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Potential due to a the sum of point charges?

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Q15. Eight identical spherical raindrops are each at a potential V, relative to the potential far away. They coalesce to make one spherical raindrop whose potential is:
    A. V/8
    B. V/2
    C. 2V
    D. 4V
    E. 8V


    2. Relevant equations
    V = sum( Vi )


    3. The attempt at a solution

    OK so if you just sum the potentials you get 8V right? But the answer's D, 4v. Can someone make sense of this to me, please? I don't get what's cutting the value in half.
     
  2. jcsd
  3. May 5, 2008 #2

    alphysicist

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    Hi Strawberry,

    You can't treat these as point charges; treat them as spheres of charge. Each raindrop originally had a charge q and a radius r. When they come together, you are right that the new large drop will have a charge of Q = 8q, but what will the new radius be?
     
  4. May 5, 2008 #3
    V = 4/3 pi r^3, so it would take 8 spheres coming together to make the radius twice as large? I thought r referred to some reference point and not the radius of a sphere. I can't even find a formula in my book for electric potential of a sphere.

    I'm sorry, but could you explain the concept of how to find potential to me? Am I supposed to convert potential to an electric field and then use Gauss' Law?
     
  5. May 5, 2008 #4

    alphysicist

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    If you are talking about a point that is outside of a sphere of charge (with everything spherically symmetric), then the potential at that point is

    [tex]
    V = \frac{k q}{r}
    [/tex]

    where q is the total charge of the sphere and r is the distance from the point you're talking about to the center of the sphere. So the formula is just like that of a point charge is you're outside the sphere.

    (One way to get this result is to find the electric frield from gauss's law and then use that to find the formula for V--just like you would do for a point charge. But I think it should be in your book somewhere.)

    So here when they say the raindrop is at a potential V I believe they mean just outside the raindrop the potential is V. So V = kq/r for these drops.

    You're right the radius is twice as large, because we could say:

    [tex]
    \frac{4}{3} \pi R^3 = 8 (\frac{4}{3} \pi r^3)
    [/tex]

    and solve for R (new radius) in terms of r (old radius).
     
  6. May 5, 2008 #5
    I dont get it either. Lets assume you are talking about gravitational potential. If the raindrops dont change their potential due to their movement in coming together, and I think the "far away" qualification is intended to exclude that, the potential of the single drop (same mass as 8 indivudals) will remain the same as the sum of the individuals. If not, I've really forgotten a lot of physics.
     
  7. May 5, 2008 #6

    alphysicist

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    Hi jimvoit,

    In terms of gravity, we could put it this way. The acceleration due to gravity on the earth surface is 9.8 m/s^2, which is also given by

    [tex]
    g=\frac{G M_{\rm earth}}{R_{\rm earth}^2}
    [/tex]

    If we had two identical planet earths and we combined them together to make one spherical planet, would the acceleration on the surface of the combined planet be twice as much? No, because although there is twice as much mass, the radius is larger so you are farther from the center.

    For the spherical raindrops, the potential at the surface of the drop is due to the charge and the radius of the drop, and both of these will increase when the eight drops combine.
     
  8. May 6, 2008 #7
    A misunderstanding of the problem

    I misunderstood the problem. I assume now that we are talking about comparing the potential of 8 identical spheres, each of volume v, and each containing charge Q, with the potential energy of 1 sphere of volume 8 v, containing charge 8 Q.

    Some algebra determines that If the identical spheres have a radius R, then the radius of the larger sphere is 2R.

    A formula from one of my textbooks gives the potential energy W of a sphere of radius R containing charge Q as: W=Q^2/(8 π ϵ0 R)

    A little more algebra determines that the potentials of the 1 sphere is 4 times the sum of the potentials potentials of the 8 individual spheres.
     
  9. May 6, 2008 #8
    I think I understand the problem now...

    I misunderstood the problem. I assume now that we are talking about comparing the potential of 8 identical spheres, each of volume v, and each containing charge Q, with the potential energy of 1 sphere of volume 8 v, containing charge 8 Q.

    Some algebra determines that If the identical spheres have a radius R, then the radius of the larger sphere is 2R.

    A formula from one of my textbooks gives the potential energy W of a sphere of radius R containing charge Q as: W=Q^2/(8 π ϵ0 R)

    A little more algebra determines that the potentials of the 1 sphere is 4 times the sum of the potentials potentials of the 8 individual spheres.
     
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