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Finding Electric Field of Two Charges

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data

    There is a charge Q=1 μC located at x=0 and a charge -Q/2 located at x=4 m. What is the E-field.

    At what value of x is the E-field zero? Express your answer in meters and do not approximate decimals digits.

    2. Relevant equations

    [itex]E=\frac{q_1 q_2 k_c}{r^2}[/itex]

    Removing the test charge:

    [itex]\frac{E}{q}=\frac{q_n k_c}{r^2}[/itex]

    3. The attempt at a solution

    I have this solved, but I am going wrong somewhere and can't see my mistake. Here is my work and the correct final answer.

    To find the E field at x=100, E1-E2=0. E2 is being subtracted due to q2 being negative.

    This gives E1=E2

    [itex]\frac{q_1 k_c}{r^2}=\frac{q_2 k_c}{(r-4^2)}[/itex]

    [itex]\frac{(r-4)^2}{r^2}=\frac{q_2 k_c}{q_1 k_c}[/itex]

    [itex]\left(\frac{(r-4)}{r}\right)^2=\frac{q_2 k_c}{q_1 k_c}[/itex]

    [itex]\frac{r-4}{r}= \pm \sqrt{\frac{q_2}{q_1}}[/itex]

    [itex]1-\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}}[/itex]

    [itex]-1\left(-\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} -1\right)[/itex]

    [itex]\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} +1[/itex]

    [itex]\frac{r}{4}= \pm \sqrt{\frac{q_1}{q_2}} +1[/itex]

    [itex]r= \left(\pm \sqrt{\frac{q_1}{q_2}} +1\right)4[/itex]

    [itex]r= \pm 4\sqrt{\frac{q_1}{q_2}} +4[/itex]

    [itex]r= 4 \pm 4\sqrt{\frac{1}{0.5}}[/itex]

    [itex]r= 4 + 4 \sqrt{2}[/itex]


    But that isn't correct. That isn't the correct answer. The correct answer is:

    [itex]r= 8 + 4 \sqrt{2}[/itex]

    Where does the extra factor of 2 come from making the 4 an 8?
     
  2. jcsd
  3. Feb 21, 2013 #2

    ehild

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    You made a mistake. The red line is wrong.

    ehild
     
  4. Feb 21, 2013 #3
    I am assuming I can't take the inverse of that thing like I did?

    Would the more appropriate solution to cross multiply everything to get r alone?
     
  5. Feb 21, 2013 #4

    tiny-tim

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    hi erok81! :smile:
    that'll still leave you with the problem of what the inverse of the RHS is :redface:

    hint: 1 = √q1/√q1 :wink:
     
    Last edited: Feb 22, 2013
  6. Feb 21, 2013 #5

    ehild

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    You can not. Try it with numbers: if 1/r = 2/3 +1
    =5/3
    is r=3/2 +1
    =5/2?
    ???

    You can take the reciprocal of both sides of an equation if neither sides can be zero. If

    4/r =a+b

    then

    r/4=1/(a+b)

    Multiply both sides by 4:

    r=4/(a+b)

    Your equation is

    [tex]\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} +1[/tex]

    Substitute the numerical values of the charges q2=Q/2 and q1=Q.

    [tex]\frac{4}{r}= \pm \sqrt{1/2} +1[/tex]

    Take the reciprocal of both sides.


    ehild
     
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