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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 ...


0

I'm not completely clear what you are asking, but isn't this just an instance of elastic scattering? i.e. the dipole oscillates in phase with the electric field of the incoming wave and emits dipole radiation with the same phase. The example I'm familiar with would be Thomson scattering from free electrons, which oscillate like classical electric dipoles. ...


0

Using the Gauss's law (see this Link), the solution is as follow, $$ \Phi (r) = \left\{ {\begin{array}{*{20}{c}} {\frac{q}{{4\pi \varepsilon {r_1}}}\,\,\,\,for\,\,\,r \le {r_1}}\\ {\frac{q}{{4\pi \varepsilon r}}\,\,\,\,for\,\,\,{r_1} \le r \le {r_2}}\\ {0\,\,\,\,\,\,\,for\,\,\,r \ge {r_2}} \end{array}} \right. $$


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 ...


0

You have quite a discontinuity in the potential in two of the corners of each of the squares. I am not speaking only of the shape, but the value of the potential is discontinuous. We usually assume that the fields can have discontinuities, but the potential is always continuous. I suspect there is no solution.


0

The charge on the second plate is an induced charge due to the first plate, induced charge cannot cancel out the original charge therefore the flux linked with a capacitor of charge Q is Q/(8.85*10^-12)


0

From Classical Electrodynamics by JD Jackson Chapter 1 Section I.2 The inverse square law is known to hold over at least 25 order of magnitude in length! Earlier: The laboratory and geophysical tests show that on length scales of order $10^{-2}$ to $10^7$ m, the inverse square law holds with extreme precision. At smaller distances we must turn to ...


0

This is a very broad question, so my answer will necessarily be more generic. Electric fields are much easier to shield than magnetic fields. This is because most materials have low permeability and hence do a poor job of shielding magnetic fields. The same is not true for electric fields as any grounded metal will do a pretty good job. Many electronic ...


0

It depends. Mostly there is very little effect because most parts of the circuit are either equi-potentials or under low impedance control with the electrical energy coupling in capacitively , which tens to be a very weak effect. It's only when you have very fast edges and high amplitudes that there is a noticeable effect - i.e. EMP - from a nuclear blast. ...


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 ...


1

As for large distances - it is hard to tell whether Coulomb's law applies with any correction or not. A main restriction on precision tests of Coulombs law at large distances is basically the inverse square distance fall-off of the physical effects. If we take a too small charge, it's strength falls of very quickly beyond measurability. On the other hand, ...


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 ...


-2

Coulomb's Law describes the force between two electrically charged particles. $$|F|=k_e{|q_1q_2|\over r^2}\qquad $$ This equation is valid for ANY distance and the force goes to zero at infinity. This means that theoretically, the Coulomb force exists between all charged particles. Note that two conditions must be satisfied for this equation to hold ...


0

Well... you don't really measure electric/magnetic forces at distances much larger than several meters, but that's because electric potentials are difficult to build up. I guess on the small end, it's a little more difficult, but the strong force is essentially the only force that matters inside of nuclei.


0

You're so close to the answer I'm not sure how to nudge you along without basically just giving you the answer. I'll try anyways. There are some troubling conceptual mistakes you've made in an otherwise straightforward derivation. For starters : The reason I don't set $E_0 x_0$ to zero is because there is no $1/x$ term; otherwise I'd be able to make the ...


0

You write, "In the case of the magnetic field we are yet to observe its source or sink." If you mean "we are yet to observe a source or sink", you're correct. However, consider the magnetic vector field (ignoring units/speaking qualitatively): $$\vec{B}=(0, \frac{z}{(1+r^2)^2},\frac{y}{(1 + r^2)^2})$$ This is a valid field because it's the curl of the ...


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} ...


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 ...


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 ...


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 ...


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 ...


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 ...


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. ...


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 ...


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. ...


-2

Displacement current - D. Have another look at Maxwell's equations.


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 ...


0

Dielectric have bonded charges. thoses bonded charge oscillate when an electric field is applied. And thoses oscillations can propagate the electric field thought the dielectric media


0

We can always use superposition principle. I will explain it in the case of capacitor. First you must understand that a positive plate will create an "out" electric field as showed in the picture. In the contrary, a negative plate will create an "in" electric field (the arrow is reversed). If you draw the electric field of 2 opposite-charged plates ...


0

The principle of superposition is always applicable. The picture below shows to capacitor plates which are oppositely charged and their respective electric fields. The positively charged one (blue) has its electric field pointing outwards which the negatively charged plate (red) has it's field pointing inwards. As you can see, the superposition of both field ...


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: ...


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 ...


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 ...


0

With respect to your test charge, there will always be an equal number of charges on your plane in all directions because the plane is infinite. So for every charge "in front" of your test charge there will be a charge "behind" your test charge. And for every charge to the left of your test charge will be a charge to the right. What this means is that no ...


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 ...


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 ...


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 ...


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 ...


0

Let's consider each sphere have positive charge of amount $3\times10^8\text{ C}$. Obviously, the smaller sphere has more charge density. Now when the two sphere are being brought close to each other,obviously electric field will be from lower sphere surface to larger sphere surface. So the potential will be large at the small sphere point with respect to the ...


0

In addition to Qmechanic answer, you might be interested to look this page too http://physicspages.com/2011/11/14/dirac-delta-function-in-three-dimensions/.


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 ...


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 ...


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 ...


0

For a convex surface, take any arbitrary point inside the closed surface. Now consider a cone with infinitesimal angle from that point so that it cuts the surface on both side.now you calculate the electric field for these infinitesimal surfaces say $\mathrm{d}s_1$ and $\mathrm{d}s_2$. You will find electric field as zero. as these two surfaces make same ...


0

From your conclusion, which is correct, $$\large q=\frac {Q_1+Q_2}2$$ and thus as charge is conserved the charge on inner surface of plate $\mathrm{I}$ is $$Q_1-q=\frac {Q_1-Q_2}2$$ and similiarly on inside of other plate $\mathrm{II}$ is $$Q_2-q=\frac {Q_2-Q_1}2$$ Note that these two are just equal but negative in charge, which poses a similiarity to a ...


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 ...


0

I really don't know what these integrals are This issue seems very broad, and I don't know what is really meant by this statement. Maybe the info below will help. I think my professor intended this to be a ⋅ in the integrals but has missed them out. Agreed. Just look at the definition of the electric potential to see this. I don't know [...] ...



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