# Tag Info

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For electrical engineering purposes, Voltage is a potential difference with respect to a reference point. Absolute electrical potential is meaningless in electrical engineering contexts because circuits involve electrical current flowing from a high-potential point to a low-potential point and not to a no-potential point. Therefore electrical engineers talk ...

9

You want a gas so you don't need to expend energy vaporising the propellant. You also want the gas to be as dense as possible so you can get as much impulse per unit volume of propellant as possible. It's also nice if the gas is inert and non-corrosive so you don't need to worry about it degrading or corroding whatever you're storing it in. Finally it's nice ...

4

Noble gases have the advantage of being chemically inert, so that they are less likely to react with atoms in the electrostatic grids. Since ion thruster to date have been deployed on unmanned transports, regular maintainance is not an option. Because of that, Noble gases are favoured over, say, hydrogen One reason to pick Xenon over Argon though, might ...

0

Potential is never discontinuous. Sure, without fringing what you say is correct. Fringing is what makes the potential continuous in this case.

3

It's unlikely. Stack charge builds up when two insulators rub against each other. This could be between your shoes and the carpet, for instance. You may be wearing shoes with soles with different materials, or walk in a way that picks up charge more easily. Silver is a conductor, can so doesn't cause static build-up. The charge is later discharged when you ...

0

Ok, the final problem was a bug in the code which calculate the forces. Repository -> glan.cpp -> line160 < I added the Vectors of the Distances between the points not their force-vectors. Here are some results: 2Charges: 3Charges: 9Charges: 16Charges: All charges have equal electrical charge .

1

If you try to concentrate separate electric charges repelling each other with the Coulomb forces, you would never succeed because that would require infinite amount of energy. That means if electron is charge concentrated into a point, it is not made of smaller parts but is an elementary particle and its rest energy is not electrostatic. This is the common ...

5

I want to complete the other answer by addressing the difference of a charge distribution made statistically up by a huge number of electrons , and what an electron means: At the level of elementary particles, one of which is the electron, there are no charge distributions, as elementary particles are point particles, and charge is a quantized quantity ...

4

The rest mass energy of an electron is 511 keV. So this is how much energy is required to create an electron, at rest, and which is not in an electromagnetic field. You need not be concerned about the point-like nature of the electron, that just makes it simpler to calculate the difference in ths energy (if necessary) caused by any external electric ...

1

Nice question! As has been said in the comments, the problem is that as you get very close to the sphere, you have to take into account all the charge present. More precisely, the field inside is zero no matter what, as opposed to a plate, in which the field is the same on both sides. This explains the factor of $2$. We might expect that taking the limit ...

1

This really interesting, I have never seen this reproduced in text books on electrostatics. I feel like the problem may be due to the fact that you are multiplying by an area $A=4\pi R^2$ as this will count the surface of the sphere on the opposing side as well as the surface you are immediately adjacent to. If you consider "flattening" the sphere out you ...

0

Going from the field to the force will be difficult, because you don't know a priori how a dipole reacts to a field, particularly if the field is not homogeneous. The way to do this is the same as you find the field for the first dipole: find the net force on the two charges of the second dipole, and then take the limit as their separation goes to zero ...

