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Finding a neutral point, electric fields. Simple algebra.

  1. Jan 1, 2010 #1
    This is a question that I'm doing as part of revision for my January mocks. Being bad a algebra in general, I'm stuck. The answer is a worked example and I still don't understand it, so I'm feeling very silly.

    1. The problem statement, all variables and given/known data
    Charges of +4C and +8C are placed 1.00m apart. At what distance from the +4C charge is the electric field strength zero?

    2. Relevant equations
    [tex]
    E = \frac{Q}{4\pi\epsilon r^2}

    [/tex]


    3. The attempt at a solution
    I picture it like this:

    <-----x-----><----------- 1-x ----------->
    +4C -------- E=0 ----------------------- +8C

    <----------------- 1m ----------------->

    So, E due to 4C = E due to 8C
    [tex]
    \frac{+4}{4\pi\epsilon . x^2} = \frac{+8}{4\pi\epsilon . (x-1)^2}
    [/tex]

    Simplifying:
    [tex]
    \frac{4}{x^2} = \frac{8}{(x-1)^2}
    [/tex]

    Square rooting & cross multiplying:
    [tex]
    \frac{\sqrt4}{x} = \frac{\sqrt8}{(x-1)}
    [/tex]

    [tex]
    2(x-1) = \sqrt8(x)
    [/tex]


    It is at this point the textbook worked example says:
    Which simplifies to
    [tex]
    x = \frac{2}{(\sqrt8 + 2)}
    [/tex]
    This is my problem, I don't know how on earth it simplified to that. As the textbook hasn't explained the steps I feel silly as it's something probably really easy.

    So I tried this:
    [tex]
    2x- 2 = \sqrt8(x)
    [/tex]
    And I can't get any further...

    I'd really appreciate a simple explanation, thankyou. :)
     
  2. jcsd
  3. Jan 1, 2010 #2
    Your mistake lies in the square root you took!
    Notice that [tex]1-x[/tex] is positive by definition, while [tex]x-1[/tex] is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
    Either solve with [tex]1-x[/tex] in the denominator[/tex] instead of [tex]x-1[/tex] or just don't take the square root and solve the quadratic.
     
  4. Jan 1, 2010 #3
    Oops! My bad, I'll look out for that in future. Okay, so if I do it right, I get:
    [tex]
    2(1-x) = \sqrt8(x)
    [/tex]

    And then
    [tex]
    2 - 2x = \sqrt8(x)
    [/tex]

    ... Uh, I still don't know how to go on from there to get [tex]x = \frac{2}{(\sqrt8 + 2)}[/tex], as the textbook says I should end up at.

    The "solving the quadratic" bit seems to be my main problem. A giant chunk of GCSE Maths has escaped my memory haha. D: I feel rather daft as the Christmas holidays seem to have wiped my brain of how to do these!
     
  5. Jan 1, 2010 #4
    [tex]2 - 2x = \sqrt8(x)[/tex]

    Isolating all the [tex]x[/tex] terms on the left, and all the numbers on the right:

    [tex]\sqrt{8}x+2x=2[/tex]
    [tex](\sqrt{8}+2)x=2[/tex]

    Dividing both sides by [tex]\sqrt{8}+2[/tex] yields:

    [tex]x=\frac{2}{\sqrt{8}+2}[/tex]

    Just some basic algebra, grab a book and get to practicing if you're rusty. ;)
     
  6. Jan 1, 2010 #5
    I can't believe this didn't occur to me. I'm feeling pretty silly :f Thankyou so much!
     
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