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6

If you are talking about point charges then, as explained above, the answer is no. But in the case of non-uniform charge distributions, it is possible for same-charge particles to attract, if they are sufficiently close. As an example, the following two particles are identical, each having a net charge of -1. Plotted below them is their potential energy as ...


3

In spherical coordinates, the volume element becomes $$ dV = r^2 \sin\phi\, dr \, d\theta\, d\phi $$ where \begin{align} r^2 &= x^2 + y^2 + z^2 \\ \tan\theta &= \frac xy \\ \tan^2\phi &= \frac{x^2 + y^2}{z^2} \end{align} If the density is uniform over a sphere of radius $R$, then the radial integral becomes $$ \int_0^R \frac{\rho\,\hat r}{r^2} ...


3

I think the easiest way to visualise the various multipoles is through their connection with the spherical harmonics. These are probably best known as the functions that give the shape of the atomic orbitals in the hydrogen atom. The $p$ orbitals correspond to a dipole, the $d$ orbitals to quadrupole and the $f$ orbitals to an octapole, that is a charge ...


3

You have miscalculated, an easy mistake to make. In your first line of math, $$F=\frac{315\times 10^{-6}}{4*10^{-14}}$$ not $$F\ne\frac{315\times 10^{6}}{4*10^{-14}}$$ So you are off by 12 orders of magnitude, but this is still a very large force. Keep in mind what you are suggesting. Electrons have a charge on the order of $10^{-19}$ coulombs, so you are ...


3

The object for which you need to find the electric field is a uniformly charged sphere. Uniformly charged means that at every point on the sphere the charge density is same. Suppose someone blind-folded you and then he rotated the sphere in some arbitrary fashion about the origin(assuming your sphere has origin as the center). Then he takes off the blind ...


2

Rotation is a relative quantity. If you rotate by an angle $\theta$ relative to me that means I rotate by an angle $-\theta$ relative to you. So rotating a system of charges by an angle $\theta$ is exactly the same as leaving the charges stationary and rotating yourself by an angle $-\theta$. If you leave the charges stationary then obviously the electric ...


2

Your fillings are either gold / palladium inlays or silver-mercury amalgam. The composite or enamel fillings don't apply here. Welding fillings together will require that you have fillings in opposing teeth. It is certainly possible to weld metals together with a brief electrical current - it's done by robots in car factories every day. However, it requires ...


2

Half of the energy is lost to the battery's internal resistance (or other resistances in the circuit).if you try to consider an ideal battery with 0 internal resistance, the notion of charging the capacitor breaks down.since the capacitor and the battery are connected by a (0 resistance) wire, their voltages are the same the instant they are connected, no ...


2

To understand Purcell's point, you should give a read to the first chapter of Griffiths' Introduction to Electrodynamics. At the end of that chapter, Griffiths deals with this very issue in some detail. The crux of his treament lies in the definition of divergence of this vector function: $\mathbf{v}=\frac{\hat{r}}{r^2}$ Before I proceed with the rest of ...


2

Let's maybe clear up a (possible) misunderstanding first. Mathematically speaking, to say that $V$ has no dependence on $z$ implies the following: $V(x_1,y_1,z_1) = V(x_1,y_1,z_2)$. The simple and intuitive way to see this is to observe the symmetry along $z$: There is no distinguishable difference between the points ${x_1,y_1,z_1}$ and ${x_1,y_1,z_2}$; ...


2

You have two questions, and they have different answers. First of all, let's be clear about what Gauss's law is in integral form: $$ \int \vec{E} \cdot \mathrm{d} \vec{A} = \frac{Q_\mathrm{encl}}{\epsilon_0} $$ In words: the total flux integrated over a closed surface is equal to the charge enclosed in that surface, divided by the permittivity of free ...


1

This is best understood by approximating the dipole as a pair of finite charges $\pm q$ separated by a finite distance $d$. In a uniform electric field, the electrostatic forces on each of the charges will cancel out exactly, but in a non-uniform one the forces on the two will be slightly different, leading to a slight imbalance and therefore a non-zero net ...


1

A capacitor is assumed to be self-contained and isolated, with no net electric charge and no influence from any external electric field. The conductors thus hold equal and opposite charges on their facing surfaces. As electric field is established which originate from the positive plate and end on the negative plate. Also, the field is uniform so is the ...


