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Homework Help: Maximum angle which a field line starting from Q will end at infinity

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I guess a) is 1/3 (probably wrong, because I assumed all the field lines that end at -q originates at +3q which is not necessarily true) but I cant figure out b)
  2. jcsd
  3. May 21, 2013 #2


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    Hello, Wxfsa. Welcome to PF!

    That sounds pretty good to me. Recall that for an electric dipole (where the positive and negative charges have equal magnitude), then every field line that ends on the negative charge starts on the positive charge. (You can forget about the one pesky line that comes into the negative charge from infinity.) In your case, the positive charge has "extra" charge and the field lines that go to infinity must originate on the positive charge.

    Consider a very tiny sphere centered on the positive charge. Can you visualize the electric field lines piercing through the surface of that sphere? How are the lines distributed over the surface?
  4. May 21, 2013 #3
    I can't, unfortunately.
  5. May 21, 2013 #4


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    You know that the electric field of a point charges varies with distance as 1/r2. So, when you get very close to a point charge, the field of the point charge becomes very large.

    If you have a system of point charges and you get very close to one of the charges, then the field of that point charge will be much larger than the field of any other point charge in the system. So, to a very good approximation, the field very near the point charge will be the same as that of the point charge alone.

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  6. May 21, 2013 #5


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    If the sphere is VERY tiny like TSny said, they pretty much all go straight out. What fraction of them terminate on the negative charge? Look at the picture in problem hint again.
  7. May 21, 2013 #6
    As far as my visualization abilities go, I would say theta is somewhere around 110-120 but I have no idea how I can actually calculate it.

    EDIT: I think I got it arccos(-1/3)=109.5 Thank you for your help.
    Last edited: May 21, 2013
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