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10

This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law \begin{equation} \nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0} \end{equation} If we integrate this over some volume $V$ and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} ...


9

Coulomb's law becomes invalid at distances of the order of the electron Compton wavelength and smaller, due to vacuum polarization. To first order in the fine structure constant, the electric potential due to a charge q at the origin is given by: $$V(r) = \frac{q u(r)}{r}$$ where $$u(r) = 1 +\frac{2\alpha}{3\pi}\int_1^{\infty}du ...


8

The answer by @NowIGetToLearnWhatAHeadIs is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation. Start with your charge distribution and a "guess" for the direction of the electric field. As you can see, I made the guess have a component upward. We'll see shortly why ...


8

You have to realize that the system is invariant under rotations about the normal to the plane. Then then electric field must also be invariant under these rotations. An electric field component in the plane does change under such a rotation, so such a component must not exist if we have this invariance. Thus the electric field is purely along the normal to ...


7

It doesn't hold for arbitrary shapes. The reason it works for spheres is that when you have a spherical charge distribution and a concentric spherical Gaussian surface, the whole system is invariant under rotations around the center of the spheres. If the electric field were different at different points on the Gaussian sphere, you could rotate the whole ...


6

I) Right, the differential form of Gauss's law $$\tag{1} {\bf\nabla} \cdot{\bf E}~=~ \frac{\rho}{\varepsilon_0} $$ uses the relatively advanced mathematical concept of Dirac delta distributions in case of point charges $$\tag{2} \rho({\bf r})~=~\sum_{i=1}^n q_i\delta^3({\bf r}-{\bf r}_i).$$ Note in particular, that it is technically wrong to claim (as ...


3

In theory, if you had a sphere encased in a great dielectric than maybe you could charge it up to the breakdown limit of the dielectric. However, for a VdG generator, you need access to the sphere for the support and charging system, which then becomes the breakdown path. As noted by @UncleAl, real accelerators use high pressure gas (not always SF6 since is ...


3

Short answer - no. If field lines are curved and parallel, then the path integral along one line or the other will either give a different potential difference, or require different field strength. In either case you cannot call the field "uniform". A little picture to clarify: The dotted lines represent equipotential lines (at right angles to the ...


3

The validity of Coulomb's Law over large distances is equivalent to bounding the mass of the photon. In quantum field theory, where one derives Coulomb's law, if the photon had a mass $m$, then the Coulomb potential gets replaced by the Yukawa potential (in natural units where $\hbar=c=1$ and Gaussian units): $$ \frac{e^{-mr}}{4\pi r}\ , $$ where $m$ is the ...


2

If we ignore the inner section, we have a box with 3 sides held at V = 0 and the top edge at V = V1. I'm pretty sure this is easily solvable by separation of variables using an oscillatory solution in x with a decaying solution along y. Using superposition we can then treat the inner box as a separate problem of similar geometry/boundary conditions. The ...


2

All that is required to "store" a charge is "not make it run away". As you know a positive charge is attracted by a negative one and vice versa - so in a capacitor you bring positive and negative charge "close without touching". They are attracted to each other but can't reach - the insulator is in the way. Compare this to the situation of two lovers in ...


2

First, I think it's right to revert your first sentence. The Gauss' law is more natural and elementary – and the $1/r^2$ Coulomb inverse square law is a consequence of the Gauss' law. In spacetimes of different dimensions, one naturally gets a different power law for the electrostatic force; Gauss' law is always right, however. You don't have electric ...


2

Let us ask the question the other way round. Given a conductor closed shape and a zero field inside, what are the possible surface charge distributions and when is one of them constant ? The potential is solution of the Laplace equation and can written as $$V(r,\theta,\phi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \left(A_{\ell m}r^\ell+B_{\ell ...


2

First of all, I'm not an expert, but that can be an advantage in trying to explain the equations in lay terms... Maxwell's equations are these, in differential form: $$ \nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}$$ $$ \nabla \cdot \mathbf{B} = 0 $$ $$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t} $$ $$ \nabla \times ...


2

When you take a brass plate of considerable thickness and place it in between two charges, say positive and negative, induction takes place in the brass plate since it is a conductor: the electrons shift to the end near the positive charge while the cations stay near the negative charge. Now, induction occurs in order to make the field outside a certain ...


