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5

Take a look at the conventional form of Maxwell equations. They tell us that Gauss's law actually applies every time. However, to get the field $\vec{E}$ from the charge distribution by the usual methods, we also need to know that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$$ Because otherwise the field could not be generated by the ...


4

Using cylindrical coordinates with the origin at the center and the $\phi = 0$ direction 'down' (the OP says the image should be rotated CCW 90 degrees), the electric field appears be have only a radial component with a sign change for $\phi = \frac{-\pi}{2}$ and $\phi = \frac{\pi}{2}$ $$\vec E = E(\rho,\phi)\hat\rho $$ $$E(\rho,\phi) = ...


3

I'll try to make a simple derivation. Suppose you have a unit point charge located at position $\vec{r}'$. Then the associated charge density is $\delta(\vec{r}-\vec{r'})$, which is a Dirac distribution. The electrostatic potential produced by this charge is given by the Coloumb's law: $$G(|\vec{r}-\vec{r}'|) = ...


3

These quantities are all infinitesimals. $dx$ and $dl$ are often used to denote infinitesimal line elements, while $dS$ and $dA$ are conventionally used for infinitesimal surface (area) elements. The list is completed with $dV$, the infinitesimal volume element. On the application of these infinitesimals: It is often simple and intuitive to think about ...


3

It will help you understand the quantum mechanical picture if you read up on atomic orbitals. These are the loci around the nucleus where the electrons have a probability to be found. You will see that the orbitals have a shape, which depends on the angular momentum of the state. The electrons carry the charge and thus you can interpret the plots as ...


3

The only way to have the charges on a charged conductor evenly spread is to make sure that there is no horizontal component of electric field in the case of "evenly spread". The only two solutions for that are things with infinite dimensions: planes, cylinders spheres For any other shape of conductor, there comes a point where the evenly distributed ...


3

If no charge is moving, there is no magnetic field. A point charge at rest has only an electric field, from "its" point of view. However, electric and magnetic fields are not seperate, since someone moving with respect to the resting charge would see a magnetic field due to the behaviour of the fields under Lorentz transformations. You may (for some ...


2

Yes indeed. To get the result you stated from Gauss' Law you must asume that the charge is distributed in such a way that you have a spherically symetric field. How you get to that field doesnt matter though. So the charge might all be concentrated at the center or all lying on the surface of a sphere: it doesnt matter as long as the field is spherically ...


2

Classical electrodynamics generally makes a continuity approximation for bulk materials. We're not interested on variation in the fields at the scale of the distances between atoms, so we just average them away.1 With that approximation, the conduction electrons are acted on by the mean field. They also contribute to the mean field but we treat the two bits ...


2

Wouldn't the field line end at the charge +q, and so the field lines that enter don't necessarily leave, making +2q contribute a negative flux? The quantity of field lines that terminate on the charge -q is unchanged by the presence of charge elsewhere. Imagine -q field lines terminate on the -q charge. Then, the net electric flux outward through ...


2

Metals consist of a lattice of positively charged atoms in a sea of electrons that can move around freely. You can almost think of a metal as a vessel for a charged liquid made from electrons. Just like with other vessels, one can take part of that liquid out and put it in another similar vessel. That's what happens when we are moving charges around. So ...


2

Do you understand that the electric field within a conductor is zero? The charge is mobile, so the internal charge rearranges itself until there is no longer any force to move them: there is no field in the interior. If you understand that, then you will realize that a test particle within the conductor will feel no force, so no work will be done in ...


2

As we know [...] in any frame of reference, its radius is a finite number Er ... do we know that? Relativity--as we understand it--does not allow for fundamental particles of finite size, and there is no experimental hint at all of any size associated with the electron. Pointedly the "classical electron radius" is about $2.8 \times 10^{-15} ...


2

Given a volume with finite charge density $\rho$, the surface charge density $\sigma$ of an embedded surface will be infinitesimal. The charge of the volume is the integral of the infinitesimal charges of the embedded surfaces. Conversely, a finite surface charge density would give you an infinite charge density there - specifically a delta function which, ...


2

The energy density of an electrostatic field is given by, $$\rho =\frac{1}{2}\varepsilon |\mathbf{E}|^2$$ where $\varepsilon$ is the permittivity of the medium. Hence, the total energy stored in a particular volume $V$ is given by a volume integral of the density, i.e. $$U = \frac{1}{2}\varepsilon\int_{V}\mathrm{d}^3 x \, |\mathbf{E}|^2$$ If all the ...


