Finding distance between two charges given their potential difference.

In summary, the conversation discusses writing a program in MATLAB to find the distance between two charges based on inputted x and y coordinates. The formula being used is V=k(q1)/r1+k(q2)/r2 and the conversation also explores different values for V and the resulting equations. The goal is to simplify the equation for use in MATLAB and find the distance between the two charges.
  • #1
rslewis96
5
0

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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  • #2
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.
 

1. How do you calculate the distance between two charges given their potential difference?

The distance between two charges can be calculated using the formula d = Q / V, where d is the distance in meters, Q is the magnitude of the charge in coulombs, and V is the potential difference in volts.

2. What units should be used for the distance and potential difference when calculating the distance between two charges?

The distance should be measured in meters and the potential difference should be measured in volts when using the formula d = Q / V to calculate the distance between two charges.

3. Can the distance between two charges be negative?

No, the distance between two charges cannot be negative. It is a physical distance and therefore must be a positive value.

4. Is the potential difference between two charges affected by the distance between them?

Yes, the potential difference between two charges is inversely proportional to the distance between them. This means that as the distance increases, the potential difference decreases.

5. How does the sign of the charges affect the calculation of distance given potential difference?

The sign of the charges does not affect the calculation of distance given potential difference. The formula d = Q / V takes into account the magnitude of the charges, regardless of their sign.

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