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"Colliding" two particles does not mean to bring two classical point-like particles on top of each other.
Even a macroscopic, Newtonian collision of two spheres does not consist of the two spheres somehow occupying the same space - their circumferences "touch", and upon trying to move into each other they repel each other and then ...
5
You should not try to think of the $\delta$ as a "usual" function when saying that (2) is undefined at $r \rightarrow 0$, because, in fact, $\delta$ is not a function. It is a distribution. You can view them as kinds of weights to integrate test functions, as in the following formula:
$$f(x) = \int\delta(y-x)f(y)dy$$
From this definition, two ...
5
This comes down to a question of what it means for particles to collide. This seems obvious because we know that two billiard balls can collide, but actually even for two macroscopic objects like billiard balls what it means for two objects to touch each other is surprisingly complicated. For more on this have a look at the question What does it mean for two ...
4
Note that Coulomb’s Law is experimentally determined, though because Maxwell’s equations describe all electromagnetic phenomena you can consider doing it this way (so you can see how the constant $k$ comes about):
Consider this equation which is one of Maxwell’s equations,
$$\oint_S \vec E \cdot \vec{dS} = \frac{Q}{\epsilon_0}$$
where this surface integral $...
3
This is the binomial approximation, where you can approximate $$(1 + x)^\alpha \approx 1 + \alpha x, \quad \text{(if $|\alpha x|\ll 1$)}.$$
The squared distance is in the denominator, as you'd expect: $$E_- = \frac{q}{4\pi\epsilon_0 \cdot AB^2}.$$
If you substitute for $AB$, you can easily show that it's just $$E_- = \frac{q}{4\pi\epsilon_0 r^2} \frac{1}{(1 -...
3
Isn't Gauss' Law for electric fields $=\frac{q_{enc}}{\epsilon_0}$derived based on the assumption that the $r^2$ (one from $d\vec{S}_{spherical}$ and the other from $\vec{E}$) cancel, otherwise the flux would depend on the radius of the sphere and we would need to know the relation between $k$ and $q_{enc}$?
It is easiest to see that $\Phi_E = \oint \vec{E} ...
2
You must note that the electrostatic force equation is :
$$F = \frac{kq_1q_2}{r^2}$$
The term $r^2$ matters the most.
The positive charge on the comb (let's just assume it as positive) is closer to the negative charge on the paper and farther from the positive charge on the far end of the paper.
Beacause of this separation differences the paper bit gets ...
2
I would say this is not quite the right approach. A perhaps better illustration is to show, instead of how big the number is, how not-big it is, and yet how, despite that, it has the potential to be extremely impactful. Sure, you can count grains of sand and pennies and imagine big mountains of these things, but electrons are also really really tiny, so it's ...
2
So one options is to go with the number of (dry) sand grains on earth's beaches. According to the link below, it's at $7.5 \times 10^{18}$.
Here's the link: https://text.npr.org/161096233
According to the next link I dug up, there are about $3 \times 10^{13}$ cells in a human body. Did they already check cells under a microscope in 7th grade? Anyway, the ...
2
To keep things manageable, I'll interpret the question like this: Given that quarks have their special pattern of electric charges with magnitudes $2/3$ and $1/3$ in units of the electron's charge, why do all hadrons (particles made of quarks) have integer electric charges in units of the electron's charge?
I'll use these inputs: Quarks are bound together by ...
2
Yes, you got it exactly right! If en electron from one side moves to the other side, it leaves behind a positive "hole" and this is where the positive charge there comes from.
2
Absolutely you can do this. It is how a Van-de-Graaff generator works. There is an energy cost to carrying the charges to the dome, so you are not getting enegy fro free, but once inside you can deposit the charge onto the dome and so build up enough charge on the dome, until the voltage is sufficiently high that a long spark occurs and discharges it.
2
Assuming that you mean to say that flow of energy can only be radially outwards due to symmetry, and since $\vec E$ already points in that direction, you can't have $\vec S$ pointing in that direction, so $\vec S$ must be $0$, and hence $\vec B$ must be parallel to $\vec E$, but this violates divergenceless-ness of $\vec B$, then I think your method is ...
