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General Electric Field Question

  1. Jun 1, 2008 #1
    I'm seeing a lot of problems in my textbook where it asks you to find the point near two charges where the total electric field is zero. Whats the logic behind figuring out if the point will be on the left side, the right side, or between the two charges?

    For example, one problem says there are two charges -7uc and 28uc seperated by 2 m. It asks you to find the point where the electric field is zero. How do you tell if the point will be on the left side of the -7uc, the right side of the 28uc, or in between?

    I have no problem solving these problems once I know where the point is going to be, but thats the major hurdle.
  2. jcsd
  3. Jun 1, 2008 #2
    Let d=distance separated between the two charges.
    Let E=electric field
    By Coulomb's Law, [tex]E=\frac{kQ}{r^2}[/tex], where r is the distance

    Let [tex]r_{1}[/tex] be the distance from one charge to the zero field, then
    [tex]r_{2}=d-r_{1}[/tex] be the distance from the other charge to the zero field.

    So, for two charges, the distance where the electric field = 0 is when
    [tex]E=\frac{kq_{1}}{r_{1}^2}=\frac{kq_{2}}{r_{2}^2}[/tex]. Since you are given d, you can solve for [tex]r_{2}[/tex]
  4. Jun 1, 2008 #3
    But if the point is not in between the two charges, then wouldn't the distance be 2 m from one charge, and 2+x from the other charge? See I just want to know how you decide if the point is in between the two charges, or if it is outside the two charges. After that I can solve the problem.
  5. Jun 1, 2008 #4
    Ah, good question. You are give that one of the charge is negative, which is the opposite sign of the second charge. Hence, it must be in between so that the electric field is zero.
  6. Jun 2, 2008 #5


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    Hi uday28fb! :smile:

    Same charges … between.

    Opposite charges … outside, on the side of the weaker charge. :smile:
  7. Jun 2, 2008 #6
    These are just equilibrium problems. For equi we just require that the forces on the Q of interest are equal in magnitude and opposite in direction. You can check each region to see if this condition can be met - you don't need to do any calculations for this, just think about the net force on the Q of interest. The post above gives the rule of thumb.
  8. Jun 2, 2008 #7

    (a) charges have same sign-- your test charge will be inbetween them, else it will not be.
    (b) your test charge must be closer to the weaker charge so that the 1/r^2 can make up for the charge being less (so that the forces are still equal in magnitude).

    That's it, just those two rules will give you any case.
  9. Jun 2, 2008 #8
    Thanks guys, I'll rep you guys. Is the logic the same for electrical potential?

    edit: how do you rep people?
    Last edited: Jun 2, 2008
  10. Jun 2, 2008 #9
    It's even easier for electric potential (V). Because V is a scaler quantity, the net V, at a point, due to a group of point charges is just the *algebraic* sum of individuas potentials. All you have to worry about are the signs. Relative to a zero V at infinity, V due to a negative point Q is negative, and vice-versa for positive Q.
  11. Jun 2, 2008 #10


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    Hi uday28fb! :smile:
    Your book is wrong.

    One charge is pulling and the other is pushing.

    E = 0 can't possibly be in between. :smile:
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