104

The title of the question asks if Maxwell's equations are mathematically precise. The answer is certainly yes. Maxwell's equations are well-posed differential equations. There is no ambiguity about the symbols, and one can unambiguously check if a given set of fields and charge/current distributions satisfy the equations. In fact, Maxwell's equations enjoy ...


45

As far as we know, electrons are fundamental particles and have no internal structure or components. Also, an electron cannot decay into other particles (unless it has a very high kinetic energy) because there is no lighter charged lepton for it to decay into. It can, however, annihilate with a positron to produce gamma rays.


34

An electron is an elementary particle in the standard model of particle physics. . The table axiomatically assumes that elementary particles are point particles in the QFT of the model, i.e. have no constituent parts. Depending on the quantum number conservation rules and if there exist consistent lower mass particles to decay to, elementary particles can ...


29

The concept of a dipole moment, and other moments such as a monopole, quadrupole, etc, comes from the process of writing a field as a sum of components called multipoles. This is known as a multipole expansion of the field. The reason we do this is that it can make calculations quicker and easier because it allows us to approximate a complicated field by a ...


25

Neutral charge just means there is no net charge when considering the entire atom/molecule. It doesn't mean there can't be a non-zero or non-symmetric electric field. This is especially true for molecules that are neutral yet still polar, such as water. you would think that there is more negative charge concentrated at the 'perimeter' of the atom, and the ...


24

This comment It was a "one-sided" fridge magnet. is a key piece of information. Your magnet is not a single dipole bar magnet; instead, it is a Halbach array, i.e., a carefully-engineered arrangement of individual dipole magnets which has been optimized to have a strong magnetic field near the magnet which becomes weak as you go away from it. ...


22

For two electrons separated by distance $r$, we have $$F_g = \frac{Gm^2}{r^2}$$ and $$F_e = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}$$ The ratio is $$\frac{F_e}{F_g} = \frac{e^2}{4\pi\epsilon_0 G m^2}$$ Now choose a unit system in which $4\pi\epsilon_0 G = 1$, yielding $$\frac{F_e}{F_g} = \frac{e^2}{m^2}$$ So the constant Feynman is referring to is the charge-...


20

Without a magnetic field existing, $\oint{\vec{E}\cdot d\ell=0}$ is always true. Your mistake is that you don't consider the edge effect of a parallel plate capacitor. As the picture you can see, when you approach the edge of the parallel plates, the horizontal component of the electric field cannot be neglected. Counting this effect in, you can have two ...


19

A proton is stable because of the strong force between quarks, which is not there in electron So you suggest a proton must rip itself apart, or has the ability to, as it is made up of quarks. But do the quarks also need to rip themselves apart? Its the same for the electron. We consider quarks and electron, both to be elementary - experimentally and ...


18

Unless I miscalculated, this does not seem possible for the following reason. Let us assess the electric field required to suspend a man with a mass of $m=70kg$. For simplicity, let us consider a plane capacitor with the plate area of $A=1m^2$, which seems reasonable for "one side of a human body". The force $F$ between the plates of the capacitor ...


16

When electric charges were first studied, scientists described effects just as you want to: electric charges exerting forces on each other. It wasn't until Michael Faraday came up with the idea of lines of force around 1855 that scientists began to rethink interactions that take place between separated objects. When one thinks about forces that affect ...


14

You are right, $\mathbf{B}$ is a vectorial quantity. Furthermore, it depends on position $\mathbf{r}$. Therefore we need to write $\mathbf{B}(\mathbf{r})$. The magnetic field around two current-carrying wires (with the currents flowing in the same direction) looks like this: (image from schoolphysics - electromagnetism - forces between currents) Notice ...


14

If you then add a charge outside of the constructed Gaussian surface, the Q enclosed remains the same. Therefore the calculation and the E calculated would be the same as well. This is incorrect. Gauss’ law only tells you about the net flux. It does not tell you about the E field. So, indeed, adding a charge outside the surface will not change the net flux, ...


13

Maxwell's equations are sometimes referred to as point equations, meaning that they apply at every point in space. That is true for $$\nabla\cdot\mathbf{D}=\rho , $$ because the divergence is only nonzero at a point in space if there is a nonzero charge density at that point in space. The fact that charges are physically point charges therefore means that ...


12

Ahhh, the delights of experimental science! You have imperfect data, and a particular model fits. But does the model make physical sense? Could there be another model which (a) is similar to the one you used (to within the noise) and (b) follows from first-principles? Would this unknown alternate model be better? Should you lie awake at night worrying that ...


