# Tag Info

0

Note that the Gauss law comes from the the Maxwell Eq. DivE=constant X charge density. When you place a charge in a dielectric the polarization so developed gives a bound charge opposite to that of that embedded. The total charge therefore decreases. The electric flux will therefore decrease since the total charge will now give lesser E. In other words the E ...

0

Electric flux is the surface integral of the electric field along or over a given surface . By Gauss's law , it depends upon the charge enclosed inside the surface . However, I guess what you mean by electric flux is not the total flux through the surface but just the flux through one of it's surfaces . Let us consider a slab made up of a material having a ...

1

For a Gaussian surface between the two plates, the total flux through the surface is zero. For the particular surface you give, all of the electric field lines crossing one of the circular faces cross the other face but in an opposite sense. This is because the outward normal vector for one face of the cylinder is opposite in direction to the outward ...

5

Your equation is wrong. The flux is not $2EA$. Rather, is $-EA + EA = 0$. The reason is because Gauss's law involves the dot product of $E$ and $dA$. The direction of $dA$ is the way out of the cylinder. When the electric field goes through the bottom of the cylinder, the electric field is in the opposite direction of the area vector, so the electric flux is ...

1

I don't see any problems up until maybe the very very last step (where you lost some constants at the very least and I can't tell what you were trying to do or why you think there is a problem). So you were at: $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx.$$ I'm not sure if you then tried a u-substitution or just found an ...

0

Why is that? Where am I getting wrong? The proper approach is to note that the integrand for the $x$ component is an odd function of $x$. But first, recall that $$\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_{0}^a f(x)dx = \int_a^0 f(-x)dx + \int_{0}^a f(x)dx$$ Now, for an odd function, $$f(-x) = -f(x)$$ thus $$\int_a^0 f(-x)dx + \int_{0}^a ... 2 how opposite magnetic fields cause physical objects to actually move (either together or away from each other). Here is a magnet, The magnetic field direction goes from the north pole to the south pole of the magnet, and observation tells us that north poles are attracted to south poles once one has many magnets. Thus a magnet will move another ... 0 An essential feature of a conservative vector field is that the line integral of the field is path indendent, i.e., the value of the line integral between two points depends only on the two points, not the path taken between the points. This is why we can associate a scalar potential function with a conservative vector field. If the field forms a closed ... 0 If the lines of electric feild form closed loops, the Guass Law won't be suitable. Therefore you can get the conclusion that the conservation of electric charge doesn't exist anymore. The lines of magnet feild form closed loops because there is no such thing like magnetic monopoles. 1 If I assume electric field lines form closed loop that would mean electric field has non zero curl. So I cant write electric field to be gradient of some scalar function. That would imply work done by the electric field will depend on the path. we know thats not the case really. Another way to see this: Closed electric field lines would mean number of ... 0 The word "current" contains the meaning of motion, what moves in an electric circuit is the electrons. In conductors electrons get an average drift velocity and create a current. In resistances this velocity falls very low ( that is what resistance to a current means, too) . What causes these induced surface charges - Is it just the strength of the ... 1 So why have we suddenly jumped from a line integral to not a line integral For a point charge, the electric field has a radial component only$$\mathbf E = \frac{kQ}{r^2}\hat{\mathbf r}$$Thus, the dot product of the electric field and the infinitesimal displacement vector is$$\mathbf E \cdot d\mathbf l = \frac{kQ}{r^2}\hat{\mathbf r}\cdot \left( ...

1

does the field ${\bf B} = \alpha f(t) {\bf \hat r}$ satisfy the no magnetic monopoles Maxwell equation? $$\nabla \cdot {\bf B} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \ \alpha f(t) \right) = \frac{2 \alpha}{r} f(t) \neq 0$$ unless $\alpha = 0$, or $f(t) = 0$, then no, this is not divergence-less. Try and picture what this magnetic field ...

0

The change to the field is not instantaneous. If you had a big magnet on the moon and somebody moved it, our instuments would show its location change at the same time we see (through a telescope) the movement.

0

You present both a static and a quasi-static case here. Let's consider them one at a time: 1) Static case: You applied a large potential to the screw relative to ground, which charged the screw. The firebrick is a good insulator, but not perfect, so the charge on the screw should polarize the material slightly, creating a slight electric field inside the ...

0

Your question is essentially: at what speed do the fields associated with point charges propagate or respond to changes in the charge distribution. A famous version of this question exists for Newtonian gravity: if the Sun disappeared suddenly, would Earth's orbit immediately change or would there be some delay? The answer is that it would take the time it ...

0

The second charge would have to be completely shielded, i.e. no induced charge should exist. If this is true, then the answer is yes. It's not weird when you think about light in the universe. We see the light of galaxies and stars that existed billions of years ago (say, more than 5 Gyr). However, even if they would still exist today, the light of our sun ...

2

Yes, you do need to have electric field in order to get an ExB drift. As the name suggests, this will happen in the direction of the cross product of the electric and magnetic field. $\vec{v}_{E \times B} = \frac{\vec{E} \times \vec{B} }{B^2}$ This drift does not depend on the charge of particle, meaning that it will move both ions and electrons in the ...

1

I agree with both of the two answers above. But I would also suggest that the term resistance should not really be applied at all to a capacitor ( I think it leads the question astray ) Going to the more general notion of impedance ( I would use the complex formulation ) the question makes more sense to me. This immediately brings in the importance of ...

0

A capacitor has an infinite resistance (well, unless the voltage gets so high it breaks down). The simplest capacitor is made from two parallel plates with nothing but space in between - as you can guess from its electronic symbol. In a DC circuit, a capacitor acts as an open circuit and does not permit current to pass. In an AC circuit a capacitor has an ...

0

Actually, the capacitor is an open circuit. It has infinite resistance. (I will assume a DC circuit.) But since the plates are so close, charge build up on one plate induces charge in the opposite plate of the opposite sign. This means that putting a battery across a capacitor (see the picture below) will let negative charge (in the form of electrons) flow ...

1

The electric field in the problem has no $z$ component, so it quite simple to calculate the flux through a cylinder with axis parallel to the $z$ axis; then you choose a cylinder that contains the sphere you are interested in. Let $\Sigma$ be the surface of the cylinder, $V$ its volume, $\Sigma '$ and $V'$ the surface and volume of the sphere; by the ...

-4

the electric flux entering the sphere is equal to electric flux leaving the sphere so it is zero...

0

If electrons are ejected out of an atom to create a constant current in a wire, then the nuclei of the atoms that lost the electron become positively ionised, which creates a positive radial electric field. Conceptually in electrostatics theory, this field must permeate throughout and beyond the confines of the wire. However, the released electron itself is ...

1

Your derivation of $V$ is fine, the $E$ not so good. But for the potential you did something better. In fact, you computed that potential for any height $d$ above the rod. So compute $V$ with a $d=0$, then compute it with your actual $d$ and then use the two to get an electric potential difference. Use the charge of the small sphere and the electric ...

-1

From http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml, the electric field is: \begin{align}E_p&=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{b}+\frac{1}{L+b}\right)\\ &=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{0.5}+\frac{1}{2}\right)\\ &=\frac{2.5Q}{4\pi\epsilon_0}\\ \end{align} Now, for the ...

Top 50 recent answers are included