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Electric Fields, Finding the zero field point

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Two particles with positive charges q1 and q2 are separated by a distance s. Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero.

    2. Relevant equations

    [tex]E= \frac{kq}{r^2}[/tex]

    3. The attempt at a solution

    Assuming x1 is the distance from q1 to the point where the field is 0.

    This is what I get, I believe it is correct but I can't seem to solve for x1.

    [tex] \frac{kq1}{x1^2} [/tex] = [tex] \frac{kq2}{(s-x1)^2} [/tex]
     
  2. jcsd
  3. Jan 24, 2009 #2

    gabbagabbahey

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    Just "cross multiply" and expand:

    [tex]\frac{kq_1}{x_1^2} =\frac{kq_2}{(s-x_1)^2}\implies kq_1(s-x_1)^2=kq_2 x_1^2[/tex]

    [tex]\implies q_1(s-x_1)^2=q_2 x_1^2 \implies q_1 x_1^2-2q_1 s x_1 +q_1 s^2=q_2 x_1^2[/tex]

    Then collect all terms on one side of the equations and solve the resulting quadratic.
     
  4. Jan 24, 2009 #3
    Thanks, but I already got to that part above. How can I start to solve for x1 from that?
    can I do this? (q2 - q1)x^2 + 2q1sx - q1s^2 = 0. What would I do from here?
     
    Last edited: Jan 24, 2009
  5. Jan 24, 2009 #4

    gabbagabbahey

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    Yes you can do that.:smile:

    Have you not learned the quadratic equation yet?! :wink:
     
  6. Jan 24, 2009 #5
    Yea this is what it came out to, which is wrong unfortunately. Is there something I can simplify? I also tried the "+" instead of the "-", no luck.

    [​IMG]
     
  7. Jan 24, 2009 #6

    Hurkyl

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    Check your reference: you got the quadratic formula wrong. (Were you using a computer algebra system? It looks like you forgot a pair of parentheses)

    Incidentally, in order to try simplifying things, did you think about expanding things first?
     
  8. Jan 24, 2009 #7
    I was doing it by hand =), I will try and expand and see if anything simplifies. Thanks.
     
  9. Jan 24, 2009 #8

    rl.bhat

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    Easier method will be to take the square root of the equation you have written and solve for x1. If the charge are same take + sign. If they are opposite take - sign.
     
  10. Jan 24, 2009 #9
    Ok, after doing it again I got this? Does it look better? Stuff indeed canceled under the square root sign.

    [​IMG]
     
  11. Jan 24, 2009 #10

    Hurkyl

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    That looks plausible. (Note things still simplify more!) You can always try substituting back into the original equation.... (which is usually a very good idea: there is rarely a better way to check if you've found a solution to an equation than actually plugging it in)

    Incidentally, how did you decide that was the solution that lies on "the line connecting the two charges" (which I assume meant to say 'line segment'), rather than the other solution?
     
  12. Jan 24, 2009 #11
    It turns out this answer is acceptable but it does simplify more. I chose the + instead of the - in the formula because the distance is somewhere between the 2 charges. I assumed using the "-" would be somewhere left of q1
     
  13. Jan 24, 2009 #12

    Hurkyl

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    Don't assume, prove! The argument isn't difficult, once you know what the right idea is. But you have to make note of your assumptions, and cover all of the cases... what if q1 > q2?


    By the way, I note that your derivation (and your solution) isn't complete; it doesn't cover all possible cases... in particular, what if q1 = q2?
     
    Last edited: Jan 24, 2009
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