12

What he means when he says it's a fundamental constant is just that you can measure it anywhere in the universe and you'll get (as far as we can tell) the same value anywhere. The ratio of the volume of the earth to the volume of a flea can only be measured here, and also depends on the flea. If it has a name, I don't know it. Rodney Dunning's answer calls ...


11

The best way to understand this is to appreciate that except for single electron atoms the orbitals are approximations that do not really exist, and to understand what the energy of an orbital is requires an understanding of the approximation. If we take a single electron atom like the hydrogen atom then the force on the single electron is central so the ...


11

This constant is the ratio between the fine structure constant $\alpha$ and the gravitational coupling constant $\alpha_G$. It is denoted by $N$ in Martin Rees's Six Numbers, but he mostly calls it "the big number". It has pretty important effects on what kind of structures are possible in the universe. Basically it sets the size hierarchy scale ...


10

The typical order of magnitude of an electric charge is $10^{-6}$ C or even $10^{-9}$ C in everyday life. For this reason, often, we see that the default unit for electric charges in Lab is a Nano Coulomb (nC). In your example, $10^6$ tons equals to $10^{9}$ kg (or $10^{10}$ N in Earth), and this means you have set the electric charges as $q=Q=1$ C. You ...


9

Let's assume the following: The terminals (i.e., the points where the wires make contact with the resistive material) are plates having the same shape as the cross-section of the resistive material. The material obeys the microscopic version of Ohm's Law, $\vec{J} = \sigma \vec{E}$, and has uniform conductivity. We are in steady state, i.e., all quantities ...


9

Yes, the arrow representing a centripetal force should point inwards. The electrostatic force is the centripetal force here. There is only one force (the electrostatic force), the centripetal force is not something different but a way of describing what the actual force does (i.e. that it forces the object onto an orbit). If one only wants to show real ...


9

Air, especially if humid (high water content), can conduct electricity to some extent. Electrons are being repelled from the negatively charged comb and some will gradually go into the air and move away. In this way the comb gradually loses its charge. Perhaps you could try it on different days. If the air is dry it should keep the charge for longer.


9

If there are charges left in any of the terminals, why approaching one terminal from a piece of paper or an electroscope doesn't show any kind of static electricity ? It is simply a matter of scale. A battery would have 1.5 V to 12 V worth of static electricity, but the minimum detection threshold for a human is about 3 kV of static electricity. So there is ...


8

The 1982 paper "An analytic solution for the potential due to a circular parallel plate capacitor" derives two exact formulas for the potential. From them you can derive the field by taking the negative gradient. One formula is an integral, and the other is an infinite series. Let $a$ be the radius of the circular plates and $2L$ their separation. ...


8

The answer is $$Q(C,V,A)=CV,$$ which is why you're having trouble. This is like the equation for a horizontal line: $$y=f(x)=2$$ Has no $x$ in the equation.


8

Clearly, after getting separated twice the distance, the energy of the capacitor $U$ will increase Yes, and the basic reason is work has to be done by an external agent to separate the plates against the attractive force between the plates. That work is energy transferred to the electric field of the capacitor and is responsible for doubling the stored ...


8

You have created a vector field that cannot be an electrostatic field. This is because $\oint \mathbf E \cdot \text d\mathbf l \neq 0$ for the field. If you want to be more careful with your setup, finite lines of charge have fringing fields that extend past the lines of charge. If you were to take this into account then you would get a zero line integral.


8

The key word in your post is net. Only the net flux must be zero, which means as much of the field must be pointing in as is pointing out of the surface. Look at this visualisation from Wikipedia If you draw any box around this dipole, you will notice as many field lines going into the box as are coming out. That's all Gauss's law tells us.


8

The electric field at position $\mathbf r$ of a point charge $q$ located at the origin is given by $$\mathbf E=\frac{kq}{r^2}\hat r$$ By definition, the electric potential at some position $\mathbf r$ due to this configuration is then $$V=-\int_\mathcal{O}^\mathbf{r}\mathbf E\cdot\text d\mathbf l$$ where $\mathcal{O}$ is your "reference position" ...


8

I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth. General relativity is not required to answer this question. Consider (for simplicity) a parallel plate capacitor where the field is constrained within the parallel plates. In this case the field is uniform (let it be $E_0$) ...


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