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The electrostatic field depends only on the total charge distribution. If the charge distribution is known, as it is in your case, then you don't need to worry about the shape or conductivity of the structure supporting the charge. The charge on a conducting solid sphere will, as you say, distribute evenly at the surface. If you by some other method manage ...


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You're getting confused because the author you're reading has expressed Gauss' law using unusual and confusing notation, to the point where I would actually call it incorrect. I can sort of see how the confusing equations could arise from a derivation of Gauss' law from Coulomb's law. From Coulomb's law, you have $$\vec{E}(\vec{r}) = ...


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Typically, one defines a variable with respect to some observation point, P. In this case, the $\vec{r}$ is a vector pointing from the origin (defined by your chosen frame of reference and coordinate basis) to point P (which happens to be the location of your point charge $q$). The $\vec{r}$' here would be a vector from point P to the source of the field ...


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Take this as the definition of the electric flux going through some surface S in some electric field E $$ \Phi_E = \int_S {\bf E} \cdot d { \bf a} $$ This says that to find the flux on the surface we simply add up the electric field at each point on the surface. On the other hand this is Gauss's law: $$ \int_S {\bf E} \cdot d { \bf a} = ...


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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 ...


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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 ...


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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 ...


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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 ...


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Wikipedia> Electric charge: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric flux: In electromagnetism, electric flux is the rate of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a ...


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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 ...


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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 ...



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