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Homework Help: Electric field in electrostatic situation

  1. Aug 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Two particles with positive charges q1 and q2 are separated by a distance s. Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero?

    2. Relevant equations

    E = kq/r2

    where k = 1/(4 x pi x epsilon-nought)

    3. The attempt at a solution

    Ok so basically the picture is something like this


    |<---distance s-->|

    Where ive placed the point P randomly in between charges q1 and q2. Ive called the distance between q1 and p "x", and the distance between q2 and p "s-x"

    So ive basically substituted q1 and x into the equation. Then i substituted q2 and (s-x) into the equation. Then i equated E1 and E2 to obtain a quadratic equation.

    x2(q1-q2) - x2q1s + q1s2 = 0

    Ive gone ahead and solved this for x using the quadratic formula. Since there were two answers cause of the plus minus, i just eliminated the minus part of it to get

    x = [2q1s + root(4q1q2s2)] / 2(q1 - q2)

    I cancelled out the 2's on the top and the bottom. However when i input my answer its not correct. Maybe i made a mistake somewhere? I dont know i cant seem to find one for now.
  2. jcsd
  3. Aug 7, 2011 #2
    I think yours is an algebraic problem:

    why did you eliminated the solution with the minus?
    Which are the physical conditions on x?
  4. Aug 8, 2011 #3
    The entire question is posted above.

    I forgot to add this in before, but i opened up a hint on "solving the quadratic and choosing the correct answer"

    "If you set the magnitudes of the fields due to q1 and q2 equal at point P, you should end up with a quadratic equation for x. This equation will have two solutions, but since you know that the zero-field point must be between the two charges, you should be able to eliminate one of the results. Also if you assume that q1 is not equal to q2 then you can use the relation

    a - b = (root[a] + root)(root[a] - root)

    to simplify your answer."

    Hope that helps a bit more
  5. Aug 8, 2011 #4
    Yes,but how do you eliminate the extra solution? (I wanted you to answer this :D)

    To choose the right x,you have to pose the condition you posted, that is the x must be between the two charges [itex]\Rightarrow 0\leq x\leq s[/itex]

    So, you must solve this inequality
    [itex] 0\leq \frac{q_1\cdot s \pm s\sqrt{q_1\cdot q_2}}{q_1-q_2}\leq s[/itex]

    First you put [itex]q_1>q_2[/itex] and solve, and then you put [itex]q_1<q_2[/itex] and solve.

    Hope it helps...
  6. Aug 9, 2011 #5
    Ok so when you say solve the inequality, im guessing you mean try the conditions where q1>q2 and q1<q2 and see if the value still lies within zero and "s".

    So first i started off with the - solution which is

    s(q1 - root[q1q2]) / (q1 - q2)

    For the q1 > q2 condition i found using logic that the value remained within zero and "s"

    For the q1 < q2 condition i found that the value was actually larger than "s" so it didnt satisfy the requirement.

    Then i did the same thing for the + solution

    s(q1 + root[q1q2]) / (q1 - q2)

    For the q1 > q2 condition i found that the value was larger than "s"

    For the q1 < q2 condition i found that the value was a negative number

    So according to all this, it means that the only solution that works is when q1 > q2 for the solution using the negative sign

    s(q1 - root[q1q2]) / (q1 - q2)

    Is this what you wanted me to do? Did i get it right? =P
  7. Aug 9, 2011 #6

    Thnx heaps DiracRules
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