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You get a rough estimate by comparing the energy lost by the dipole as it aligns with the field $\sim d E$ with the thermal energy $\sim k_B T$. Only when the former is much larger than the latter will the alignment be stable against thermal fluctuations. Taking this website's value for the dipole moment of water $$d \approx 1.85 D \approx 1.85 \times ... 0 I guess that by line integral you mean integral along the path whose differential element is d\vec \ell. Well, in fact you don't need to mention the path, it's \theta that "knows" the path. If the path between the points a and b is along \vec F, the \theta = 0. If the path is some undulated line, \theta will vary along the path. Now, if a ... 8 Ideally, test charge should not affect the charge distribution of the source. An infinitesimal charge will ensure, for example, that the electric field it produces does not redistribute charges on any conductors in your system. A large test charge would polarize nearby objects, thus affecting the field you're trying to measure in the first place. 1 A good database with several materials could be found here RefractiveIndex.INFO . It could be accesed online por download un different formats. 0 For what concerns most calculations, the two forms are equivalent. Infact, I'd say that you could safely use the identification$$ \tag{1} \int d^3 x \rho(\textbf x) \sim \sum_i q_i,$$in all those circumstances in which it makes sense to talk about localized charges (with some exceptions though, see the last paragraph). You can of course mathematically ... 1 Imagine a hollow metallic sphere which is not charged . Now we have a point charge outside of sphere at distance r from centre of sphere we have potential say V at center of sphere , now we take another point inside shell which is not its center , Will potential change value here? Since there is no charge inside, the potential inside satisfies ... 1 For any (perfect) conducting shell the electric or magnetic fields will induce some charges to move through the shell such that they cancel the fields inside. This is called a 'Faraday's cage'. So it doesn't matter what field's or currents you apply outside of the shell, the fields inside will always cancel. -3 electron-affinity of the material. Condition of the surfaces of the materials. The intensity of the rubbing procees-hard or gentle 0 I can't figure out your question, and wonder if it contains a typo. However, consider a very simple system (simple in teh sense that it's made of material you should know, and operates according to principles you've seen before). First use a conducting rod pointing in the x direction, and being moved at a steady speed in the y direction in a region with a ... 0 I have to say that I don't fully understand your question, and there still seem to be inconsistent factors of 4\pi. But I want to get at the heart of the Green function for your boundary conditions, and then you can always ask another question about how to handle it for other boundary conditions.$$G(r,r')=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-a)^2}}$$is ... 0 To get the biggest effect, make the direction of the field, the wire, and the motion of the wire be mutually orthogonal. The effect is that the magnetic field can produce a force with a component along the direction of the wire (which is different than the direction of the velocity of the charges), and the person pulling the wire can do work on the ... 0 You have an expression for the total energy$$U_E = \frac{1}{2}\int_V \rho(\vec{r})\phi(\vec{r})\mathrm{d}^3x.$$And now you can break the charge density into two parts \rho=\rho_1+\rho_2 and the potential into two parts \phi=\phi_1+\phi_2, where each is due to just one sphere (so the additions hold by linearity of charge and superposition of ... 1 There is an important fact for this problem, namely that the spheres are very very far away. If they are close together, but connected by a long wire then it's a different issue. In isolation a sphere of charge Q and radius R has a potential of kQ/r relative to a zero potential at infinity, and so it's potential at r=R is V=kQ/R. If the spheres ... 0 Electrostatics refers to that class of electrical problems where the time derivative of all quantities is zero. You can have a current as long as it doesn't change with time. 5 Can anyone provide me with a physical interpretation For an electric field satisfying the equation, the work associated with (slowly) moving a test charge around a closed path is zero. To see this, recall that the electric force on a charge is$$\vec F = q\vec E $$The work associated with moving a particle along a closed path is$$W = \oint \vec ...

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If you hook up a battery, then at first electrons might rush in one side (and leave a charge imbalance in the wire they came from) and electrons might rush out of one side (an thus leave a charge imbalance in the wire they entered). But these charge imbalances cause electrons to move into or away from these regions. Even though the conduction electrons ...

2

Electrostatics refers to slow-varying fields with time (i.e. constant fields) and it is not in contradiction with particle movement. Think about a constant field acting upon an electron, we would be in the electrostatic regime (since the field is constant) and yet the electron would move. Electrostatic or electrodynamic only refers to the time evolution of ...

2

The electromotive force in a wire is the line integral $$\mathcal{E}=\int_\text{wire}\vec E\cdot d\vec\ell$$ (There may or may not be a negative.) So for a closed circuit the induced EMF is the loop line integral $$\mathcal{E}_\text{loop}=\oint_\text{loop}E\cdot d\vec\ell$$ The Maxwell-Faraday equation $$\nabla\times\vec E=\dot{\vec B}$$ leads to, upon ...

2

It says (physically) that if you measure the field vector around a closed curve in space that you will get a zero sum. (You can think of this as moving a detector along the curve and measuring the total amount of work (the vector dot-product of force times delta-x) done along that path.) It implies that a field that does not change over time will not produce ...

2

Given a vector field $\mathbf F$ in $\mathbb R^3$ and a closed path $\gamma:\mathbb R\to\mathbb R^3$, the integral $$\int_\gamma \mathbf F\cdot\text d\mathbf r$$ represents the work done by $\mathbf F$ along $\gamma$ (this is true more generally, when $\gamma$ is not necessarily a loop). If this quantity is zero, then it can be proven that the integral ...

