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11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.


8

There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy $W$ of a uniform spherical charge distribution of radius $a$ and total charge $Q$ $$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$ Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ...


7

The lightning is just electricity, a huge burst of electrons that try to find the path of least resistance through the molecules of the atmosphere to the ground. The electrons come from many different places in the clouds and the atmosphere is not homogeneous: there are differences in humidity, temperature, density, particle count, velocity, etc., and this ...


5

There's no minimum distance. Yet, as the two particles get closer to each other, they will either scatter off each other (in the quantum mechanical sense of interacting via Feynman diagrams) or form a bound system - if we're talking electron-positron (which is as close to point charges as it gets), they might become positronium, but that won't last long, ...


5

The electric potential $\phi:\mathbb{R}^3\to\mathbb{R}$ is the solution to Laplace's equation and therefore a harmonic function. Harmonic functions enjoy several nice properties, some of them listed on the Wikipedia page. Concerning OP's second point, let us mention that there is a theorem similar to Liouville's theorem from complex analysis that a bounded ...


4

Why is it never just a straight line? I think it can be interesting to answer both questions and add a few interesting details to the answer already given. Electrons move in a straight line only in vacuum, where they meet no obstacles. What happens here is the same that happens with a river: the water of a waterfall goes in a straight line because it ...


4

I feel like it will be where the voltage is zero Imagine that, instead of voltage, the height is the same along any closed contour. If you think clearly about this, you'll realize that wherever the lines of equal height are closely spaced, the height is changing rapidly - the slope is large there. Where the lines are spaced far apart, the slope is ...


4

In this case, the image method can be used to calculate the potential (and hence the electric field) in the region $z>0$, with a negative charge $-q$ located at $(0,0,-d)$, since the potential would be $V(x,y,0)=0$, in this case. But for points in the region $z<0$, the potential is given by the solution of Lapalace equation $\nabla^2 V=0$, with ...


4

I think you are reading a lot into what is a minor distinction. Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what ...


3

This is a typical case of a problem which is clear enough physically speaking, but mathematically messy. Where rigorous results are folkloristically employed to achieve some result which, actually, would need much more care in deriving it... But presumably, mathematical details would not change the physical picture. Here the difference between theoretical ...


3

I too was confused by this difference between gravity and electromagnetism. Hopefully the following clears things up. The gravitational potential a distance $r$ from a mass $M$ is $$ \phi_g=-\frac{GM}{r}, $$ the gravitational field is $$ {\bf g} = - \nabla \phi_g, $$ and the gravitational potential energy (of two masses $M$ and $m$ separated by a distance ...


3

Your feeling is correct, since in this case $V$ is a continuous function as it is harmonic. So the constant value in $int(B)$ implies the same constant value for $\partial B$.


3

A charge radiates every time is accelerated. The power radiated is given by the Larmor formula. Putting this into the introductions to the motion of a charge in electromagnetic fields would be a meaningless complication, as much as considering air friction. But yes, a charge in a magnetic field would not spin indefinitely, even in vacuum.


3

The definition of current density is $J = \frac{I}{A}$, or more precisely, $J = \frac{\mathrm{d}^2 I}{\mathrm{d}^2 A}$. It is always true, by definition. $J=\sigma E$ is a different equation: it's equivalent to Ohm's law, which you know better as $V = IR$. Ohm's law is not universal; it only works for certain materials, called ohmic materials. For an ohmic ...


3

$\oint E\cdot dS = \frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV $ if $\frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV$ then $\frac{\rho}{\epsilon} = \nabla \cdot E$ if $\rho=0$ then $\frac{\rho}{\epsilon} = 0 = \nabla \cdot E$ Is this what your looking for? $\rho$ would be zero say, ...


3

Given that $\rho$ is the charge density, the integral, $$\frac{1}{\epsilon_0}\iiint_{V} \rho\, dV = \frac{Q}{\epsilon_0}$$ Now, Gauss' law states that, $$\iint_{\partial V} E \, dS = \frac{Q}{\epsilon_0}$$ Hence, we arrive at your 'global form' by simply equating: $$\iint_{\partial V} E \, dS = \frac{1}{\epsilon_0}\iiint_{V} \rho\, dV$$ By the ...


