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Homework Help: Where is the electric field is zero?

  1. May 18, 2004 #1
    two charges of 1.5X10^-6 c and 3.0X10^-6 c are 0.2 m apart. Where is the electric field between them equal to zero?

    heres what i got....

    the electric field will be 0 when the field strength of the first charge minus the field strength of the second charge equals 0.

    therefore,

    q(1)-----x------P-------(0.2 - x)-------q(2)

    where...
    P is where the electric field equals 0
    q(1) is the first charge
    q(2) is the second charge
    x is the distance (in metres) from the charge

    [ kq(1) / (x)^2 ] - [ kq(2)/ ((0.2-x)^2) ] = 0

    from here,
    i cancel out the k's
    find the common denominator and cancel it out once my numerator is expanded
    try and use the quadratic equation to solve for x. however, when i try to solve for x i get a complex number...what am i doing wrong? the book says the answer is 0.08m (approx.) if you know a faster and much easier way, please do tell...thanks a bunch...
     
  2. jcsd
  3. May 18, 2004 #2

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    Let the neutral point be a distance r from the smaller charge Q and a distance R-r from the bigger charge 2Q. Note that

    kQ/r^2 - 2kQ/(R-r)^2 = 0 expresses the neutrality of the electric force along the line between the charges, by Coulomb's law.

    After some algebraic manipulation you get

    r^2 + 2Rr - R^2 = 0.

    Applying the quadratic formula,

    r = -R +/- sqrt(2)R. (I am too lazy to figure out how to stack the plus or minus symbol, so I wrote it as +/-.)

    Discard the root that does not lie between the two charges. This leaves you with

    r = [sqrt(2) - 1]R.

    Here R=0.2 m, so you have

    r = [1.414 - 1] 0.2

    which works out to about 0.08 meters, that being the distance from the smaller charge.
     
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