# Tag Info

## Hot answers tagged electrostatics

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 ...

4

Worth noticing the difference between capacity and capacitance. You want the latter to have the meaning of the former - when instead it just describes how much the voltage increases when you add charge. It doesn't tell us when the capacitor is "full" - for that you need to know the rated voltage as well as the capacitance. When a capacitor is used in a ...

4

The integral you wrote integrates $\rho$ over the whole space. This is impossible to calculate if $\rho$ is not known in the whole space. For example, when the charge $\rho$ is known only inside some finite region enclosed by a metallic shell, the shell is known to have constant potential $\phi$ on its inner surface. This information is useless in ...

4

I think there is some trick? The electric field is horizontal thus the electric potential varies in the horizontal direction only, not the vertical direction. I wouldn't call this a trick but it does appear that the question tests your conceptual grasp of the relationship between the electric field and electric potential. In particular, you should be ...

4

This is a nice example of one of the foundational issues in relativity: how do we know that energy-momentum transforms like a four-vector, or, essentially, how do we know that $E=mc^2$? A historical overview is given in [Ohanian 2008] and [Ohanian 2009]. As Ohanian points out, there are logical problems if one tries to do what Einstein did in 1905 and prove ...

3

I think that electrons do escape. For example, Electrostatic Ion Thruster: Electrons are emitted from a separate cathode placed near the ion beam, called the neutralizer, towards the ion beam to ensure that equal amounts of positive and negative charge are ejected. Neutralizing is needed to prevent the spacecraft from gaining a net negative charge. I ...

2

What keeps electrons on a negatively-charged conductor from leaving? It is a quantum mechanical phenomenon. Wherever there exists an electric field potential there exist energy levels , i.e. stable orbital locations which can be occupied by an electron. How does this happen? Even the simple Hydrogen atom has a negative ion state, an anion. This is ...

2

Why do not you try this? The shell is always an equipotential object. When the charge is inside the shell, the potential of the shell is $kq/R$ with $R$ the radius of the shell, so the energy of the charge is $kq^2/R$. When the charge moves to infinity, the potential energy of it is zero. Therefore, the work is $kq^2/R$.

2

Consider that the electrostatic potential $\varphi$ isn't directly observable. The potential $\varphi+C$ where $C$ is a constant gives the same electric field, and so the same physics. Because changing $\varphi$ by a constant should give the same physics, you cannot conclude that the charge on the shell is $4\pi\epsilon_0 a_1 \varphi_1$. The charge is a ...

2

What Jim is talking about in the comments is a limit: for $d$ becoming increasingly large, $F$ becomes increasingly small. For example, if $q_1 = q_2 = 1\,\text{C}$ and the charges are a distance $d = 1\,\text{m}$ apart, we find that the force is $$F = k \sim 10^{10}\,\text{N}.$$ If the distance is made 1000 times bigger, the force is $$F = ... 1 r_{12} is the distance between the charges. r_{12} is not the distance that the charge has travelled. To calculate the work done to move the two charges together we would use \int_\infty^a F.ds which is like the formula you have in your question - your formula works if F is constant - here we need to use an integral as the force changes as r ... 1 Think of it like this. Electric fields point in the directions that positive charges will tend to go, which is towards a decreasing potential. So, if Q is positive, it will have radial electric field lines that will point along decreasing regions of electrostatic potential; that is, point A will have a higher potential than will point B, and so the ... 1 Dust sticking to things is a complex process but can be broken down into several stages and analyzed. First though lets define our dust. Dust Size The aerodynamics of dust are most easily approximated by pretending all of the particles are spheres with a density equal to water (1000 \frac{kg}{m^3}). Each particle is assigned an aerodynamic diameter that ... 1 Your whole derivation is correct. Even in the presence of two different dieletric materials, the \mathbf{D} field will not be affected, but for the free charge density that you already dealt with. So the field will be$$\mathbf{D}=\begin{cases}\dfrac{1}{5}\beta r^3\hat{\mathbf{r}},&\text{if $r<a$},\\ \dfrac{1}{5}\beta ...

1

The equation for capacitance is Q=CV or V=1CQ. I don't understand what is the physical meaning of this "C": Does the charge in a system changes linearly with voltage under all circumstances? This first part is the statement of the behavior of an "ideal" capacitor. Because it is an idealization, it is easy to characterize as a linear component whose ...

1

TEM waves do exist in multi-conductor waveguides such as coaxial guides, in fact they exist in any homogeneous waveguide with more than one conductor. There are no propagating TEM, TE or TM modes if the cross section is inhomogeneous but at cutoff frequency the hybrid modes degenerate to the respective transversal modes.

1

A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is $K$ and the thickness of the dielectric is $t$, then for calculating the force it behaves as if the thickness was $t\sqrt{K}$. To see this let's take the example we know about where the dielectric fills the space between the charges: In (a) the thickness of ...

1

Can we just calculate by using 1/2 CV^2? Yes. If $v_C(t)$ is the instantaneous voltage across the capacitor, the instantaneous stored energy is just $$U_C(t) = \frac{C}{2}v^2_C(t)$$

1

We start with the integral $$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$ Since we are integrating over $\vec{r}$, we can without loss of generality, arrange for $\vec{r}'$to lie along the $+Z$ axis, so that $\vec{r}'=r'\hat{z}$. Then the angle between $\vec{r}$ and $\vec{r}'$ is the standard angle (in spherical coordinates) ...

1

In answer 1) Yes, charge will flow to make the potential of the wire the same as the + terminal of the battery given that the wire is neutral or at earth prior to making contact. The ammount of charge that would flow would depends on the difference in potential between the wire before you attach it to the battery and the battery - we could calculate that ...

1

The lighting shape and path is a non-linear phenomenon. Why? Because it involves a multitude of interactions with the medium (air molecules, atmosphere) and these depending on range of interecations are also non-linear. However there are compatibility conditions (or conserved quantities), as such this is a good example of the fractal geometry of nature.

1

The reason is that Coulomb's law is only directly valid for point charges, i.e. for charges, sizes of which are much smaller than distance between them. For particular symmetry reasons it appears to also be applicable to spherically symmetric balls of charge - but note that the distance you should take is not between the surfaces - it's between the centers ...

1

Let us first consider a capacitor that is charged and not connected to a battery or other electric power source. I think you need to take into account that the Electric field in the capacitor is reduced by inserting the dielectric and also the voltage drops between the plates by inserting the capacitor. The dielectric is pulled in to the cap as the ...

1

Electrons do leave surfaces either due to electrical potentials pulling them as you describe or as a result of a combination of heat plus electrical potentials. THe electrons are held in because the Fermi level is lower than the vacuum level. Furthermore, the difference in potential between the Fermi level and the vacuum level is 'felt' by the electrons ...

1

"Vacuum" actually means the ground state of space. When there are real particles in space, "it" is in an excited state, and no longer the vacuum. But we usually think of a small region of empty space as approximating a vacuum, for reasons of locality, etc. If you have two real particles in space, and talk about their mutual interaction, then you necessarily ...

1

Your boundaries are at $r=a$ and $r=b$. Notice that the potentials at these two surfaces are independent of $\theta$ (they are spherically symmetric). Look at a list of the first few Legendre Polynomials $P_{l}(\cos{\theta})$. For what value of $l$ does $P_{l}(\cos{\theta})$ not depend on $\theta$? Further, notice that $V(a) = V(r=a,\theta) = V_{0}$, and ...

Only top voted, non community-wiki answers of a minimum length are eligible