7
Because the plates are connected by wires, they remain an equipotential surface. We can therefore safely assign both left plates (-) with $V=0$ and both right plates (+) with $V=V_0$. This MUST be true in both pictures A and B.
Now in what follows let $u$ denote the upper plates separated by distance $d_u$ and $l$ denote the lower plates, separated by ...
4
The charge of individual photons is indeed zero, but light, or microwaves, contain oscillating electric (and magnetic) fields.
Charged particles and electrically polarized molecules such as $\rm H_2O$ will experience forces when they experience these electric fields.
When the microwave oven is turned on, and the microwaves hit the water,
the polar molecules ...
4
It is the energy for the entire system. It is the energy required to put $+Q$ on one plate and $-Q$ on the other plate.
3
The number of field lines on a diagram has no quantitative meaning. Electric field lines are simply a useful tool to provide primarily qualitative information on the nature of the electric field. (See below for the information that field line drawings are intended to convey). Consequently, two different diagrams from different authors are likely to show a ...
3
Not in the same way as a black hole - at least, not for the electric force itself.
You see, a big part of the reason that black holes have an event horizon is that they "act on everything equally": that's the famed weak equivalence principle of gravity, that gravitational "charge", or gravitational mass, is strictly and exactly ...
3
Here is another question: What is an electron?
An electron is a particle with certain properties. It has mass $m_e$, charge $q_e$, and spin $1/2$.
So what is mass? If a force is applied to an electron, the mass tells you how quickly it accelerates. $a = f/m_e$. If the electron is near the Earth, it tells you how strong the force of gravity is. $f = G \space ...
3
If you look at Jefimenko's formulation of Maxwell's equation, the two sources of a magnetic field at $(\vec r, t)$ are:
$$\frac{\vec r-\vec r'}{|\vec r-\vec r'|^3}\times \vec J(\vec r', t_r)$$
$$\frac{\vec r-\vec r'}{|\vec r-\vec r'|^2}\times \frac 1 c \frac{\partial J(\vec r', t_r)}{\partial t}$$
with the retard time being $t_r = r-\frac{|\vec r-\vec r'|}c$....
2
Charge, mass, energy, these are just properties of physical objects. You could compare them to colors. If you say "protons are clouds of charge", it sounds similar to "cherrys are spheres of red". "red" is not some kind of material which you could use to constitute a cherry, similarly, charge is not some kind of material you ...
2
Physics uses mathematics. Mathematics is a discipline where everything can be defined and accurately predicted , and questions within a purely mathematical theory are answered by reference to theorems, and theorems by reference to the axioms of the mathematical theory. A why question in a purely mathematical theory ends up on the axioms, and the answer ...
2
When you state "Photon and Higgs boson have an electric charge equal to zero" you are probably thinking about experimental observations, however the model you describe is not aiming to be a phenomenological model but a pedagogical one. The Higgs boson in the Standard Model (which is aiming at describing the real world) is not exactly like the one ...
2
The direction in which the field lines point between a pair of opposite charges is simply a matter of convention. The little arrows could be reversed throughout the universe and the physics would stay the same.
2
tl;dr
Actually, it is not necessary to take the gaussian surface while computing field due to charged conducting plane as a cylinder with one end burried down in the plane, while the other in the free space. Let me prove to you what I just said.
Conductor
Let's imagine a really long conducting plane charged with surface charge density $\sigma$. Along with ...
2
It is the total energy stored in the entire system.
Consider that you have an uncharged parallel plate capacitor and you decide to charge it up such that the charge on one plate is $+Q$ and the charge on the other plate is $-Q$ where $Q>0$. At any point in time, the charge on the positive plate is $q$, where $0\leq q\leq Q$, which gives a potential ...
2
The theory perspective
I know that for a $0Ω$ wire of uniform cross sectional area, the potential difference across its ends is zero
Yes but only trivially so. Why should there be any potential difference across you resistor at all? Is it in a circuit? Is there a potential across it? Is there a current flowing through it? You haven't mentioned any of this. ...
1
The electrons don't move indefinitely on a straight line inside a wire, they colide with the nuclei repeatedly at a scale given by their mean free path, which is of the order of dozens on nanometers in metals. A change in the cross section that happens in scales longer than that should not affect resistance then.
1
In order to keep electric charges inside an ideal conductor of varying cross section, you certainly have to apply a force to the charges, because they can't all move on straight lines (inertial movement). And for that you need an electric field, as you have correctly pointed out. However, this accelerating electric field has nothing to do with Ohm's law (nor ...