6

The vanishing of closed line integrals means that the field is conservative. Since $\oint \vec E \cdot \mathrm{d}\vec l$ is equivalent to $\vec \nabla \times \vec E = 0$, the "physical interpretation" is the the electric field is irrotational, i.e. it has no "vortices". The, more valuable, mathematical implication is that there is a scalar potential whose ...

1

I think the answer is this: You are right that there is no resultant force on Q from the external field E, because it is inside a closed, conductive shell (C), which acts as a faraday cage. However, even if there was no external electric field E, the charge Q produces its own electric field and that will interact with the conductive shell C. If Q is ...

1

Gauss' law for dielectric materials is (takes into account charge displacement): $$\int_{Volume} \rho_{free} dV = \int_{Area} \epsilon_0 \epsilon \vec{E} d\vec{S}$$ Taking a sphere concentric with the ball and with radius $r < R$, we find the electric field at $r$ from the centre: $$\frac{4}{3}\pi r^3 \rho_{free} = 4 \pi r^2 \epsilon \epsilon_0 E ... 2 For this simple example, you're assuming that the polarization of the material is zero, and therefore, it's dielectric constant is 1, just like for vacuum. It's true that real materials initially with charge density \rho would self-polarize, and would pick up a surface charge density. The self-polarization would mess up the simple linear relationship ... 11 The statement "electric field inside a conductor is zero" is true only after charges have distributed themselves in the most optimal way on the surface - it is an electrostatic result. Starting with an arbitrary charge distribution, there will be forces that cause a redistribution of the charge until, for a sphere, they are distributed uniformly. At that ... 2 Good question, but normally there are more than two charges. Note that the electric field inside conductors is zero because the charges on the outside move to an arrangement where it is zero - or as close to zero as possible. Consider a sphere with a uniform density of charge on the outside. If the contributions of all the charges on the surface are ... 2 The electric field due to a charge surface element dq does have to go inside the conductor, but it is cancelled out by other charges elsewhere on the sphere. So the field inside the conductor is zero, and there is no contradiction. The precise nature of this cancellation can be seen by doing a surface integral, but it seems mysterious if you do it that ... 0 If they are conducting shells then they charge distribution on the outer shell will be absolutely independent of the position/shape of the inner surface. There is a similar question and answer here which is about the position of a single charge inside a sphere - the reasoning is exactly the same here, though the problem is slightly different. First point ... 1 When you charge an insulator, you typically put charge just on the outside. The insulator presumably needs to be connected to the "earth" in some way, and it is important that you use a good insulator for that mechanical connection. You will find that the main leakage mechanism is usually moisture in the air. As long as the charged surface of the insulator ... 0 I assume you have already considered using a cathode ray tube, right? If not, you can apply a constant/static electric field and deflect the electron beam, which is a nice example of electrostatics. The electron will undergo constant acceleration and the problem can be nicely linked with the equations used in kinematics from the previous semester. You can ... 0 to use Gauss Law, you should have a system which is symmetric. I mean there should not be any boundary conditions and for such systems Gauss Law can be used in any problem. But the problems which you should use image methods or any further like Green's formula etc., are not symmetric. A point charge near a conducting sphere break the symmetry of electric ... 1 Its because the charge does not move - it is stationary -> static. So it was given a name which is very descriptive: static electricity. 3 On point 2: While vacuum itself, being composed of nothing at all, is not expensive, a capacitor structure able to maintain a vacuum when surrounded by air is impractically expensive. On point 3: Don't think of higher-value capacitors as requiring less voltage. Rather, a higher-value capacitor allows us to "store" more charge at the same voltage. In a ... 0 Outside of charge distributions, the general (non-\phi-dependent) solution of \Delta\Phi=0 takes the form that you described. Now, supposing that the solutions for r \lt R and r \gt R take the form that you mention, then the electrical field \overrightarrow{E} = -\nabla\Phi will have the following values in spherical coordinates:$$r \lt R: ...

2

It's just a matter of rationalisation and mathematical convenience. $4\pi$ corresponds to the whole solid angle, which usually simplifies when you deal with Gauss' theorem. In other words, you can simply redefine a constant $k$ to be any multiple of another constant, $1/E_0$ in this case.

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If you are asking about what happens to the electrons in a metal when placed in an electrostatic field, the simple answer is they get attracted to the end of the metal closes to the 'positive' side of the electric field, leaving an excess of 'positive' charge on the other side, the net effect of which is to 'cancel' out the applied electric field within the ...

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