# Tag Info

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 $E_0$) ...

7

Yes, the electric fields in a capacitor add to its weight. But not so that you’d notice with anything so crude as a balance. Suppose you had a one-farad “supercapacitor” that you could charge up to one kilovolt. The energy stored in the electric field would be $$U = \frac12 C V^2 = \frac12\times10^6 \rm\,J$$ This is an awful lot of energy for a capacitor, ...

4

Here is a simplified intuitive picture: No matter how far you are from the infinite plane, it looks the same. Zoom out, still infinite. For a finite plane, the further you get, the smaller it looks. So the field gets weaker. You can also do an integral. The electric field at distance $d$ perpendicular to the center of a circular plane will be proportional to:...

4

The edge of the universe doesn't affect a local experiment in the same way that the arbitrary constant you can change to $\phi(x)$ doesn't affect a local experiment. Boundary conditions really just constrain the constant in $\phi(x)$, the formula: $\phi(x) = \int \frac{\rho(x)}{|x-x'|}d^3x'$ will always give you a correct answer even if you don't treat ...

3

You seem to mix up two things here: Which charge distributions generate what multipole fields and how multipole-expansion works. Yes, you can reach a pure quadrupole field by a limiting procedure involving two dipoles (just take two anti-parallel dipole sources and let their distance go to zero from above). But that doesn't mean you can expand the quadrupole ...

3

The reason your hand doesn't quite touch the cube is that the electrons around the atoms in your hand are repelling the electrons around the atoms in the cube. Specifically, photons exchanged between those electrons mediate the electromagnetic repulsion. The actual distance depends a lot on how hard you press, and even if you press really really hard, the ...

3

One way to realise that the field won't decrease for an infinite plate is to imagine the field lines. The spacing between the field lines shows the strength of the field. For a positive sphere all the lines point away from the centre. The field lines will get further apart as we go away from the sphere, so the field strength decreases. For an infinite ...

2

I think it's easiest to do this in Cartesian coordinates, then convert back to cylindrical at the end. The potential at position $\mathbf{r}$ is $$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_{-a}^{a}{\left[\iint_{\mathcal{R}}{\frac{k\delta(x')\delta(y')}{\sqrt{(x - x')^2 + (y-y')^2 + (z - z')^2}}\,dx'dy'}\right]\,dz'},$$ where $\mathcal{R}$ is a region in $\... 2 I think you are reading too much into the significance of negative numbers. We need to use numbers in physics to quantify things. Take distance, for example- if I walk from my desk I can qualify the extent of my movement by saying that I have walked 10 metres, say. If I walk another 10 metres to return to my desk, I have walked 20 metres in total, but my ... 2 When the spheres exchange charge, they do this by exchanging electrons. Hence, the negatively charged sphere will have more electrons than the positively charged sphere, and should have more mass. It's useful to note here that the mass change will be very difficult to detect in practice. The electron mass-to-charge ratio is$5.7\times 10^{-12}\,\text{kg}/\...

2

"We say, Electric Flux, 𝜙𝐸=𝐸𝑆cos(𝜃) where 𝜃 is the angle between E and S. However, there is another definition for the same..." No. There is not another definition. The electric flux, $d\Phi$, through a surface element, $d\mathbf S$, is defined by $$d\Phi=\mathbf E.d\mathbf S$$ Gauss's law states that the net flux emerging through a closed ...

2

The electric flux through a small surface is defined to be $$\Delta\Phi=\mathbf{E}\cdot\Delta\mathbf{S}=E\ \Delta S\ \cos(\theta).$$ For larger surfaces the more general definition is $$\Phi=\int \mathbf{E}(\mathbf{r})\cdot d\mathbf{S}.$$ This definition applies to arbitrary surfaces (open or closed, plane or curved) and arbitrary electric fields (homogenous ...

2

A dipole, by definition, has a fixed separation between its positive and negative part - throughout all of the dynamics, the extra $-kq^2/d$ is an irrelevant constant that does not alter any of the physics. The electric potential $U$ is only defined up to a constant - its numerical value at a point is irrelevant, all that matters is how it changes in space.

1

$$\int \nabla \cdot E dv = Q enclosed / \epsilon_0$$ This is an equivalent form of gauss law. This states that the flux is dependant on the divergence of E about my volume enclosed by my gaussian surface. suppose I have a charge Q located at the center, the divergence of E is zero everywhere apart from the center at which It is $\rho / \epsilon_0$ If I were ...

1

Nope! The electric flux depends on only the charge enclosed, regardless of the geometry of your Gaussian surface. Here is Gauss’s Law in integral form: $$\oint \textbf{E} \cdot d\textbf{A} = \frac{Q_{enc}}{\epsilon_0}$$ Altering the surface would make computing the integral more difficult, but the net flux would still be the same regardless because some of ...

