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    Hey all...
    I have a quick question
    how does
    become
    if 2aƴ = Q
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    (Original post by valentine_strongrod)
    Hey all...
    I have a quick question
    how does
    become
    if 2aƴ = Q
    \displaystyle \frac{Q}{4\pi \epsilon x \sqrt{a^2 + x^2}} = \frac{2ay}{4\pi \epsilon x \sqrt{a^2 + x^2}} = \frac{ay}{2\pi \epsilon x \sqrt{a^2 + x^2}} = \frac{y}{2\pi \epsilon x \frac{1}{a} \sqrt{a^2 + x^2}}

    Taking the \frac{1}{a} on the bottom into the square root will then give you your final answer.
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    (Original post by james.h)
    \displaystyle \frac{Q}{4\pi \epsilon x \sqrt{a^2 + x^2}} = \frac{2ay}{4\pi \epsilon x \sqrt{a^2 + x^2}} = \frac{ay}{2\pi \epsilon x \sqrt{a^2 + x^2}} = \frac{y}{2\pi \epsilon x \frac{1}{a} \sqrt{a^2 + x^2}}

    Taking the \frac{1}{a} on the bottom into the square root will then give you your final answer.

    Thanks man!
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    I know this is a Physics problem..
    but the math of it is confusing me..
    so here goes:
    2 positive charges (q1,q2) are separated by distance (s). At wat distance from charge q1 is the electric field zero...

    Equation for electric field = kq/r^2

    so i set them up: q1/s^2 + q2/ (s-x)^2 = 0
    so q1/x^2 = -q2/(s-x)^2
    - q2/q1 = (s-x)^2/ x^2
    sqrt root both sides to get - sqrt q2/ sqrt q1 = (s-x) /x
    now i do - sqrtq2/sqrtq1 = s/x - 1
    so 1 - sqrt q2/sqrtq1 = s/x
    final answer = sqrt(q1)s / (sqrt(q1)- sqrt(q2))
    and this is wrong! =(
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    (Original post by valentine_strongrod)
    I know this is a Physics problem..


    so i set them up: q1/s^2 + q2/ (s-x)^2 = 0

    =(

    Are you sure about this? If the're both positive charges won't they act in opposite directions?
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    (Original post by ian.slater)
    Are you sure about this? If the're both positive charges won't they act in opposite directions?
    hmm
    wudnt there electric field add up to make zero tho?
    Crap u are right i think.. let me try that again
 
 
 
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