1
Your dilemma is resolved if you start by defining the system under consideration.
System - test charge $Q$ only
Net force on charge is the sum of the external force due to you and external force due to the charges.
These two external forces are equal in magnitude and opposite in direction so, indeed, the net force of the charge is zero and the net work don ...
1
The total current $I$ through a surface is a global quantity (with respect to the surface). The current density $\bf J$ is a vector and it is a function of the point, in general. The integral over the surface of $\bf J \cdot n$ is the total current $I$, $\bf n$ being a unit vector orthogonal to each surface element. If the surface is so small that variations ...
1
$R^2\sin(\theta)\mathrm d\theta\mathrm d\phi$ is the area element on the sphere. $\theta$ is the polar angle, so $\theta\in[0,\pi/2]$ defines the upper hemisphere. The extra $\cos(\theta)$ arises because you are taking the $\hat z$ component of the force.
1
If you just have a spherical shell of constant charge density, there will be no electric field on the interior. You can see this by applying Gauss' Law to a sphere inside your shell,
$$\int_\text{over the sphere} E \cdot dA \simeq \int_\text{inside the sphere} dq = 0$$
But the same equation, for spheres outside the spherical shell, will yield $Q$ for the ...
1
Let the capacitance of the electrometer be $C_e$ and that of the ball be $C_b$. If the potential of the electrometer before the ball touches the electrometer for the $n^\text{th}$ time is $V_{n-1}$ and the potential of the electrometer when the ball is in contact for the $n^\text{th}$ time is $V_n$, then since charge is conserved and the ball, while in ...
1
As it is pointed out in the comment that $1/r^2$ is for a point charge. And a charge distribution can make of any kind of field like in this case that goes like $r^5$.
As an example, consider potential due to a change distribution $\rho(x',y',z')$. If you recall the multipole expansion that looks like
$$\phi_A=\frac{1}{4\pi\epsilon_0}\left[\frac{K_0}{r}+\...
1
Imagine an electron at rest, i.e. with zero kinetic energy. This electron is then accelerated and then stopped again. Both, the acceleration and the deceleration, would cause the emission of a photon according to your assumption. So this procedure would create two photons just like that. The electron has the same energy as initially. Therefore it would ...
1
It is not useful to think about it in terms of acceleration, because you don't know how it accelerates, you don't know if it has had a weak acceleration for longer time, or a strong acceleration for a shorter time. Fortunately, that's quite irrelevant...
You should think if more like a collision. Wen you study the impact of two billiard balls, you don't ...
1
You can evaluate the electric energy of a sphere $U_{el}$ in this way:
first consider that the charge is $Q=\rho \frac{4}{3}\pi R^3$.
Let's start from the general formula: $$U_{el}=\frac{1}{2}\int_{Volume}VdQ$$
Then, using Gauss'Law you can calculate the electric potential of the sphere for radiuses smaller then the radius of the sphere: $$V(r<R)=\frac{\...
1
Is there some theoretical reason for the lack of composite particles that would result in fractional charges?
It is an experimental fact that there are no fraction of the electron's charge particles in the data of the large number of experiments in hight energy physics.
So, a theory had to be developed that would fit mathematically this experimental ...
1
Imagine some volume filled with point-like charges at the location $\mathbf{R}$, the charge density is then
$$\rho = \sum_{n=1}^N q_n \delta(\mathbf{r} - \mathbf{r}_n - \mathbf{R}) $$
where $\mathbf{r}_n$ is the vector that shows from the centre of the volume to the nth charge. The macroscopic charge density however is an average over this microscopic charge ...
1
Your assumption that the repulsive and attractive forces are equal in magnitude is misguiding you.
When the comb is rubbed against dry hair, elections from hair jump into the comb. This gives comb a negative charge, leaving your hairs with slightly positive charge on them. When the tiny bits of paper are brought in vicinity of the negatively charged comb, ...
1
If you were to count 1 electron every second, it would take you around 20 billion years, more than current age of the universe, to count all the electrons in 1 Coulomb.
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