2

First, one of the implications of the electrostatic potential satisfying Laplace's equation is that the extremes are at the boundaries. If the potential of the surface of the sphere is zero and the potential at infinity is zero, the only solution for the potential outside the sphere is the trivial solution, i.e., the potential is zero everywhere outside the ...


2

Here's a simple way of looking at it: If you are close to an infinite plane, you may be feeling stronger attraction by every individual part of it, but "more" of those parts are pulling you at a significant angle. This way, a lot of the attraction is canceling out. As it happens (this is anything but coincidence though), these two opposite effects exactly ...


2

The electron in the $n$ semiconductor and the hole in the $p$ type semiconductor are delocalised and not bound to any particular atom, so arguments based on completing octets aren't useful. This diagram shows roughly how the depletion layer forms: In the $n$ type semiconductor the doping creates donor states in the band gap, and electrons from these ...


2

We have $$\tan \alpha=\frac{a-x}{b}\implies\sec^2 \alpha \,d\alpha=-\frac{dx}{b}\implies\frac{r^2}{b^2}d\alpha=-\frac{dx}{b}\implies rd\alpha=-\cos\alpha\,dx,$$ which is the desired relation with a minus sign, that must be present, since an increase of $x$ decreases $\alpha$.


2

Perhaps looking at it like this will help clarify things: You have two triangles, one with angle $\alpha$ and one with angle $\alpha + \mathrm{d}\alpha$. The sides opposite those angles differ by $\mathrm{d}x$, but the hypotenuses are essentially the same, both equal to $r$. Write an equation expressing the fact that the triangles' opposite sides differ ...


2

Here is your picture, with a couple of additional angles and segments drawn: Do you see it now?


2

That's probably for charged solid sphere, not a cylinder. In any case, setting the potential at infinity as zero, we have for $r>R$: $$V(r)-V(\infty)=-\int_\infty^rE(r')dr'\implies V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$$ For $r<R$, we got: $$V(r)-V(\infty)=-\int_\infty^rE(r')dr'=-\int_\infty^RE(r')dr'-\int_R^rE(r')dr' \implies \\ ...


2

The safety of a low voltage DC power supply is not established by the voltage on its output, but by the isolation between its input and output terminals. For example, a defective 12V power supply may have a short between the 120V AC input terminal and its negative output terminal. A user who would be connected to ground would then experience a 120V AC ...


2

It's not the case. In the second, the electron will radiate. This is how light species lose energy, and cool in Penning traps, and one of the factors that limit the energies of particles in circular particle accelerators. For a reference see: http://en.wikipedia.org/wiki/Cyclotron_radiation


2

If I understand correctly you are asking how observer dependent is electromagnetic radiation. The first thing is that non uniformly accelerated charges are described in a inertial frame by Larmor's formula and Abrahm-Lorentz force which take into account the radiated field and the recoil on the particle. Now in Newtonian mechanics and special relativity ...


1

Just for the sake of having an answer: In $\phi \nabla ^2 \phi$, consider the term $\phi \frac{\partial ^2 \phi}{\partial x^2}$. By the product rule, this is equal to $\frac{\partial}{\partial x} \left(\phi \frac{\partial \phi}{\partial x}\right) - \left(\frac{\partial \phi}{\partial x}\right)^2$. Combining all three terms, we get $\phi \nabla^2 \phi = ...


1

In every negative acceleration electron loos energy, and this of course in the form of photons. This is not surprising because negative acceleration could be only after positive acceleration, the electron has to move befor he could be stopped or declined. And how the electron can be accelerated? By electric fields where the electron get the kinetic energy ...


1

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. So, if there is force acting on a unit charge, then electric field does exist. It is the way by which we can prove the existence of electric field (as per definition demands). I don't ...


1

In electrostatics, the force on a charged particle is caused by the presence of an electric field $\mathbf{E}$, so that $\mathbf{F} = q\mathbf{E}$. Expressed in terms of the potential $V$, $\mathbf{E}=-\nabla V$. Physically, what that equation means is that in order to give rise to an electric field and therefore to a force field, you need a potential that ...



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