V=[k(q)/r] + [k(-q)/r] Finding the points where v is equal to zero.

  • Thread starter rslewis96
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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...

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.
 

Answers and Replies

  • #2
haruspex
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To continue with your method, you could simply invert each side. Then multiplying out will lead to a lot of simplification. Since the charges are equal and opposite, it is evident that the solution will be the locus of points equidistant from the two charges, i.e. the perpendicular bisector of the line segment joining them.
 

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