2

The first bullet is correct, the outer shell does not contribute. This easily follows from Gauss' law. For this you use the fact that the electric field must be radial and any cylinder inside the cylindrical shell does not enclose the charge density $-\lambda$. You might think that close to the negatively charged shell there is an additional electric field ...


2

I think you just missunderstood the textbook article. It says, There are ice particles in the clouds, which grow, collide, fracture and break apart. The smaller particles acquire positive charge and the larger ones negative charge. Not the clouds grow, collide, fracture and break apart, but the ice particles. In fact, the article is a bit ...


1

The method of images is useful when you have a point charge near a $V=0$, a.k.a. grounded, surface; usually a plane or a sphere. In this configuration you can substitute the surface by an additional point charge (of opposite charge) and the problem becomes finding the potential due to two point charges. In these other problems, you don't have this ...


1

Maxwell's equations state $$ \nabla \cdot \vec E = \frac{\rho}{\epsilon_0}$$ $$ \nabla \cdot \vec B = 0 $$ If we accept Maxwell's equations as true, there is no source/sink of the magnetic field, since the divergence of the magnetic field is zero no matter what. Yet, no matter how you feel about the Dirac delta, where there is charge, there is non-zero ...


1

The brass plate is a conductor, so the potential will be the same on both sides. The thickness of the brass plate therefore subtracts from the effective distance between the two charges, making the electric field strength higher in the remaining open space between the charges. This stronger field will cause more force to be experienced by each charge. ...


1

Equi-spaced field lines are used to denote a uniform electric field. That means the field strength does not vary in the region, there is no potential difference between any two points on a surface normal to ${\vec E}$, no work is done in moving along a closed loop, or between two points on such an equipotential surface. Further, you expect that if you place ...


1

There are various layers one could address your question. As all observables in QM, the charge exist because there exist a self-adjoint operator associated to it. This operator corresponds to a (class of) experimental instruments that the observers use to make measurements on the system, the collection of possible numerical outcomes being known as charges. ...


1

Electrical conduction is charge carriers (most often electrons) moving through a conductor. In an insulator there are no free charge carriers to allow conduction. However electric field effects can still propagate as the material is made up of atoms, consisting of positive protons and negative electrons. In the simplest case these will act as a dipole. These ...


1

I agree with the result, but I would like explain another more general and rapid approach. Because of the radius of the hole is negligible with respect to the radius of the sphere, and the only posible direction for E compatible with the symmetry is the z axis, and finally having in mind that the tangencial components of E are continuous, the solution is ...


1

Pick a point above the plane. From a point in the plane directly under the point above, draw a circle of some radius. Consider the contribution of the charge elements along the circle to the electric field at the point above the plane. Since the charge density is uniform, the horizontal components of the electric field from charge elements on opposite ...


1

An answer connected to Gauss law (I hope everything is correct, since it's long ago for me ... so no warranty): An infinite plane of uniform charge for example in the z-plane has the charge distribution: $\rho=q\,\delta(z)$ Thus, the electrostatic potential should be $\Phi=\frac{q\,|z|}{2\pi}$. Hence, the electric vectorfield is: ...


1

If the majority of the drone is a "gasbag", you may want to install a fairly heavy-gauge wire running from the top to the bottom to act as a sort of lightning rod. The sensitive electronics should be fine if they are within a Faraday Cage. As long as the lightning has an easy path to take through your aircraft, it shouldn't damage anything else. That ...


1

In addition to BMS answer, I want to point out the integration part as I have seen,in the comments, you have some problems in the integration part. First you should have written the unit vectors in the expression of the electric field. The electric fields are $\vec{E}_{\text{in}}(\vec{r})=\dfrac{Q}{4\pi \epsilon_0 R^3}r\hat{r}$ and ...


1

If you look at the two particles in the centre of mass frame you'll see something like: Your task is to find $b$. In the COM frame both particles move in an effective central potential, and the two trajectories are symmetric. I'm going to ignore the green one and just show how to calculate $b$ for the red trajectory. The closest approach is then $2b$. In ...


1

OK, I will try to say something a bit more useful than my other response. I am still on the skeptical side of getting a closed-form, simple solution to this problem, specially using separation of variables. I think that the problem is that the potential outside of a square with that border conditions cannot be attacked using separation of variables. My ...



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