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Electric Field Strength Between Two Charges

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Two isolated point charges, -7 μC and +2 μC, are at a fixed distance apart. At which point is it possible for the electric field strength to be zero?

    ................................-7 μC.......+2 μC

    I don't know how to insert image
    sorry for messy drawing (I had to put little dots because it wouldn't make spaces although I spaced them out)

    2. Relevant equations
    F= k(q1)(q2)/r^2
    E = F/q

    3. The attempt at a solution
    The actual answer is (D).

    By these two equations, I got
    E = kq/r^2

    I thought, to make the field strength 0, I had to find the distance at which the field strength of these two charges are equal.

    ∴ E = E
    kq/r^2 = kq/r^2
    k(-7 μC )/r^2 = k(+2 μC)/r^2
    I know this is way off,,, since I will cancel out the distance!!
    I actually did choose the right answer (D) simply because -7 μC has larger magnitude that field strength would be much smaller at (D)

    Please tell me how I can solve this algebraically. Thank you :D
  2. jcsd
  3. Oct 31, 2013 #2


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    hi rnjscksdyd! welcome to pf! :smile:
    correct :smile:
    no, it isn't the same r, is it? :wink:
  4. Oct 31, 2013 #3
    Try and us the Principle of Superposition. That might lead you to the answer, pal! :)
  5. Oct 31, 2013 #4
    Let d represent the distance between the two charges, and let x represent the distance to the right of the 2 μC charge at which the field is zero. In terms of x and d, how for is the -7 μC charge from the point where the field is zero? You will be solving for x in terms of d.
  6. Oct 31, 2013 #5
    I thought 'r' was a distance between the two charges. So, 'r' should be different in that, say, 'r1' is the distance between -7μC and the point where electric field strength is 0 and 'r2' is the distance between +2μC and the point where electric field strength is 0, right??

    Could you explain further please :( I thought superposition was for the waves, like constructive and destructive interference :( how do i apply this concept in electric field?

    would that mean:
    -7μC/(d+x) = +2μC/(x) ?

    If it is so, how do I know that the answer is D? From the above equation I got:
    -7μC/(d+x) = +2μC/(x)
    -7μCx = +2μC(d+x)
    x = -2d/9

    Also, how do I know that 'x' is the distance to the right of +2μC supposing I did not know the answer?

    Sorry I'm just bad at physics :(
  7. Oct 31, 2013 #6
    That's not the equation I get. If I do a force balance on a test charge, I get:


    Note the squares in the denominators. Next, take the square root of both sides of the equation.

    It doesn't matter. The algebra would take care of that anyway.

  8. Oct 31, 2013 #7
    Sorry, forgot about the squares :D
    Then, is this right?
    [tex]\frac{-7μC}{(d+x)^2}= \frac{+2μC}{x^2}[/tex]
    [tex]-7μCx^2 = +2μC(d+x)^2[/tex]
    [tex]-7μCx^2 = +2μC(d^2 + 2dx + x^2)[/tex]
    [tex]-7x^2 = 2d^2 + 4dx + 2x^2[/tex]
    I took out μC for now (it looks better :D)
    [tex]0 = 9x^2 + 4dx + 2d^2[/tex]

    [itex]x=\frac{-4d \pm \sqrt{16d^2 - 72d^2}}{18}[/itex]

    but then I get a complex number since I get negative square root. Where did I go wrong?
  9. Nov 1, 2013 #8


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    hi rnjscksdyd! :smile:

    (just got up :zzz:)
    it does! … and you could have taken it out from the start :wink:
    your work (and result) is perfect except

    can you think of a situation in which it isn't (d + x) ? :smile:
  10. Nov 1, 2013 #9
    You still have that minus sign in front of the 7, which is incorrect. It should be a + sign.
    [tex]\frac{7μC}{(d+x)^2}= \frac{+2μC}{x^2}[/tex]
    Taking the square root of both sides:
  11. Nov 2, 2013 #10
    Sorry for late reply, my internet was down -_-;; this country :(
    :( what do you mean when it isn't (d+x)?

    Does that mean... when that value is negative, the field will be 0 on the left side and on the right if the value is positive?
  12. Nov 3, 2013 #11


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    hi rnjscksdyd! :smile:
    if it's d+x on the left, isn't it d-x on the right? :wink:
  13. Nov 4, 2013 #12
    'According to this principle the total amount of force on a given particle is the vector sum of the individual forces applied on that charged particle by all the other charged particles. The individual forces of the particles is calculated using the formula of Coulombs law and is not influenced by the presence of other particles.'

    Basically, Find the Coulomb's Force acting on each particle, one at a time, and then add it vectorially.
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