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16

Charge is a fundamental conserved property of particles. It is, if you like, a measure of how much a particle interacts with electromagnetic fields. A particle with charge can produce and be affected by electromagnetic fields. This is what we mean when we say a particle has charge. Its a simple quantised way to measure the coupling strength of particles with ...


15

Electric field lines are a visualization of the electrical vector field. At each point, the direction (tangent) of the field line is in the direction of the electric field. At each point in space (in the absence of any charge), the electric field has a single direction, whereas crossing field lines would somehow indicate the electric field pointing in two ...


12

This is a good example of a procedure that happens in many areas of physics. In general, physical laws - and particularly conservation laws - tend to be most naturally phrased in integral form, or even in mixed integro-differential form. For an example of the latter, consider the integral form of Faraday's law: $$ \oint_{\partial S}\mathbf{E}\cdot\text ...


11

If there was a closed field line a particle following that line would eventually return to the same place but having a different energy so the field would not be conservative.


11

If you put your rod in a ultra high vacuum it will stay charged almost forever, but since you probably keep it exposed to air, this is where the electron excess slowly migrates (and the same for the electron defect in the silk). Since the charge exchange requires an hit between an air molecule and a spot of the rod where an electron excess is present, and ...


11

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


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} ...


9

In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where $\mathbf{E}=\mathbf{0}$. Letting the charges be located at $(\cos(2\pi k/N),\sin(2\pi k/N))$ for $k\in\{1,2,...,N\}$, we can generate plots of $|\mathbf{E}|^{-1}$ for various $N$. The zeros of ...


9

Although this is quite an old question, I have to disagree with answer by LuboŇ°. First, Pauli exclusion principle says that no two fermions can share the same state. But, if the electrons have different spins (i.e. are in so called spin-singlet state), then they can be in the same positional state. Next, indeed, in classical case, two charged particles ...


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 ...


8

One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...


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 ...


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 ...


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 physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a ...


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 ...


7

Field lines draw all of their validity from Gauss's law for the electrostatic field, $$ \nabla\cdot \mathbf{E}=\frac1{\epsilon_0}\rho,\ \text{or equivalently}\ \oint_{\partial\Omega}\mathbf{E}\cdot\text d\mathbf{S}=\frac1{\epsilon_0}Q_\Omega, $$ where $Q_\Omega=\int_\Omega\rho\,\text d\mathbf{r}$ is the electric charge in a volume $\Omega$ whose surface is ...


6

If the rods were really far apart then the positive charge would be equally distributed throughout each rod. If you push the rods together then the new equilibrium involves fewer charges bunched around the closer points and more of them at the far ends; the energy you exert is the energy it takes to move these charges around. Even if we didn't consider the ...


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 ...


6

Charge is a quantity which arises from Noether's theorem, due to continuuous global symmetries (up to a total derivative) of an action, and as such we have many types of charge, other than electric. For example, consider the Dirac Lagrangian, $$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi$$ which describes fermions. It is invariant by a ...


5

The electric field at any point is the sum of all the fields due to each individual charge in the system. The field has a magnitude and a direction. The field lines are a representation of the magnitude and direction of the field over an illustrated area. The field lines point in the direction of the field. If lines from two sources were to cross, we could ...


5

There is indeed a connection. The holomorphy is easily seen in the electrostatic potential. In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified ...


5

Let's draw the setup: The expression for $V(r)$ is simply (I'll set $\kappa$ to 1 for convenience): $$ V(r) = \frac{20}{r} - \frac{40}{r+1} $$ so the potential is zero when: $$ \frac{20}{r} = \frac{40}{r+1} $$ Only this isn't quite right because the potential for each charge is symmetric so the potential due to charge $A$ obeys $V_A(r) = V_A(-r)$ and ...


5

$$A=\pi r^2$$ $$\frac{dA}{dr}=\pi\cdot2r$$ $$dA=2\pi rdr$$ Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$ You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit ...


5

for an infinite sheet of constant charge, the text says that the electric field is constant on any one side of the sheet. But that seems intuitively wrong to me, since I would think the field should be stronger the closer a point is to the sheet. There's a geometric scaling argument at hand, and you probably need to appreciate Gauss' Law to get a real ...


5

In physics and engineering, we often abstract and idealize a physical problem to gain insight into the physics, e.g., infinite plane of charge, infinite line of charge, point charge, etc. Now, it goes without saying that if these idealizations didn't represent good approximations of relevant physical systems, they wouldn't be used. With regards to your ...


5

It will never form a stable structure as both electrostatic and gravitation forces decreases at the same rate with distance. And if one dominates at some distance, then it will continue to dominate forever, as the charge to mass ratio of the star will still be same as that of the electron.


5

The only property of metals used in deriving $C=\varepsilon A/d$ is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter. But if you use something other than metal, then it will of course change the capacitance.


5

While it's good to be skeptical of approximations, it should be absolutely understandable for you to ignore this effect in the case of, say, the Earth's E/M field acting on a tiny electron beam. You couldn't find a simple general formula for the force on a particle due to the field it creates, and so things are dealt with on a case-by-case basis. One case ...


5

The electrical field of a line charge is a potential field. In the following we use cylindrical coordinates $\vec{r}(z,\alpha,r)$. If the line charge lies along the $z$-axis from $z=-L/2$ up to $z=L/2$ then the potential is $$ \varphi(\vec{r}(r,\alpha,z)) = \frac\lambda{4\pi\epsilon_0}\ln\left(\frac{\frac L2-z+\sqrt{\left(\frac L2-z\right)^2+r^2}}{\frac ...



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