Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Distance between two charges where field strength = 0

  1. Apr 28, 2008 #1

    TFM

    User Avatar

    [SOLVED] Distance between two charges where field strength = 0

    1. The problem statement, all variables and given/known data

    Two particles having charges q1 = 0.500 nC and q2 = 9.00 nC are separated by a distance of 2.00 m.

    At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

    2. Relevant equations

    [tex] E = \frac{1}{4 \pi \epsilon _0}\frac{q}{r^2} [/tex]

    I have reduced this two:

    [tex]\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}[/tex]

    3. The attempt at a solution

    putting in the variables:

    [tex]\frac{0.5*10^{-9}}{r_1^2} = \frac{9.0*10^{-9}}{r_2^2}[/tex]

    I can rearrang to get:

    [tex]\frac{r_2^2}{r_1_2} = \frac{9.0*10^{-9}}{0.5*10^{-9}}[/tex],

    and:

    [tex]\frac{r_2^2}{r_1_2} = 18 [/tex]

    But I am not sure how to get the right distance to the actual point?

    TFM
     
  2. jcsd
  3. Apr 28, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    From this point,
    let x be the distance from the first charge to the point where E=0 such that,

    [tex]\frac{q_1}{x^2} = \frac{q_2}{(2-x)^2}[/tex]

    Do you follow?
     
  4. Apr 28, 2008 #3

    TFM

    User Avatar

    That makes sense, so now:

    [tex]\frac{q_1}{x^2} = \frac{q_2}{4-4x+x^2}[/tex]

    Rearranging:

    [tex]\frac{q_1}{q_2} = \frac{x^2}{4-4x+x^2}[/tex]

    And, inserting q1 and q2:

    [tex] 0.0556 = \frac{x^2}{4-4x+x^2}[/tex]

    How does this look?

    TFM
     
  5. Apr 28, 2008 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Looks good to me, although I would be tempted to leave the ratio of the charges as a fraction.
     
  6. Apr 28, 2008 #5

    TFM

    User Avatar

    So:

    [tex]\frac{0.5*10^{-9}}{9*10^{-9}} = \frac{x^2}{4 - 4x + x^2}[/tex]

    Okay, so where should I go from here?

    TFM
     
  7. Apr 28, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How about multiplying through by the denominator of the RHS, then solving for x?
     
  8. Apr 28, 2008 #7

    TFM

    User Avatar

    So:

    [tex] (0.5*10^{-9})(4 - 4x + x^2) = 9 x 10^{-9}x^2 [/tex]

    [tex] (2*10^{-9} - (2*10^{-9})x + (0.5*10^{-9})x^2 = 9*10^{-9}x^2 [/tex]

    Doubling everything:

    [tex] (4*10^{-9} - (4*10^{-9})x + (1*10^{-9})x^2 = 18*10^{-9}x^2 [/tex]

    [tex] (4*10^{-9} - (4*10^{-9})x + (1*10^{-9})x^2 = 18*10^{-9}x^2 [/tex]

    Taking everything over to the LHS:

    [tex] (4*10^{-9} - (4*10^{-9})x + ((1-18)*10^{-9})x^2 = 0 [/tex]

    [tex] (4*10^{-9} - (4*10^{-9})x + (-17*10^{-9})x^2 = 0 [/tex]

    Then using the formula:

    [tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

    where:

    [tex] a = 4*10^{-9} , b = -4*10^{-9} , c = -17*10^{-9} [/tex]

    Giving:

    [tex]x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{8*10^{-8}}[/tex]

    Which gives x as being either 2.62 (too big) or -1.62?

    TFM
     
  9. Apr 28, 2008 #8
    You're correct except for the bit where you start using the quadratic formula.

    The quadratic formula as you have written it corresponds to:
    [tex]ax^2 + bx + c = 0[/tex]

    However, you're equation is in the form:
    [tex]c + bx + ax^2 = 0[/tex]

    (Which is obviously the same, but you are switching a and c!)
     
  10. Apr 28, 2008 #9

    TFM

    User Avatar

    Thanks, Nick89, This should be:

    [tex]x = \frac{4*10^{-9} \pm 1.697*10^{-8}}{2*-17*10^{-8}}[/tex]

    This gives -0.61 and 0.38, which look better

    TFM
     
  11. Apr 28, 2008 #10

    TFM

    User Avatar

    I just tried putting in 0.38, an it is the correct answer :smile:

    Thanks everybody,

    TFM
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook