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17

Gauss's law is always fine. It is one of the tenets of electromagnetism, as one of Maxwell's equations, and as far as we can tell they always agree with experiment. The problem you've uncovered is simply that "a uniform charge density of infinite extent" is not actually physically possible, and it turns out that (i) it is not possible to express it as the ...


4

The best interpretation I can see of this fact is related to Gauss' theorem. Poisson equation is of the form $$\nabla^2\phi = \rho,$$ but if you set $\mathbf F = \nabla\phi$, this is also an equation for the vector field $\mathbf F$, $$\nabla\cdot\mathbf F = \rho.$$ Let us consider a density distribution given by a total mass $Q$ centered at a single point, ...


4

but I dont understand how $$\frac{1}{2}d(\vec v \cdot \vec v) = \frac{1}{2}\left(\vec v \cdot d\vec v + d\vec v \cdot v \right) = \vec v \cdot d\vec v$$ Also, I thought that $\vec{PM}\cdot d\vec{PM} = PMdPM$ if the angle between the two vectors is zero... While it's true that $\vec u \cdot \vec v = uv\cos\alpha$, it's also true that $$dv = ...


3

Yes, a discontinuity in the electric field is always associated with a charge distribution with infinite density. That's a necessary implication of Gauss' law. To think of it qualitatively, where there is a discontinuity in the field, there must be field lines either starting or ending. (If it's not obvious why that is, think about the fact that the ...


3

Let's be really clear about the exact kind of mistake you are making. Consider the numbers $1,-1,1,-1,1,-1, \dots$. If you wanted to sum them you could argue that $1-1+1-1+1-1+\dots=(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0$. Your friend could argue that $$1-1+1-1+1-1+\dots=1+(-1+1)+(-1+1)++\dots=1+0+0+\dots=1.$$ The problem is that an infinite series isn't ...


3

It is safe to sit under a car rather than to stand under a tree during lightening It is safe so sit inside a car during a thunderstorm, but I don't think sitting under a car is a good idea. The car behaves as two resistors in series. The car body has a very low resistance (assuming it's metal bodied) so the current flows through the car body with very ...


2

If you don't specify the boundary, there will be many solutions (including the unsymmetric ones) The problem of this question is that solving Maxwell's differential equations necessarily involves specifying the boundary conditions which usually can be chosen the obvious ones, however here such a boundary simply does not exist. Usually, it is ...


2

You smell ozone ($\mathrm{O_3}$, from the Greek word ozein for "smell"), and maybe nitrous oxide - the reaction product of oxygen and $\mathrm{N_2}$. There is a nice description of the formation and action of ozone at this link. Briefly: Oxygen molecules ($\mathrm{O_2}$) can be dissociated (broken into atoms or ions) by either UV light, or electrical ...


2

Yes there is a net charge on the earth; no you can't use that to generate energy. For a current to flow, charge has to be moved by an electric field. In the case of Earth, how do you propose to create a circuit to tap into this field? You could move a charge from the ground to the clouds - and maybe do a very small amount of work while moving it. But then ...


2

It is not clear to me whether the total charge is also scaled, or you want that to be conserved. So I will work under the assumption of the former case and show that the potential scales as $n^2$. The new density $\rho'$ is related to the old density by $$\rho'(\mathbf r) = \rho(\mathbf r/n).$$ Denoting by $\Omega$ the support of $\rho$, we have $$V_0 = ...


2

No, that is the simplest way to solve the problem. As mentioned in the comments, this is in the absolute scale of things a very easy problem: the spherical symmetry allows you to even have an integral to calculate, and the exponential is not only exactly integrable, but easily so. If you allow general spherically symmetric charge densities $$\rho(\mathbf ...


2

In this case it is more accurate to say that V is proportional to Q because the potential, V, (at some distance from the charged insulator) is dependent on how much charge, Q there is on the charged insulator setting up the electric field. It is not the source (charged insulator) that acquires potential--the potential we are referring to is at a point away ...


2

When the voltage is high enough so that the speed becomes relativistic, two effects precludes becoming superluminal. First you must now use the relativistic form of newton's second law, with 4-vector velocity and acceleration. Second you must keep in mind that an accelerating charged particle would radiate with a power proportional to the square of its ...


2

Distribution of charge within the region where the field is located, is obviously uniquely defined, because it's just $$\rho=\epsilon_0\nabla \vec{E}$$ However, if you cut out a region of space, and want to predict the contents of this region based only on the field outside, you can't do it in a unique way. The reason is that you have too many degrees of ...


2

Static electricity is a subfield of classical electromagnetism, the theory of electrically charged particles as well as electric and magnetic fields. Photons only enter the picture in the quantum version of the theory, Quantum Electrodyanimcs (QED). EM waves are not mediated by photons, but EM waves and (real) photons are one and the same thing. Static ...


