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Homework Help
Introductory Physics Homework Help
V=[k(q)/r] + [k(-q)/r] Finding the points where v is equal to zero.
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[QUOTE="rslewis96, post: 4544696, member: 491641"] Hello! First of all, thank you for your time in reading my post. Secondly, I am not looking for people to freely give me the answers. With all that said... [h2]Homework Statement [/h2] I have a test point randomly selected on a graph, but what I am trying to figure out is where on this graph would this test point or potential difference would be equal to 0 between two charges. I picked three points, q is at (x,y), (-q) is at (x1,y1) and my test point is at (x2,y2). So with this I should be able to set the equation up and go from there. Where: v is volts, k is 8.99*10^9 (Nm^2)/c^2, q is charges in coulombs and r is the distance between the two points or the distance formula √(x2-x1)^2+(y2-y1)) [h2]Homework Equations[/h2] V=[k(q)/r] + [k(-q)/r] distance formula √(x2-x1)^2+(y2-y1)) [h2]The Attempt at a Solution[/h2] set v equal to zero you get: 0=(8.99*10^9)(q)/(√((x2-x1)^2+(y2-y1)^2) + (8.99*10^9)(-q)/√((x2-x)^2 +(y2-y)^2) move (8.99*10^9)(-q)/√((x2-x)^2 +(y2-y)^2) to the right side. -(8.99*10^9)(-q)/√((x2-x)^2 +(y2-y)^2) = (8.99*10^9)(q)/(√((x2-x1)^2+(y2-y1)^2) from here 8.99*10^9 cancel out. -(-q)/√((x2-x)^2 +(y2-y)^2) = (q)/(√((x2-x1)^2+(y2-y1)^2) multiply the negative sign on the left. (+q)/√((x2-x)^2 +(y2-y)^2) = (q)/(√((x2-x1)^2+(y2-y1)^2) Now the q cancel out. 1/√((x2-x)^2 +(y2-y)^2) = 1/(√((x2-x1)^2+(y2-y1)^2) I then square both sides to get: 1/((x2-x)^2 +(y2-y)^2) = 1/((x2-x1)^2+(y2-y1)^2) and from here I am lost. Any advice on where I should go from here would be great. [/QUOTE]
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Introductory Physics Homework Help
V=[k(q)/r] + [k(-q)/r] Finding the points where v is equal to zero.
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