1

I have came up with this: Charges are the sources of the electric field. So, whatever the point that field lines are "created" or "destroyed", must be a charge. Then, if there are a charge, then must be on the center. Calculating the electric flux: $$ \phi = \iint_S\ \mathbf E\cdot d\mathbf s = \frac{Q}{\epsilon_0} $$ Let's pick a sphere as gaussian ...


1

Here is an intuitive explanation: Basically $C = Q/V$ means, if I have an capacitor such that store the same amount of charge $Q$ requiring a lower potential, then it has more capacitance. So, would be nice to build device which store huge amount of charge with ridiculous low voltage. This would require a huuuge capacitance. That's why $Q/V$ is important. ...


1

Typically, one defines a variable with respect to some observation point, P. In this case, the $\vec{r}$ is a vector pointing from the origin (defined by your chosen frame of reference and coordinate basis) to point P (which happens to be the location of your point charge $q$). The $\vec{r}$' here would be a vector from point P to the source of the field ...


1

You're getting confused because the author you're reading has expressed Gauss' law using unusual and confusing notation, to the point where I would actually call it incorrect. I can sort of see how the confusing equations could arise from a derivation of Gauss' law from Coulomb's law. From Coulomb's law, you have $$\vec{E}(\vec{r}) = ...


1

The electrostatic field depends only on the total charge distribution. If the charge distribution is known, as it is in your case, then you don't need to worry about the shape or conductivity of the structure supporting the charge. The charge on a conducting solid sphere will, as you say, distribute evenly at the surface. If you by some other method manage ...


1

Voltage has absolutely nothing to do with charge. I can "move" an infinite amount of charge trough a superconductor with zero voltage. Are you asking about the relationship of charge to voltage on a capacitor? That's a linear relationship: Q=C*U. The charges, in that case, are not "created" but merely separated. If you want more charge for the same amount of ...


1

It is very hard to use a voltage source to induce charge on an insulator. The reason is that by definition, an insulator does not conduct electricity - so if you apply an electrode at one place, you will not move electrons elsewhere, and so you cannot induce a net charge (the best you can hope for is to create polarization, and maybe pull off a handful of ...


1

The image charge only serves for the construction of the field in the half space $z> 0$. The problem with the two point charges is only a model problem to calculate the field for $z>0$. If the field in the real problem is static then it is zero in the conductor half space $z < 0$. Therefore, the construction with the pillbox really delivers a ...


1

This can be done in five steps (four integrals). Start with the force of two point charges: you know this equation $$F=\frac{Q_1Q_2}{4\pi\epsilon_0 r^2}$$ Integrate this force over an infinitesimally thin ring of charge: now you have the force of a ring on an off-axis point (hint: you only need the axial component - the radial components will cancel due ...


1

you don't need two charges, you can use only one charge and the answer will be the same, you only should note to add a second image charge in the centre of the sphere to make the sphere neutral. in both cases, when you take the limit, you get same electric dipole in the centre of sphere. but because the symmetry of two charges, calculations are somehow ...


1

The Triboelectric effect is the process through which materials can become electrically charged through friction when they come in contact with other different materials. These materials do not have to be insulators for this effect to take place however if they are good conductors the charge will usually flow away. There is a series of materials ranging ...


1

When you put a charged insulator in air, the reason you lose charge is mostly due to humidity in the air. I gave details of this mechanism in a recent answer to a related question: http://physics.stackexchange.com/a/130988/26969 The curves in the referenced paper (some of which I reproduce in that answer) show you how the leakage current is a function of ...


1

A charged conducting material in the form of a sphere or an infinite plane can only be uniformly charged in the absence of external charges. Any other shape of charged conducting material can be induced to be uniformly charged by the placing the right external charge density around it. A conducting infinite cylinder is also uniformly charged (in the absence ...


1

Well, I have a way of remembering if you live in the UK - you always drive your motor on the left! The diagram is correct - it shows what force will be exerted by the external magnetic field on the current carrying wire. The direction of the force is found from the left hand rule: splay your thumb and first two fingers out so they are mutually at right ...


1

The answers already in here are good; unfortunately the integrals that arise are quite nasty, and don't have solutions in terms of elementary functions. Here is some more detail, in the special case when the tubes have zero length (so they are just charged circular loops), and further they have the same radius $b$, with separation $d$. You'll see that this ...



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