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1

Before we talk about the term "spherical shell" and "thin sphere", let us talk about the possible cases of the sphere itself. Generally speaking, it depends on what is given to you. If you have been given $\rho _s$, then probably it will be hallow. If you have been given $\rho _v$, then definitely it will be solid. Keep in mind that whether it was ...


2

Wouldn't the field line end at the charge +q, and so the field lines that enter don't necessarily leave, making +2q contribute a negative flux? The quantity of field lines that terminate on the charge -q is unchanged by the presence of charge elsewhere. Imagine -q field lines terminate on the -q charge. Then, the net electric flux outward through ...


1

The net flux crossing the Gaussian surface (according to Gauss's law) is irrespective of those charges outside it. In some detail about your case: The total flux leaving the surface drawn is -q (so it is actually entering the surface not leaving it). So although the 2q charge outside the surface do contribute to the flux crossing the surface, the net ...


1

In a conducting sphere, like charges repel each other - so when there is a net charge, it will all appear on the surface (they try to get as far away from each other as possible). In a non-conducting material, charge will stay wherever you put it - so if you have a solid material with a net charge per unit mass (not sure how you achieved that), it will not ...


3

It will help you understand the quantum mechanical picture if you read up on atomic orbitals. These are the loci around the nucleus where the electrons have a probability to be found. You will see that the orbitals have a shape, which depends on the angular momentum of the state. The electrons carry the charge and thus you can interpret the plots as ...


2

Yes indeed. To get the result you stated from Gauss' Law you must asume that the charge is distributed in such a way that you have a spherically symetric field. How you get to that field doesnt matter though. So the charge might all be concentrated at the center or all lying on the surface of a sphere: it doesnt matter as long as the field is spherically ...


0

By Gauss' Law, $E\cdot A=\frac {q}{\epsilon_0}$ (assuming that the Electric Field is constant at every $dA$ and that it is always parallel to $dA$, which it is in this case) Let us define the charge contained in the original problem cylinder as being $Q$ whereas the charge in the smaller Guassian cylinder as being $q$. Therefore, the charge in the ...


1

Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is $$E=\frac{ρr}{(2ε_0)} $$ Now, to calculate the potential difference between the surface and axis of the cylinder, $${\Delta ...


0

Actually, using a cylinder for your Gaussian surface is your best approach. The fact the area is infinite should not matter, if you expression the infinite length of the cylinder as a variable, say $l$. Noting that the Gaussian surface area, $A = 2\pi Rl$, and that $Q = \rho l$, the $l$ term should eventually cancel out in your working out.


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In a conductor (perfect conductor or imperfect conductor) in steady-state there is no macroscopic electric field, i.e. if you average the electric field over a volume that is big enough to include many atoms, the result is zero. That's what the textbook meant. You are certainly right that there are nonzero microscopic electric fields, e.g. the large field ...


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Your textbook is talking about a perfect conductor, a model of a real conductor. In a perfect conductor charge is considered to be chopped up into infinitely tiny portions. Charge in a perfect conductor is a continuum, not quantized as we know charge in real conductors to be. There are no discrete charges ... electrons or protons ... to get close to.



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