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Grounded/Connected Conductors

  1. Oct 29, 2013 #1
    Hello,

    This is a sequence of my two last posts on this theme(i'm clearly messing this up in my head) and I got no clarifying answer on neither, so i'm trying here in the homework threads! Help if you know 100% sure otherwise its just going to get more and more confusing in my head because there is alot of misleading information in this internet of ours!

    Best regards!
    ---------------------------------------------------------------------------------------------
    Hello, I've been trying to understand how the fact of grounding a conductor affects its charge distribution.

    So, for example, lets assume there are three spherical shells with radius R1 R2 and R3. Supose I charge the R1 shell with q and the R3 shell with -q , and I connect the R2 shell to the ground and now I want to find out what is the charge distribution over the inside and outside of these shells.

    My quick answer would be:

    As far as I can see, the middle shell acts like a perfect faraday cage, and the distribution of charge of R3 of -q will scatter through the outside surface.

    so the Electric field should only exist betwen R1 and R2 and out of the bigger shell (r>R3)

    The charge distribution would be:

    R1- = 0
    R1+ = q
    R2- = -q
    R2+ = 0
    R3- = 0
    R3+ = -q

    Based in the fact that if the cage acts and contains the electric field inside the larger (R3) shell then the Gauss law would work (* now that I see it I think here is the problem, because the potential must somehow go back to zero *)

    Although I've seen people say that the distribution should be:

    R1- = 0
    R1+ = q
    R2- = -q
    R2+ = q'
    R3- = -q'
    R3+ = -q+q'

    and then be solved (knowing that The potential from the midle sphere must be zero)
    calculating the potential from infinity to R2 must be zero, so The potential from infinity to R3 plus the potential from R3 to R2 must be zero, and this 2 potentials must be simetric.

    In the meantime, while I was writting all of this down, I came to the conclusion thaat the second option is the correcto one, but I was hoping I could have someone else's aproval on this.

    Thanks in advance,

    Pedro
    -------------------------------------------------------------------------------------------
    Hello, again!
    I have been doing some research on my own and I found that my first answer is the correct one.
    (and yes, its suposed to assume R1<R2<R3)

    My second theory was wrong because of the fact, wich is still a bit confusing to me, that when we ground the second conductor, then it becomes the reference, being then impossible to assume that the potential of the radii equal to infinity is zero. So the mistake was to assume there were two reference points, and that the potential in infinity is zero because it might be not in reference to our second conductor.

    So, if i got it straight, the inner conductor is charged Q, so, in order for the Electric field to be zero inside the middle conductor, it has to "produce" or "find a way" to get its inner surface charged with -Q. the part that still makes me a little confused its that i was always taught that if we charge a conducting shell, the charge will tend to spread in the outer surface, where the potential is lower. But assuming that "nature finds a way" then i'll accept that this is possible without creating a mirrored charge in the outside surface of the middle conductor (if anybody can explain this to me I would be in life debt!!!:) ). So the charge distribution is now Q in the outside of the tiny shell and -Q in the inside of the middle shell, and zero charge in the outside surface of the midle shell, because somehow, as the negative charges get attracted towards the tiny shell the ground stops the +Q charge from creating in the outside of the middle shell by giving away electrons.

    If the Faraday cage blocks evry electric field coming from the inside of it, and throws away no electric field, then evrywhere inside the big (and last) shell the electric field is zero, so the charge on it is, just like we would expect from any regular conducting spherical shell, zero inside and -Q outside.



    But now, I have another problem that has been troubling my mind:

    Supose I have now, 4 spherical shells with R1,R2,R3,R4 and R1<R2<R3<R4.

    I connect the spherical shell of R3 to the ground, I connect with a very tiny wire the last shell (R4) and the second shell (R2), and then I charge up the inner of all spheres with Q and the second shpere with -Q. What happens now?(I am making it a little confusing and probably a little harder than usual, but is in order so I get it straight into my head straight and for all)

    Here we have 2 important concepts -> the spherical shells (2) and (4) have the same potential
    -> the spherical shell (3) has potential zero.

    I have learned that when two conductors are connected ( spherical conductors) the excess amount of charge flows to the outer sphere because the charge repells itself to the point where its all in the lowest potential, so my guess is that, at start, the charge is distributed in the outside of (1) (Q) and in the outside of (4) (-Q).
    Now I know that (2) inside must be -Q in order to the E- Field to become zero inside(1) but now I don't know if the +Q charge outside of (2) stays outside of (2) or if it stays outside/inside of (4).
    Need Help please!

    Thanks in advance

    Pedro
     
  2. jcsd
  3. Oct 30, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    With the inner shell grounded and the outer one given some charge, you get a spherical capacitor. Is the charge zero at the inner surface of the outer shell?

    At the same time, the outer shell makes also a capacitor with infinity, like an isolated metal sphere. So you have two capacitors, connected parallel. The potential is zero both at infinity and at the inner shell.

    ehild
     
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