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College Physics 2 Electricity

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Two charges are separated as shown. Where would you place a -3 microCoulombs (mC) charge such that the net force on this charge is 0?

    + +
    4mC 8mC
    Distance between charges are 80cm apart.

    2. Relevant equations

    F= (k*q1*q2) / (r*r)

    3. The attempt at a solution

    I found out the Force on q1 ( i made q1 the 4mC charge and q2 the 8mC charger) to q2
    it came to be 4.5*10^(-1) N. Same for the F for q2 on q1. Then i manipluated the equation to slove for r. Once that was established i but the -3mC charge in the equation with the 4mC charge and F being 4.5*10^(-1) giving me an approximation of .5m aka 50 cm. But that doesnt seem right, what equation im not using to get this answer?
     
  2. jcsd
  3. Oct 23, 2008 #2
    Because the net force acting on the charge is 0N, we know that the electric field is 0 N/C.

    Now, through simple arguements we know the point we are looking for is inbetween the two charges and that the field strength from charge 1 has to equal the field strength from charge 2 (in magnitude).

    *Let a be the distance from q1 to the point of E = 0 and b be the distance from q2 to the point of E = 0.

    So, E1 = E2

    k*q1/a2 = k*q2/b2

    Cancel k to get,

    q1/a2 = q2/b2

    you can reorganize the equasion to be simplier to use later on to,

    q1*b2 = q2*a2

    Now you also have a + b = 80cm

    By substituting into another, you can solve for a or b, and thus solve for the other, giving you the point in which there is zero force on the charge.
     
  4. Oct 27, 2008 #3
    I subsituted b for 80-a

    so using the equation you gave q1= 4mC q2=8mC

    4mC*[(80-a)^2] = 8mC*[a^2]
    = 4mC* (6400-160a+[a^2]) = 8mC*a^2

    which seems more confusing then just doing problem itself
     
  5. Oct 27, 2008 #4
    you now have a simple equasion to solve for the value of "a"

    4mC* (6400-160a+[a^2]) = 8mC*a^2
    (6400-160a+[a^2]) = 2*a^2
    6400 = a^2 + 160a

    a = 33.137 (approx)

    so the charge must be placed 33.137cm from the 4mC charge.
     
  6. Oct 28, 2008 #5
    Thankyou. The force of the 3mC charge from the other two chargers were about .02 difference. Basically one was 9.7*10^(-11) and the other was 9.9*10^(-11). Close to zero as possible, but thankyou.
     
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