1
First we can solve direct problem with fixed voltage $\pm 1$ on the capacitor, divided capacitor and divided capacitor with moved part. As result we can define electric charge on every part of capacitors (we use FEM to solve these problems). Distribution of potential, electric field and electric charge on the capacitor
Distribution of potential, electric ...
1
As a guide to dealing with this, use the following:
Capacitance $C=\frac{\epsilon_0 A}{d}$
where d is the distance between the plates and A is area.
The situation is like two capacitors in parallel and the total capacitance $C_T = C_1+C_2$
Also the electric field between the plates does depend on distance, $E=\frac{V}{d}$
The voltage will be the same for ...
1
The answer is current actually does flow and charges do accumulate at ends. That's why arcs form in open circuits at high voltage. And as you mentioned the flow would be a momentary one and not continuous.
When we connect a battery to an open wire, the potential of the battery does push charges around. But as soon as the charge reaches the end and accumulate,...
1
Current (or more properly, conduction current) can flow into an open wire and charge accumulate at the end IF there is displacement current to continue the circuit. Displacement current is not the flow of charge, but a change in the electric field. Two situations where this might happen is in an antenna, where the displacement current is provided by an ...
1
The electric field is not zero at point P according to Gauss's Law. The electric flux, through the entirety of the surface $ 4\pi R^2 $ (assuming it to be a sphere of radius R enclosing the point charges) is zero.
Mathematically, Gauss's Law states that, $$ \int \vec{E} \cdot d\vec{S} = \frac{Q_{enclosed}}{ \varepsilon_{o}} $$
It doesnt imply in any way ...
1
When we use a Gauss's Law to find the field from a point source (or other spherically symmetric charge distribution), we use a sphere as the Gaussian surface because we can see from symmetry that the field component normal to the surface must be equal at all points on the sphere.
In this problem you have two point sources, and no spherical symmetry. You can'...
1
The electric field from coulombs law
$$|\mathbf{E}|=k\left(\frac{q}{a^2}-\frac{q}{(r+a)^2}\right)$$
The flux through the given gaussian surface from the gauss law
$$\oint_\mathcal{S}\mathbf{E}\cdot d\mathbf{S}=0$$
Note that you can't take out $\mathbf{E}$ from the integral so you can't conclude $\mathbf{E}=0$.
1
It's more or less just a mathematical construct. If you want to solve the problem of the electric field created by a charge density $\rho$, then you have $\vec{\nabla} \cdot \vec{E} = \rho$. If we assume that both $\vec{E}_1$ and $\vec{E}_2$ satisfy this equation, then their difference satisfies
$$
\vec{\nabla} \cdot (\vec{E}_2 - \vec{E}_1) = \vec{\nabla} \...
1
How does matter have mass if energy doesn't have mass and matter is just made of particles that themselves are just formless clouds of energy?
But energy does have mass associated with it, about 1.1e-17 kg per Joule. This is due to the famous Einstein relation between energy $\Delta E$ acquired by a body when absorbing EM radiation and the corresponding ...
1
The volume $V$ consists of the free space between the outer boundary and the conductors; it does not include the conductors themselves. This means that the boundary of $V$ is not just the outer boundary, but also includes the surfaces of the conductors. Even if you define your potential such that $V_3 = 0$ on the outer boundary (we can always do this by ...
1
Eq. (4) comes about by construction of the model, and should be trivial to check by inspection: it is the reason behind constituting the multiplets this way. The generators of su(3), normalized like all generators, must, therefore, be the
Gell-Mann matrices halved,
$$
T_3=\operatorname{diag} (1/2,-1/2,0), \qquad T_8=\operatorname{diag} (1,1,-2)/2\sqrt{3}, ~~...
1
Well the thing is that for a spherical distribution of charges the potential and hence the electric field intensity at any point on the surface or out of the body can be calculated by assuming the charge on the spherical body to be concentrated at the centre of the sphere.This can be proved by Gauss's Law as I have illustrated below:
The net flux through the ...
1
Whether the items collide doesn't depend on the equality or unequality of the charges. Collision (or not) depends on the
impact parameter (line of undeviated velocity), which must be smaller than the combined radii of the particles ($r_1+r_2$), and
the initial kinetic energy of the moving particle, and
the product ($Q_1Q_2$) of the charges.
The easiest to ...
1
The repulsion between them is what mediates the collision.
Your second scenario isn't much different from the first. The incoming charge slows, there is a distance of closest approach, then they begin to separate. If the interaction between them is the Coulomb force, then the incoming charge goes slower and slower and slower, but never stops. This is ...
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