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Homework Help: Capacitance - sphere

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Two solid conducting spheres have radius r1 = 3 cm and r2 = 42 cm. The two spheres are connected by a thin conducting wire. Assume:

    - the wire is very long and thin, with negligible surface area, so it does not affect the capacitance of the system.

    - the switch is closed.

    - the spheres are very far apart, so they do not affect each other's charge distribution.

    If charge Q is placed on sphere 1, it distributes between the two spheres. Find the capacitance of this system, in pF.

    Two solid conducting spheres have radius r1 = 6.98 cm and r2 = 30.9 cm. The two spheres are connected by a thin conducting wire, with the initially switch closed. Assume the wire is very long and thin, and the same assumptions apply. Charge 7.11 μC is placed on spheres, and distributes between the two spheres.

    Now the radius of sphere 2 begins to decrease at the rate of:

    dr2/dt = -2.54 cm/s.

    Find dq1/dt the rate at which the charge on sphere 1 is changing, in μC/s. The sign indicates if the charge is increasing or decreasing.
    2. Relevant equations

    3. The attempt at a solution
    1) Since the switch is closed, the charge remains the same on sphere1 and sphere2?
    C1 = Q/V = Q/(Q/4*pi*E0*r1) = 4*pi*E0*r1
    C2 = Q/V = Q/(Q/4*pi*E0*r2) = 4*pi*E0*r2
    then since it is in series Ctotal=1/(1/C1+1/C2)
    Am I on the right approach?

    2)Can anyone list steps that I can follow to solve it. I am really suck with integral rate

    thank you for the help
  2. jcsd
  3. Sep 20, 2015 #2


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    Where is the switch?
    Why do you think that the capacitors are connected in series? If you take the spheres as capacitors where is their other terminals connected? Is any battery present?
  4. Sep 21, 2015 #3
    the switch is between spheres, no battery present. So I see I was wrong.
    what I confuse is it states "the spheres are very far apart, so they do not affect each other's charge distribution."
    but charge only place on sphere1, so there is no charge on sphere2. I only need to figure out V1 for sphere1 then use Q/V1?
    ps: I upload the diagram for problems, two problems use the same diagram.

    Attached Files:

    • 001.png
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  5. Sep 21, 2015 #4


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    'They don not influence each other's charge distribution" means that the charge distribution becomes homogeneous on both spheres when equilibrium is reached. Otherwise, when two charged spheres are close together, the opposite charges attract, like charges repel each other, so the charge distribution becomes inhomogeneous.


    Charge is placed on sphere 1 initially, but it will distribute on both spheres as they are connected through a wire. What determines how much charge goes over to sphere 2 and remains on sphere 1?
  6. Sep 22, 2015 #5
    thank you for your explanation. is it electric potential? electric potential on sphere 1 must be equal to electric potential on sphere 2?
  7. Sep 22, 2015 #6


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    Yes, exactly!
  8. Sep 23, 2015 #7
    thank you!
  9. Sep 23, 2015 #8
    I'm working on the same problems and for the second one this is my solution attempt (Not sure that is right though):

    Because V1 = V2 = V you can work out Q1 in terms of r2 which is the variable you are looking for.

    → kQ1/r1 = kQ2/r2 (1)

    Then since Q = Q1 + Q2Q2 = Q - Q1 (2)

    ⇒ substituting (2) in (1) and playing with your algebra a little bit, you get Q1 = Q/(r2/r1+1) the equation we are looking for.

    The variable here is r2

    by taking d/dt you will get something like this: [Qr1/((r2 + r1)2]dr2/dt

    from here is just a matter of plugging in the known quantities and also don't forget the result has to be in μC/s so review your unit transformations up to here. Hopefully this works! :nb)
  10. Sep 23, 2015 #9
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