1

Well, if you have an extremum (say local maximum) of $W$ at $p$, then you have a small open ball $N$ centered in $p$ such as $W(x)< W(p)$ for some $x\ne p$ in $N$. Therefore $$ \frac{1}{{\rm vol}(N)}\int_N W(x)dx < W(p)~~.\tag{1}$$ assuming we have taken the ball $N$ small enough for $W(x)<W(p)$ for some $x\ne p$ in $N$. But if $\nabla^2W=0$, by ...


1

Laplace's equations, as well as all other kinds of fields equations, always describe a local property, whether or not the right hand side is non-zero. In particular, it restricts all the possible fields configurations to the ones obeying $$ \nabla^2 \phi(x,t)= f(x,t). $$ It describes the local properties of the field $\phi$ upon the space in the region where ...


1

Ok, here is an answer in some simple words. Harmonic functions obey the following property: if you draw a circle around any point $x_0$, and take the average of the function over that circle, that average will be equal to the value of the function at $x_0$. Now, if $W$ is not constant, it must have a maximum somewhere. Let's call this maximum point $x_0$. ...


1

The E field inside the conducting wire is 0, so what is it really doing? The potential difference between two points is related to the electric field along the path between them: $$V_{ba}=\int_a^b \vec{E}\cdot{}\vec{dl}$$ So the fact that the high-conductivity material forces a (near-)zero electric field is exactly why the two ends of the conductor ...


1

Think of a gas. We will ignore gravitational potential, and we'll consider the situation to be at steady state. Compressing the gas takes energy, so we conclude that the entire container of the gas will be at a constant level of compression - in this case, constant density as well. You may intuitively understand that this is true no matter how oblong the ...


1

How can the energy required for squeezing be the energy required for assembling the charges? The energy required for squeezing the sphere from initial radius $r_0 \to (r_0-dr)$ is not the energy required to assemble the charges. He merely calculated the change in energy (or the work required) each time you squeezed the sphere by a displacement of $dr$. ...


1

The electric field is a vector quantity, representing the electric force per unit charge acting on a test particle at a particular position in space. Since force is a vector, the electric field too is a vector quantity. The electric potential however is not a vector. The electric potential is the amount of electric potential energy that a unitary point ...


1

A "point xxx" is almost invariably a theoretical concept intended either to simplify assumptions, or to allow you to ignore aspects which are more or less irrelevant to the aspects being considered. Any physical attribute in a point object usually would led to an infinite value of some other attribute. A point charge has infinite charge density and so ...


1

Electric field is zero in that point because the sum of electric field vectors have same intensity and direction, but are opposite. That point is halfway between two like charges.


1

Is it simply that the repulsion of the equidistant point charges creates a dead zone? It has nothing to do with repulsion or attraction. To find the electric field at some location due to a set of point charges, you have to add the electric field contribution due to each of the point charges. You also have to remember that the electric field is a ...


1

In short, while $q_2$ does exert a force against $q_1$, this force does not perform any work because $q_1$ does not move. The work performed is the product of the force exerted times the distance the particle moves against (or with) it; since $q_1$ does not move there is no work performed on it. The potential energy is defined to be the work required to get ...


1

Gauss' Law says $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ whereas the total vector area is $\iint_S \mathbf{dA}=\mathbf{0}$ for some closed surface $S$. The total vector area is taking vector sum of all the differential area vectors that are normal to the surface, whereas $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ is taking the dot ...


1

There is a much simpler way to find the charge on the conductor. Using the method of images, the image charges are just the negative of the charges above the plane. Therefore, the total charge on the surface of the plane is -Q. The first-term in your dipole field is wrong. The unit vectors have unit value and do not cancel r squared in the denominator.


1

Yes it is right to say that electric field by a finite plate is not constant and zero at infinity. But in case of capacitors,the separation between the plates is so small as compared to dimensions of plate that with respect to the separation between the plates the plate itself can be considered as infinite. It is just a relative assumption to simplify ...


1

Let $x$ be dimensionless and Using the property $\delta (ax)=\frac{1}{|a|}\delta (x)$ we see that indeed the dimension of a Dirac delta is the dimension of the inverse of its argument. One reoccurring example is eg $\delta(p'-p)$ where $p$ denotes momentum, this delta has dimension of inverse mass in natural units.


1

Consider the three-terminal device that is your stacked capacitor: A ----============================================= (dielectric medium ɛ) =============================================---- B (dielectric medium ɛ) C ----============================================= All three plates have the same area A. It's ...


1

I think, in order to generate the field $E$ there should be some charge distribution inside the cube. For example an electromagnetic wave propagates through vacuum, that is, there is no charge necessary to generate it, at least with fitting boundary conditions. But for a static electric field with non-zero divergence, there should be some charge ...



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