from $\mathbf x$.

11

You are right - potential is a scalar. But a dipole moment is a vector - it has magnitude and direction. When sodium channels open up, charge flows. Lots of charge moving a little bit causes a change in the dipole moment of the heart. This in turn induces charge to move elsewhere in response (the dielectric properties of tissue cause a propagation of the ...

0

I think your approach isn't wrong; however in your calculations you're making the assumption that the potential difference between plates, $V$, is constant: What remains constant is the charge on each plate. So the equation becomes: $$W=\int_0^d {{q^2} \over {2\epsilon_0A}} \;\mathrm{dx}={{q^2d} \over {2\epsilon_0A}}$$ Since $C=\epsilon_0A/d$ we obtain ...

0

The net charge induced is zero for an isolated spherical shell since the spherical shell can't acquire or give up charge. It's like a poker game with your friends, the money can only move around. Some might lose money some might gain money but the net change in money for all of you is zero. If the spherical shell were grounded or held at a constant ...

0

From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d. That's not the ordinary understanding and, from the electrical circuit perspective, the energy stored equals the work done by the external circuit separating electric charge in the ...

0

The calculation is not right, but the result is true. You need infinite work to pull apart two oppositely charged infinite plates, they have each infinite charge! The electric field due to one infinite plate is $\frac{\sigma}{2 \epsilon_0}$. The force on an area $A$ of the second plate will thus be $\frac{- \sigma^2 A}{2 \epsilon_0}$, which is infinite for ...

0

As John Rennie marked, the result should be $\tfrac{1}{2}CV^2$. Let me deduce this for you; Let's start with an uncharged condenser & by some means you remove one an electron from one plate & transfer it to the other plate. You have to do hardly any work to transfer the first electron but as you gradually continue the process, the field that is ...

0

A wide rubber band should work fine. Both the belt and the roller must be extremely clean. EXTREMELY extremely clean. A single fingerprint can be enough to mess things up, since the contamination distributes itself all over as the motor runs. If your belt and roller aren't clean, then oil film is touching oil film, and no rubber actually touches any ...

0

There must be some asymmetry at the start. It can go either way, positive or negative. It's like balancing a razor blade on edge, and "falls" one way or the other. But, under humid conditions, the Kelvin generator gives zero output. Build it, turn on the water, and nothing happens. This occurs because a many-megohms resistance appears across all the ...

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