New answers tagged

1

When you put +ve charge on a conductor, you are really removing the same amount of -ve charge, because only the electrons move. Gauss' Law assumes that charge is infinitely divisible, and can be spread uniformly throughout a volume or over a surface. This is a good approximation when the charge is on the order of $1 \mu C$, corresponding to about $10^{13}$ ...


3

I believe the answer to your question lies in the Gauss theorem itself $$ \oint \textbf{E}d\textbf{S} \sim Q $$ and the symmetry of the system, which defines the shape of equipotential surfaces. In case of a point charge there is a rotational symmetry about any axis going through the charge, so the equipotential surfaces are spheres whose area is ...


3

I don't like the "you can't get away" explanation. There is a simple explanation with field lines: In all three cases, the field lines are straight lines from the point charge to infinity. You can easily calculate the density of the field lines for each object. For a point charge, the "number" of field lines through any sphere around the point charge is ...


0

A capacitor is supposed to have infinite dimensions.And Electric field strength of a charged plane sheet of infinite dimensions is constant over infinity i.e. distance does not matter.


0

Using Gauss theorem we measure electric flux i.e. E.dA and if you use Gauss theorem for this particular geometry of plane parallel plates then you will get the usual result where the distance from the plate of the point where the field is to be measured will not appear. Further, E is proportional to 1/r^2 and the area ds is proportional to r^2 and while ...


1

For an INFINITE parallel plate capacitor, the electric field has the same value everywhere between the 2 plates. An intuitive reason for that is: suppose you have a small test charge +q at a distance $x$ away from the +ve plate and a distance $d-x$ away from the -ve plate. The +ve plate will repel the charge and the -ve plate will attract it. Now if the ...


2

In initially writing the Einstein-Hilbert action on a spacetime M: $$S=\intop_{M}(kR+l_{m})\sqrt{-g}d^{4}x$$ Where $R$ is the Ricci scalar, and $l_{m}$ is the matter lagrangian, we have k as an unknown (constant) quantity. Taking the variation of the action, with respect to $g^{\mu\nu}$ and dropping the integral, one obtains: $$k(R_{\mu\nu}-\frac{1}{2}...


7

We have Newton's law in the form $$ F = \frac{Gm_1m_2}{r^2}$$ which is the same as the field equation for the potential $$ \nabla^2 \phi = 4\pi G \rho $$ where $\rho$ is the mass density. The $4\pi$ here does indeed come from the solid angle of an entire sphere, but by redefining $G$ we could put it in Newton's law instead. This is connected to GR by ...


1

Pi crops up in mathematics and physics in all sorts of places that have absolutely nothing to do with circles and basic geometry, such as $e^{i\pi}=-1$, which is purely a statement about numbers. Don't think of it as the ratio of the circumference of a circle to its diameter. Think of it as a fundamentally important mathematical constant, and one of its ...


2

1) Electric flux lines model helps us to understand the behavior of an electric field much simply and it's pretty easy to visualize it. The definition of electric flux is the number of filed lines passing a given area normal to it. The field lines show the direction and magnitude of electric force at some point. The density of field lines at some region of ...


0

Electric flux is a useful mathematical concept which gives us a measure of how much the field flows through an area which is positioned within the field. Consider a surface of area A which is perpendicular to an electric field E. The flux is given by EA. The larger the electric field, the more the field 'flows' and therefore the larger the flux. Similarly if ...


2

Exactly. You have a little mass below you, and far more above you. But the mass above you is also in far more distance. The result is the - for the first spot, surprising - fact, that these two exactly balance eachother. Thus, if somehow you don't feel the gravity of the Sun, then there is a weightlessness in the whole sphere. Actually, you can imagine ...


0

The pdf J.D Callen, Fundamentals of Plasma Physics, chapter 3 defines $B=\sqrt{\vec{B}\cdot \vec{B}}$ and $\hat{b}=\vec{B}/B$, and proves that as you walk along a field line (arc length segment $d\ell$): $$\frac{dB}{d\ell}=\hat{b}\cdot \nabla B=-B \nabla \cdot \hat{b}$$ (where the second equality holds from $\nabla\cdot(B \hat{b})=0$) If the field lines ...


0

Applying Gauss at a distance $r$ from the centre of the sphere you get $g(r)\;4\pi r^2 = - 4 \pi G M_{\text{enclosed}} \Rightarrow g = - \dfrac{GM_{\text{enclosed}}}{r^2}$ where $g$ is the gravitational field. So all you need to do is to find $M_{\text{enclosed}}$ for each of the regions. What this equation tells you is that you can treat the enclosed mass $...


0

Ohm's law is \begin{equation}J=\sigma E \end{equation} Substituting Ohm's law in $ \nabla . E=\frac{\rho}{\epsilon} $ gives $$ \nabla .J=\frac{\sigma \rho}{\epsilon} $$ The continuity equation (charge conservation) is $$\nabla . J=- \frac{\partial\rho}{\partial t}$$ Equating the above two equations gives $$\frac{\partial\rho}{\partial t}=-\frac{\sigma \...


3

The boundary conditions by themselves can't tell you anything about a conductor. The boundary conditions can't even tell which side of the surface has the conductor! One way to model a conductor is as an Ohmic conductor where there is a constant $\sigma$ (different than the surface charge density listed in your boundary conditions) and then you assert the ...


2

You need to use Ohms law: $J = \sigma E$ which has to be added to Maxwell's equations as a bulk observation, as explained by this answer. You can then conclude that the electric field is zero in a conductor for: perfect conductor where $\rho = 1/\sigma = 0$ and $J$ is finite static case where $J = 0$ and $\sigma$ is finite


0

As Anthony B said,the number of field lines cutting any sphere surrounding a point charge is the same(because any field line which passes through a sphere of radius 1 also Passes through a sphere of radius 200) given that, the flux = E 4pir^2 should be constant. That explains the 1/r^2 dependance theoretically


1

This is a much more deeper question then it looks in first glance. The simple logic given by @Anthony B is not enough for proving the inverse square law. There are numerous experiments that have been done to verify this law. There is a collection of the experimental works in this review. In earlier days Cavendish and Coulomb have performed experiments with ...


3

As such there is no real theoretical proof to the inverse square dependence of the electric field in classical electrodynamics. It is an experimental fact famously known as the Coulomb's law. When combined with the superposition principle, it gives us the Gauss's law of classical electrodynamics: $$\nabla \cdot\mathbf E = \frac{\rho}{\epsilon_0}.$$ But, ...


2

You can prove it using the concept of electric flux. For instance. If you surround a point charge with a sphere if r=1, or a sphere with r =10, you know that the electric flux ( field strength times area) must be the same. A sphere is easy because every point is equidistant to the charge.


3

By "the designation of 1 and 2 is interchangeable" it means the flux passing through the upper surface equals the flux passing through the lower surface. In more details, because your system has reflection symmetry about the $xy$ plane and $E$ field is a vector, $\textbf{E}(x,y,-z)$ should be the mirror image of $\textbf{E}(x,y,z)$. Hence $$\textbf{E}_z(x,...


4

For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is ...


0

Gauss law in 2D would have to be: $$\oint \mathbf{E} \cdot \mathbf{\hat{n}} dl = 2 \pi q$$ because you are reducing your surface in 3D to a line in 2D, and keep the idea of measure of the boundary and its orthogonal direction or normal. To get the expression of the field you have to make use of the fact that the electric field is isotropic. In other words,...



Top 50 recent answers are included