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Force Between Two Conducting Spheres

  1. May 17, 2013 #1
    Hi physicists here. :)
    I've just joined the forums and here's my very first question :P :


    Aakash PHYSICS JEE (Main & Advanced) Study Package - 5 & 6 (Class XII)
    Chapter -
    Electric Charges and Field

    Assignment (page 12)
    SECTION - A; Q.no - 1


    What I expected the answer to be was (1). The electric field due to a conducting sphere of charge Q is equivalent to the same due to a point charge at the center of the sphere as total charge on it appears as concentrated at the center for the points outside the charged sphere. So, the force should have remained the same.

    Any suggestions?

    Well, it's really nice meeting you all. :)

    P.S.: I didn't think it to be a homework question. Well, if moderator thinks it is, the same can move the thread. I beg your pardon for that on the grounds that I just joined the joined the forum today.
     
    Last edited: May 17, 2013
  2. jcsd
  3. May 17, 2013 #2

    Zag

    User Avatar

    Welcome to the forums, Paras Lehana! :)

    Let me try to answer your question now, and I have to say that some drawings would be helpful in explaining the situation, but I will try my best to describe everything with words. So let me know if I am not being completely clear.

    You are right about the field outside of a charged conducting sphere being equivalent to the field generated by a point charge sitting at the center of the sphere. However, this concept is applicable only when the sphere is isolated from the environment.

    In the present case, each sphere is not completely isolated from the environment because they are in the presence of each other. As a consequence they will interact. The fact that the two conducting spheres are charged implies that they will induce agregation of charges in certain regions of each other surfaces. To see this, think of both as spheres being negatively charged (and, therefore, will repel each other as required by the problem). These charges are free to move over the surface of their respetive spheres because we are dealing with a conductor. We can thus predict that the negative charge of one sphere will repel those of the other and tend to move and acumulate on the side of its respective sphere which is the furthest alway from the negative charges of the other sphere. Thus, if you have one sphere standing on the right and another one of the left, the negative charges of the sphere on the right will acumulate on the rightmost side of that sphere, and by symmetry, on the leftmost side of the sphere on the left. Now, we know that the amount of charge on both spheres remains the same during this process because charge is conserved. However, these charges will be further appart from each other now (when compared to the situation of point charges sitting at the geometrical center of the spheres) - and because Coulomb's Force decreases with the square of the distance, the force has to be smaller now.

    I hope this helps!

    Best,
    Zag
     
    Last edited: May 17, 2013
  4. May 18, 2013 #3
    You were damn clear Zag. Thanks mate, I got it! The charge would've been the same if the spheres were non-conducting. But here, as they're conducting, the charges can interact due to induction, right?
     
  5. May 19, 2013 #4
    Why would the force be less ? It seems intuitive but what is the reason ? Yes,the distance increases ,but we simply cannot apply the inverse square law of coulombs force ,since they are not point charges .
     
  6. May 19, 2013 #5

    ehild

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    Homework Helper
    Gold Member

    You can consider the charge distribution on the spheres as stack of point charges. Those on one sphere interact with the other point charges on the other sphere. The presence of the metal also influences the force , the contribution of charges on the opposite sides is much less then the force in accordance with Coulomb's Law.
    Have a look at that:http://rspa.royalsocietypublishing.org/content/early/2012/05/22/rspa.2012.0133.full

    ehild
     
    Last edited: May 19, 2013
  7. Jun 23, 2013 #6
    EDIT: Got It! Thanks, Zag!
     
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