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Charge of three small spheres when connected to a big one

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Three small metal solid uncharged spheres are put on vertexes of an equilateral triangle. Then they all one by one are connected to a big conducting solid sphere, which is at the same distance from all the small ones, by a wire. Thus the first sphere is charged to q1 and the second one - to q2. Find the charge of the last small sphere if the potential of the big one does not change.


    2. Relevant equations

    [itex] V = kq/r [/itex]

    3. The attempt at a solution

    No idea how to do this one :( I was maybe thinking about the conservation of charge, but if the potential of the big sphere does not change, I'm not able to apply that conservation law. As I understand, all the small spheres are of the same potential when connected, but I'm not sure what to do with it. And does it matter whether the big sphere is at the same distance from the three small ones or not?

    I'm completely lost. Any help appreciated!
     
  2. jcsd
  3. Feb 23, 2014 #2

    haruspex

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    It doesn't say, but I presume each connection is broken before the next one is made.
    Let the potential of the large sphere be V. When a given small sphere is connected to it, what potential does it become? What are the contributors to the potential at that sphere?
     
  4. Feb 23, 2014 #3
    Well I guess then the small sphere it becomes potential V as well?
    I do not understand the second question completely - but as I understand, you are asking what influences the potential of the second sphere? Is it charge?
     
  5. Feb 23, 2014 #4

    haruspex

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    Let's back up for a moment and just consider two conducting spheres.
    Let their centres be distance D apart, total charges Q1, Q2, and the radii R1, R2.
    Where are the charges distributed on sphere 1?
    What is the potential at the centre of sphere 1?

    Edit: I'm not sure I can answer that last question myself.... I might invite others into the thread.
     
    Last edited: Feb 24, 2014
  6. Feb 24, 2014 #5
    Okay, so the charges on sphere 1 are distributed on the surface, I think?
    And the potential must be [itex] V = \frac{kQ_1}{R_1} + \frac{kQ_2}{D} [/itex] ? Is that right?
    But how does that help in the original problem? We don't know radii or distance..
     
  7. Feb 24, 2014 #6

    haruspex

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    Not knowing those distances shouldn't matter. The same distances will arise in equations for q1, q2 and q3, so we can eliminate them and arrive at an expression for q3 in terms of q1 and q2.
    The problem, though, is that ## \frac{kQ_2}{D}## is wrong. The very presence of the three small spheres shifts the charge distribution on the large sphere and on each other. When charge flows from the large sphere to the first small sphere it all shifts again.
    I've put a request for help on the homework helpers' page, but no nibbles yet.
     
  8. Feb 24, 2014 #7

    haruspex

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    Had a response from mfb, suggesting we assume all the sphere radii are small compared with the distances between the spheres. I don't like it since we were given no basis for assuming that applies to the large sphere, but it does yield an answer. It allows us to treat all surface charge distributions as uniform. I.e. you can use ## \frac{kQ_2}{D}##.
    So apply that to the large sphere and the first small sphere. Then do the same equation for the second small sphere, but adding in the potential from the first small sphere.
     
  9. Feb 25, 2014 #8
    So then let's say the charge of the big sphere is [itex] Q [/itex] and its potential is [itex]V[/itex]. Then when we connect the first small sphere its potential now is V as well, and the eq. [itex]V_1 = V = \frac{kQ}{D}+\frac{kQ_1}{r}[/itex]. And the second sphere's potential would be [itex]V_2 = V = \frac{kQ}{D} + \frac{kQ_1}{d} + {kQ_2}{r} [/itex]. From that we can derive [itex]V_1 = V_2 => \frac{kQ}{D}+\frac{kQ_1}{r} = \frac{kQ}{D} + \frac{kQ_1}{d} + \frac{kQ_2}{r} => r = \frac{(Q_1 - Q_2)d}{Q_1} [/itex]. Is that correct? Then the rest is totally clear :)

    A huge thank to you!
     
  10. Feb 25, 2014 #9

    haruspex

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    That all looks right.
     
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