Finding distance between two charges given their potential difference.

AI Thread Summary
The discussion centers on finding the distance between two charges based on their potential difference using MATLAB. The user is attempting to simplify the equation V = k(q1)/r1 + k(q2)/r2 for inputting values to derive x and y coordinates. There is confusion regarding the formula, as it is pointed out that the equation represents the potential sum rather than the potential difference. The user is also uncertain about how to specify the coordinates for the test point and the charges. Clarification on the correct formula and simplification techniques is sought to facilitate the programming task.
rslewis96
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Homework Statement


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.

Homework 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)

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.
 
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rslewis96 said:

Homework Statement


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.

I think you meant for the test point to be (x,y)?

Homework Equations


Formula I am using to find potential difference between the two: V=k(q1)/r1+k(q2)/r2

That is not the potential difference. That is the potential sum.
 
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