- #1

rslewis96

- 5

- 0

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...

## Homework Statement

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))

## Homework Equations

V=[k(q)/r] + [k(-q)/r]

distance formula √(x2-x1)^2+(y2-y1))

## The Attempt at a Solution

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.