2

You have two distinct errors. One is claiming that because the electric field goes to zero at infinity so does the flux. The flux is the integral of the electric field over the surface. The electric field goes down as $\frac 1{r^2}$, but the area goes up as $r^2$, s the flux constant. Ask Gauss about this. The second is to claim that because (from the ...


1

Your problem is the assumption "both the charges contribute flux to their respective Gaussian surfaces only at the common surface." When you take your surface out to infinity you are also increasing its area, so even though the electric field goes to zero, the integral of the electric field over your surface will be non-zero. In fact, the total flux ...


1

If I assume electric field lines form closed loop that would mean electric field has non zero curl. So I cant write electric field to be gradient of some scalar function. That would imply work done by the electric field will depend on the path. we know thats not the case really. Another way to see this: Closed electric field lines would mean number of ...


1

If there are no charges inside the cylinder, then the potential obeys Laplace's Equation: $$\nabla^2V = 0$$ In cylindrical coordinates, that's $$\left(\frac{\partial^2}{\partial r^2} + \frac 1 r \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}\right) V = 0$$ Based on your notation, it seems ...


1

Consider a single charge $q$ (out of a given system of charges) at a point $\textbf x$ in free space. The charge $q$ will feel electrostatic forces due to all the other charges. To analyze how $q$ will behave, we have to consider the force that acts on it, which is $$ \textbf F = q \textbf E,$$ where $\textbf E$ is the electrostatic field generated by all ...


1

Why would you want to? $$\tag{1}\mathbf{D}=\int d^3r\,\rho(r)\begin{pmatrix} 3x^2-r^2 & 3yx & 3zx \\ 3xy & 3y^2-r^2 & 3zy\\ 3xz & 3yz & 3z^2-r^2 \end{pmatrix}$$ Looks rather strange. Components $$D_{ij}=\int d^3x \,\rho(x)(3x_ix_j-r^2\delta_{ij})$$ are much better. Besides, you're going to calculate (1) component by component ...


1

I understand that in fact, your problem is $$Does \ the \ energy \ density \ depend \ on \ the \ surface \ that \ we \ use \ ? \ In \ either \ case \ please \ can \ you \ explain \ why?$$ I followed the text that you indicate, and the answer is yes, the energy depends on that surface. After deriving the formula $$W = \frac {1}{2} \sum _i q_i \left( ...


1

The resolution is that in the stationary frame, in addition to the Electric field present due to the charge of the particles, there will also be a magnetic field because the moving charges are a current. Note that transforming from the particles' rest frame to the frame in which they move, not only do you have to transform their positions and velocities, ...


1

For proving that the Coulomb E satisfies the Gauss law in differential form, simply take the divergence of Coulombian E. You will need the Dirac delta and some of its properties. For deriving the divergence of E field, do just the same as above... You can also derive the Gauss law in integral form from the differential form by applying the divergence ...


1

The electric field in the problem has no $z$ component, so it quite simple to calculate the flux through a cylinder with axis parallel to the $ z $ axis; then you choose a cylinder that contains the sphere you are interested in. Let $\Sigma$ be the surface of the cylinder, $ V $ its volume, $\Sigma '$ and $ V' $ the surface and volume of the sphere; by the ...


1

The excess charge is repelled from itself (because all he charge carriers have same sign). It can only go so far away that it ends on the surface. Therefor it is actually kind of logical that net charge will be present on the surfaces (in the static situation).


1

If you are considering electrostatic then by definition all the charges do not move. A conductor is an element whose charges are not bounded, i.e. they will move if any force acts of them. In electrostatic thus we must have that the net force on any charge in the conductor must be zero. Considering only electrostatic forces this means that inside the ...


1

You're not particularly spot on with your definition of electric flux. Most fundamentally, the electric flux $\Phi$ through a given surface $S$ is defined to be $$ \Phi=\int_S \mathbf E\cdot\mathrm d \mathbf S. $$ If you introduce a well-defined model in terms of field lines, then this does end up describing the number of field lines that cross $S$, to ...


1

So why have we suddenly jumped from a line integral to not a line integral For a point charge, the electric field has a radial component only $$\mathbf E = \frac{kQ}{r^2}\hat{\mathbf r}$$ Thus, the dot product of the electric field and the infinitesimal displacement vector is $$\mathbf E \cdot d\mathbf l = \frac{kQ}{r^2}\hat{\mathbf r}\cdot \left( ...


1

The micro world of atoms and molecules is ruled by quantum mechanics. The forces controlling the interactions between them at the level affecting everyday situations, as touching, are electromagnetic. In quantum mechanics the electrons are in bound states around the atoms, and at most can be mobile in bands when in a solid, i.e a bound state due to the ...



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