- #1
rslewis96
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Homework Statement
I am trying to determine at what points the given potential difference will be using v=k(q2)/r2-k(q1)/r1.
I know the location of both q1 and q2 using cartesian coordinates, as well as, there magnitudes which I have set each to one.
q1=(1,0)
q2=(-1,0)
Homework Equations
v=kq/r,
Vba=Vb-Va
Vb=k(q2)/r2
Va=k(q1)/r1
v=k(q2)/r2-k(q1)/r1
r1=sqrt((x1-x0)^2+(y1-y0)^2)
r2=sqrt((x2-x0)^2+(y2-y0)^2)
The Attempt at a Solution
First I found that if you set v=0, then the points will be right along the y-axis.
r2=r1
Which this makes sense given that each point is a equal distance from each other along the x-axis.
Next I tried to see if I could determine the points given that v>0 or v<0. How I wanted to proceeded was to create a general form of an equation, such as x0=whatever*y0 or y0=whatever*x0. This would allow me to create an x,y table and plug in for one variable to find the other.
My first thought was to get the content within both r1 and r2 out of the square root. I started by squaring both r1 and r2. I then needed to get the equation v=k(q2)/r2-k(q1)/r1 having even powers of r1 and r2.
Here is my attempt:
(v/k)=(1/r2-1/r1) => (r1*r2)(v/k)=(r1-r2) => ((r1*r2)(v/k))^2=(r1^2-2r1r2+r2^2) => (((r1*r2)(v/k))^2)-r1^2-r2^2=-2r1r2
=> [((r1*r2)(v/k))^2-r1^2-r2^2]^2=(-2r1r2)^2
I tried to work with this equation, but got nowhere.
I am not sure if I am going in the wrong direction completely or what. I'm not looking for anyone to solve this for me, I am looking for advice on how to proceed. Any advice will be helpful.
Thank you for your time.