Tag Info

Hot answers tagged

4

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 ...


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 ...


2

It sounds like the math is not the problem. It is visualizing the geometry. A point charge fills all space with an electric field. If you have multiple charges, they all create electric fields. To find the electric field at a point, add them up. Do this with a sphere. If you are outside as shown above, all the E field contributions are more or less to ...


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

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 ...


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

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 ...


1

Matter consists of electrons, protons and neutrons. In most cases the total charge of matter is zero and one would consider the body as neutral. However the charges are distributed in the body and each of them creates a non zero electric field (which keeps the matter stable). All individual fields add to a total field which is zero for a electric neutral ...


1

Musing on this I've thought of one possible take on it. If we assume a overall neutral universe, then every electric field line that starts on a positive charge ends on a negative charge somewhere and vice versa. That means that there are charges outside the sphere in the case where we've put a change in the hollow, they've just been pushed off to Far Far ...


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

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 ...


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 ...


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 ...


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 ...



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