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Free charges on a sphere

  1. Aug 12, 2009 #1
    Ok, this question may sound stupid to some. But still, I 'll ask.

    In electrostatics we frequently discuss, potential due to uniformly charged sphere.where the whole sphere's surface has positive or negative charge.
    So, this means, free charges of the same type(+ve or -ve) have been distributed all over the surface.

    Now, my question is , won't the sphere explode because of repelling forces between the free charges which are present on the surface of the sphere(as they are all postive or all negative)?
  2. jcsd
  3. Aug 12, 2009 #2
    Well if the charges are distributed evenly over the surface, then the repulsive force are not radial.

    The forces acting on the charges are in equilibrium so everything is balanced hence if the sphere. The equilibrium exists because no matter in which direction you look from a particular charge you see the same number of charges at every distance. So the net force is the same in all directions I believe.
  4. Aug 12, 2009 #3

    Jano L.

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    If charges are distributed uniformly over the surface of a sphere, they repel each other and this manifests in a decrease of the pressure in the sphere (there is net radial outward force proportional to surface area). Try to calculate the force! For this you must find the intensity right at the radius where charges are.
  5. Aug 12, 2009 #4
    I would also like to ask, what about the free charges on one plate of a parallel plate capacitor?
  6. Aug 12, 2009 #5

    Jano L.

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    That's the same. You just find the intensity at the first plate (caused by the second plate) and then multiply it by the total charge on the first plate and you get the force. However, the sphere is a bit more interesting. This has an application in the physics of water droplets. When the droplet starts forming in the air, it is very small and thus has very high pressure inside due to capillarity. This leads to high rate of evaporation of molecules from the droplet and it is likely that such a droplet would quickly vaporize. But imagine that our droplet contains some net charge. Then this charge will distribute on its surface and will cause decrease of the pressure inside and therefore to decrease of evaporation rate. Then this droplet can more easily survive and starts to grow. That's probably how Wilson chamber works - charged particles facilitates droplet forming.
  7. Aug 12, 2009 #6
    Thanks very much for the answers , Jano L.
  8. Aug 12, 2009 #7
    There are two 1 Farad (1 million microfarad), 20 volt aluminum capacitors (3 " diameter) for sale at www.digikey.com. (1/2)C V^2 = 200 joules; not very much energy.
    [Edit] According to Wiki, the self capacitance of a sphere of radius R is
    C = 4 pi e0 R, so for a sphere of radius 0.5 meters
    the capacitance C = 56 picoFarads.
    If we took the charge (Q = CV = 20 Coulombs) on the 1 Farad capacitor above, and charged this sphere, the voltage would then be

    V = Q/C =20 Coulombs/56 picoFarads = 357 billion volts.

    So 20 Coulombs is a lot of charge, especially in a 3" diameter capacitor.
    Last edited: Aug 12, 2009
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