1

The electric dipole moment of a charge distribution about a point ${\bf R}$ is independent of ${\bf R}$ only when the total charge is zero. In that case p({\bf R})\equiv \int ({\bf r}-{\bf R}) \rho({\bf r})d^3r = \int {\bf r} \rho({\bf r})d^3r-\int {\bf R} \rho({\bf r})d^3r\\ = \int {\bf r} \rho({\bf r})d^3r-{\bf R}\int \rho({\bf r})d^3r\\ = \int {\bf ... 1 Bohr model uses "electrostatic forces" that work only for single electron atoms. Quantum theory today uses Schrodinger's equations to understand multi-electron systems. Moreover, Bohr model is actually incorrect, and that's why electric potentials are used to solve problems, since the angular momentum rule that Bohr devised with mvr = nh/2\pi isn'... 1 You are not applying G's theorem correctly in the case of the cylindrical capacitor. Take away the inner cylinder, leaving just the outer charged cylinder. Now consider a co-axial, cylindrical Gaussian surface inside the outer cylinder. According to G's theorem no net electric flux will pass through the Gaussian cylinder. So, by symmetry, the electric field ... 1 First, an electron has both an electric charge -e and an intrinsic magnetic dipole moment N-S, μ. Static electricity in the needle means that you rearrange the charges in the needle and you spatially split them in negative region (one end of the needle) and positive (other end of the needle). Negative region is the region which has an excess of free ... 1 Let's consider using a battery to create the current in the coil. To magnetise the needle, you would insert the needle into the coil, then connect the battery. Connecting the battery caused the current to change from zero to its maximum value which caused a magnetic field to build up inside the coil. When the battery is disconnected, the magnetic field falls ... 1 \vec{E}=-\nabla V means, kindly writing all the components and functional dependences explicitly, \begin{align} E_x (x,y,z) = -\frac{\partial V}{\partial x}(x,y,z) \,,\quad E_y (x,y,z) = -\frac{\partial V}{\partial y} (x,y,z) \,,\quad E_z (x,y,z) = -\frac{\partial V}{\partial z} (x,y,z) \,, \end{align} for \vec{E} a function associating a vector \vec{E} ... 1 You are asking about electrostatics first of all, so once you freeze time, there is no notion of causality. Without that notion you seem to be forgetting physics is about models. We accept the Poisson equation based on certain theoretical principles and observations and is successful. If you admit that your field is governed by a differential equation, we ... 1 The energy-momentum predicted by the Poynting vector for static fields cannot be detected by electromagnetic means. Only a divergency can be detected, or more precisely, the four divergency of the - in this case Belinfante - energy momentum tensor, which is actually the Lorentz four-force (force extended with power). It should in principle be detectable by ... 1 Not really answering your question , but abit more insight into what's going on. This is the same effect as induction pretty much. however instead of a changing magnetic field induces an electric field that moves charges INSIDE a medium, if the charges themselves are bound to the material then the total force on each of the charges will cause the ... 1 A current carrying coil experiences no net force in a uniform electric field because it is electrically neutral. For each positive charge in the coil with a force \vec F = q \vec E there is a corresponding negative charge with force -\vec F =-q \vec E. The isolated charge is not electrically neutral. The motion of the isolated charge is irrelevant as is ... 1 Yes. The whole concept of an image charge is based on this consideration actually: image charge produces the same field as the real distribution of charge. So if you, for example, call initial image charge IC1 and consider real distribution as another image charge IC2 (why not?), you'll see that they produce the same field. 1 Defining the charge density with a delta function in cylindrical coordinates just means that when we integrate over all space, we'd better recover the total charge. In Cartesian coordinates:\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty k\delta(x)\delta(y)\theta(a-|z|)dxdydz=2\int_{0}^a\int_{-\infty}^\infty\int_{-\infty}^\infty k\...

1

Someone more knowledgeable could provide a better explanation, but I believe that the electrostatic forces described are by nature "elastic" in that they exabit a behavior of wanting to return to an equilibrium position when displaced by something set on them. The atoms can be thought of as balls separated by springs, where the springs are actually ...

1

It is only applicable to cases where the fields fall of like $1/r^2$ to preserve the integral where the surface term increases like $r^2$. Gauss' law applies to any configuration of enclosed charge because the law only says that the net electric flux across the surface enclosing the charge equals the net charge enclosed divided by the electrical ...

1

Fix an arbitrary point $\mathcal O$ in 3D as the origin of the rectilinear coordinates system. This $\mathcal O$ (not denoted explicitly in your figure) is the common feet of vectors $\mathbf{r}$ and $\mathbf{r'}$ whose angle between them is $\alpha$. The observation point where you wish to find the potential $V$ is located at $\mathbf{r} = (x,y,z)$. The ...

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