The charges outside the shell affect the charge distribution on the outer shell only. Likewise, the charge inside the shell affects the charge distribution on the inside surface of the shell only. What you do to the outside charges cannot be communicated to the cavity inside because the field in the conducting material is always zero and that decouples the two regions of space.keemosabi said:... but why wouldn't the charges outside of the shell induce a charge on the shell, which would then affect the field at P?
I pretty much understand, I'm just wondering how Gauss' Law proves this. Can't a point outside of a Gaussian Surface still affect the E-field at a point on the surface? It just wouldn't be in included in the integral, right? So how would this say that the charge on the inner shell does not affect the E-field at point P?kuruman said:The charges outside the shell affect the charge distribution on the outer shell only. Likewise, the charge inside the shell affects the charge distribution on the inside surface of the shell only. What you do to the outside charges cannot be communicated to the cavity inside because the field in the conducting material is always zero and that decouples the two regions of space.
You can see this from Gauss's Law. If you draw a Gaussian surface so that it is entirely inside the conducting material, the field on it is zero everywhere which means that the flux through it is zero which means that the net charge is zero always and no matter what happens outside the shell.
Not if the Gaussian surface is entirely inside the conducting material. The integral is zero because the E-field is zero everywhere on the Gaussian surface regardless of what kind of charge you put outside or inside the shell anywhere you please. This then says that the net charge enclosed by the Gaussian surface (as defined above) is zero no matter what.keemosabi said:I pretty much understand, I'm just wondering how Gauss' Law proves this. Can't a point outside of a Gaussian Surface still affect the E-field at a point on the surface?
So it's like the conducting shell prevents the outer charges from affecting the E-field inside of the shell?kuruman said:Not if the Gaussian surface is entirely inside the conducting material. The integral is zero because the E-field is zero everywhere on the Gaussian surface regardless of what kind of charge you put outside or inside the shell anywhere you please. This then says that the net charge enclosed by the Gaussian surface (as defined above) is zero no matter what.
Yup.keemosabi said:So it's like the conducting shell prevents the outer charges from affecting the E-field inside of the shell?
Gauss' Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the electric flux through a closed surface is proportional to the total electric charge enclosed by that surface.
"Induction by charges outside a shell" refers to the phenomenon in which an electric field is induced inside a hollow conducting shell due to the presence of external charges. This effect is described by Gauss' Law and is a result of the redistribution of electric charges on the inner surface of the shell.
Gauss' Law states that the electric flux through a closed surface is proportional to the total electric charge enclosed by that surface. In the case of charges outside a conducting shell, the electric flux through the surface of the shell is equal to the charge enclosed by the shell, and therefore the electric field inside the shell is zero.
Gauss' Law is of great significance in electromagnetism as it provides a powerful tool for calculating electric fields in different situations. It allows us to determine the electric field at any point in space, given the distribution of charges in that space. It also helps us understand the behavior of electric fields in different scenarios, such as induction by external charges.
Yes, Gauss' Law can be applied to non-conducting shells as well. The only requirement is that the electric charges must be located outside the shell, and the shell itself must be a closed surface. This is because Gauss' Law relates the electric flux through a closed surface to the charge enclosed by that surface, regardless of the nature of the surface.