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Homework Help: Electric Field Zero where?

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    If we choose a point P along the X-axis, with a distance X from the origin, we get this:


    Assuming that Q=1 for simplicity:



    Basically, I've chosen P to be to the right of Q2, giving the distance between Q1 and P to be L+x and the distance between Q2 and P to be x.

    If we want to find the point in space where the sums of the Electric fields is zero, we add up the field from Q1 and the field from Q2 and set equal to zero.

    [tex]E_{P1}=\frac{k(-7)}{(L+x)^2} + \frac{k(3)}{x^2} = 0[/tex]

    [tex]E_{P1}=\frac{k(-7)}{(L+x)^2} = - \frac{k(3)}{x^2}[/tex]

    I end up with, after assuming L=1, cross-multiplying:

    -7x^2 = (-3)(1+x)^2

    But when I solve this, there is no real answer.
    Last edited: Aug 27, 2009
  2. jcsd
  3. Aug 27, 2009 #2
    You chose the correct region to look for a net 0 field, (to the right of the 3.00q charge). However, when I solved this equation, I got x==6L.

    Let L==1
    Since we let L==1, x==6*L==6
  4. Aug 27, 2009 #3

    Doc Al

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    Staff: Mentor

    Try again. You should get two real answers, only one of which is meaningful for this problem.
  5. Aug 27, 2009 #4
    The only two regions is could possibly be zero are the outer two. The inner region would push a positive test charge to the left at all points. But on the outer two, there are possibly points where it could stabilize. How can I possible determine just by looking which one to try?

    1.) To the right of the right particle.
    If a positive test particle was placed here, it would be repelled by the 3Q charge and attracted by the -7Q charge. But the 3Q charge is much closer.

    2.) To the left of the left particle.
    If a positive test particle was placed here, it would be repelled by the 3Q charge and attracted by the -7Q charge. But the -7Q charge is much closer.

    So just by looking, how can I tell which to choose?

    At the marked point, I have identical work on my paper. How did you get to the next line?
  6. Aug 27, 2009 #5

    Doc Al

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    Staff: Mentor

    The only way the field contributions can cancel is if you're closer to the smaller charge.
  7. Aug 27, 2009 #6
    I then brought the x term onto the left side of the equation, 0.333x==2, and divided by 0.333. x==6
  8. Aug 27, 2009 #7

    Doc Al

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    Staff: Mentor

    No, letting L = 1 gives you:
    7/3 = (1 + 2x + x^2)/x^2

    (You still have to solve the quadratic.)
  9. Aug 27, 2009 #8
    Okay, well...

    This time I got it down to: 4x^2-6x-3=0

    X turns out to be: -.395 and 1.90.

    is 1.9 the correct answer? I don't want to lose any more points on this stupid assignment.

    Also, why would I disregard the negative one? I know theoretically why i can't have a particle between the two and have a 0 E-field, but is this the reason why I'm disregarding the negative distance?
  10. Aug 27, 2009 #9
    Yes, since you assumed L = 1, x=-.395 will be between the two charges.
  11. Aug 27, 2009 #10
    1.90 is a wrong submission. Apparently, that is not the answer.
  12. Aug 27, 2009 #11
    You ought to count from the origin...
  13. Aug 27, 2009 #12
    I dont understand how to relate x to L in this problem.
  14. Aug 27, 2009 #13
    What are the coordinates of q1 and q2? Therefore what is the coordinate of the point with zero E field?
  15. Aug 27, 2009 #14
    As should be clear by now, the field can be zero only on the right of q2. Suppose it is zero at a distance s from it. Then, at that point the magnitude of the fields due to the two charge must be same. As such we get (after canceling [tex]\dfrac{1}{4\pi\epsilon_0}[/tex] on both side):
    That, is
    so that
    I can readily do away with the negative sign (all the quantities on the left are positive, since I called s as distance). So we get
    Finally, I get
    Plugging in numbers,
    (taking the significant figures in to account)
  16. Aug 27, 2009 #15
    I'm lost at what I should enter into the submission box. 1.90 is NOT right, as it yelled at me and threw a pie at my face when I tried. Is the problem asking for a submission of 2.90? (If L=1, s=1.9, L+s=2.90)
  17. Aug 27, 2009 #16
    Yes, it should be wanting 2.90.
  18. Aug 27, 2009 #17
    Yes the answer for the problem would be 2.90. In my last post I found the distance from q2 of the point where the field is zero. The problem asks for the coordinates, i.e. the distance from the origin which is where q1 sits. So the answer should be L+s = 2.90 L
    So u fill in 2.90.
  19. Aug 27, 2009 #18
    2.90 is the correct answer.

    These problems given on this online homework section are TERRIBLY overcomplicated. Why on Earth is it asking for an answer in multiples of L??? I've never seen a question so convoluted in a math book. Why doesn't it just ask straight up "At what point on the x-axis..."??
  20. Aug 28, 2009 #19
    They're really pretty much the same thing isn't it? If L is unknown you will have to give your answer in terms of L anyway.
  21. Aug 28, 2009 #20

    Doc Al

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    Staff: Mentor

    Yes. Don't forget that when you set up your equation for x, x was the distance to the right of q2. To get the distance from the origin (where q1 is) you have to add L.
  22. Aug 28, 2009 #21
    Indeed. But asking, "As a multiple of L...", is extremely confusing.
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