I was quite surprised to read this all over the news today:
Elusive, nearly massive subatomic particles called neutrinos appear to travel just faster than light, a team of physicists in Europe reports. If so, the observation would wreck Einstein's theory of special relativity, which demands...
And field lines from point charge come radially outwards in all direction
But:
My teacher, yesterday said:
Field due to line charges are only radially outwards because of the fact that: though the fields start in radially outward in all direction they ultimately become straight due to bending due to the presence of other point charges's field lines
@JohnRennie If you are unable to understand this^, I can show you what he meant through a diagram
field lines from an isolated point charge come radially outwards in all direction
Consider two point charges, spaced apart by some distance $d$ - it doesn't matter what the value of $d$ is.
Because the field has a $1/r^2$ dependence if we get close enough to one of the charges its field becomes arbitrarily bigger than the field from the other charge.
So if we get close enough to one of the charges it behaves as if it is isolated and its field lines emerge radially.
How close we have to get before the charge behaves as if it is isolated depends on how big the spacing between the charges $d$ is. If $d$ is very small then we need to get very, very close to the charge before it is (approximately) isolated. OK so far?
Just calculate the field using Maxwell's equations or Gauss' law or whatever method is easiest. Then the field lines are just the direction of $\mathbf E$.
For a line charge they are radially outwards (or inwards) normal to the line charge
@Abcd Yes, very near the charges that's true. But as soon as the field from the other charges becomes significant the field lines will curve towards the normal to the line.
@Abcd OK, the total flux is given by Gauss' law. We consider a cylinder centred on the line charge, then the flux though the surface of the cylinder is strightforward to calculate. yes?
Imagine the line charge was a light source and the cylinder is a screen. If you have a plate inside the cylinder it will cast a shadow on the cylinder. Yes?
And calculating the flux through the shadow is easy. If the angle subtended by the shadow is $\theta$ then the fraction of the total flux going through the shadow is just $\theta/2\pi$.
They've chosen the plate dimensions to make the calculation easy because $\tan\theta = \sqrt{3}$ so $\theta = 60º$
Calculate the flux through a part of the cylinder of length $2\ell$, then multiply by $\theta/2\pi$ to get the flux through the shadow. And that's the flux through the plate.
The length of the plate is $2\ell$ so we want the flux through a length $2\ell$ of the cylinder, and by Gauss' law that is: $$ \Phi = \frac{Q}{\epsilon_0} = \frac{2\ell\lambda}{\epsilon_0} $$
And the shadow covers $1/6$ of the cylinder so the flux through the shadow is: $$ \Phi = \frac{\ell\lambda}{3\epsilon_0} $$
Update 2017-05-01
The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.)
So here is my take. I modified the ...
@JohnRennie Answer is $\dfrac{\lambda l}{6 \epsilon_o}$
JEE2018 aspirants can try this question from AITS:
Find the time period of a satellite of mass m orbiting a pipe shaped planet of infinite length, radius R. The radius of the orbit of the satellite is r.
The surface mass density of the planet is $\sigma$
Calculate the energy released when 1000 small water droplets each of same radius $/10^-7m$ coalesce to form one large drop.The surface tension of water is 7X$10^-2$N/m
So what is happening in this question is that the total air-water surface area is changing as the droplets merge, and that means the energy of the surface is changing as well.
@Abhinav consider a line in the surface of length $h$. What we're going to do is pull this line a small distance $dx$ to create a new patch of surface. OK so far?
@Akash.B You seem to be interested in the exotic topics in physics, which is surely good for motivation. But to pursue physics at a professional level just asking and reading about popular science won't help. You need to go through all the pains of learning physics and mathematics properly.
@Abhinav I don't think it's a good way to learn, and it's outrageously stressful for the students. But that's what you're stuck with. I guess you just have to ride with it.
according to the formula, $\psi((x1,up),(x1,down))=-\psi((x1,down),(x1,up))$
that means of i add the amplitudes for this measurement, they cancel each other out, which means that i cant find them both at the same place
but their spins are in opposite directions, so shouldnt they be able to occupy the same place in space?
like in the helium atom, where they can both the in the ground state if they have opposite spins
i woukd think that i have to calculate the probability like this: P(2 electrons at x1, 1 spin up, 1 spin down)=$|\psi((x1,up),(x1,down))+\psi((x1,down),(x1,up))|^2=|\psi((x1,up),(x1,down))-\psi((x1,up),(x1,down))|^2=0$
@Akash.B get a book. Try to read it yourself. Or pay someone extra money to teach you.
@JohnRennie true. But, for someone who is preparing to pursue engineering so that they won't starve, is all that stress worth it just to be better at specific problem-solving?
was i unclear in my formulation of the question? According to the feynman lectures, whenever there are two indistinguishable ways that lead to the same measurement outcome, we have to add all the amplitudes and then square them to get the probability.
and according to the antisymmetry principle for fermions, the two amplitudes should cancel. But that isnt what we observe for example in helium.
I was quite surprised to read this all over the news today:
Elusive, nearly massive subatomic particles called neutrinos appear to travel just faster than light, a team of physicists in Europe reports. If so, the observation would wreck Einstein's theory of special relativity, which demands...
Update 2017-05-01
The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.)
So here is my take. I modified the ...
