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Finding distance between two charges given their potential difference.

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    I am trying to write a program in matlab where I input certain values and get a result of possible x and y coordinates. Writing the program is not a problem, I just need to simplify the equation for matlab. I am trying to rewrite a formula so I can find what the distance is between two charges based on outputted x and y coordinates.

    I am given two charges with unspecified magnitudes, q1 and q2, as well as, the constant(k) 9*10^9Nm^2/C^2. Based on whatever value I input for V, I would like to output possible x and y values. (x,y) are the points used for the first charge, (x1,y1) are the points used for the second charge, and (x2,y2) are the points used for the test point.

    2. Relevant equations
    Formula I am using to find potential difference between the two: V=k(q1)/r1+k(q2)/r2.
    r1=√((x2-x1)^2+(y2-y1)^2)
    r2=√((x2-x)^2+(y2-y)^2)

    3. The attempt at a solution
    I first found out what the distance would be if given that V=0, q1=(+1) and q2=(-1). Input the values: 0V=k(1)/r1+k(-1)/r2. After you move k(-1)/r2 to the left and cancel out k on both sides, you come out with r1=r2. This would mean that the distance between q1 and q2 is right in the middle or a straight line.

    So... what would it be if V=1 with charges of the same magnitudes? Well, input your values again and get 1V=k(1)/r1+k(-1)/r2. Move k(-1)/r2 to the left giving you 1V+k(1)/r2=k(1)/r1. Multiply by (r1*r2) to give you r1(1V-k(1))=r2(k(1)), then divide r2 and (1V-k(1)) to give you (r1/r2)=[k(1)/(1-k(1))]. You then square both sides to get (r1/r2)^2=[k(1)/(1-k(1))]^2. This is where I get hung up. I'm not sure I can simplify even further so is to cancel out any unnecessary values. Also, would I need to specify possible points of the test point, the first charge or the second charge so is to find the others?

    Thank you for your time.
     
  2. jcsd
  3. Oct 27, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    I think you meant for the test point to be (x,y)?
    That is not the potential difference. That is the potential sum.
     
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