Finding the potential difference given two unknown points.

In summary, the problem is to determine the points where a given potential difference will be using the equation v=k(q2)/r2-k(q1)/r1, given the locations and magnitudes of q1=(1,0) and q2=(-1,0) using cartesian coordinates. The equations v=kq/r, Vba=Vb-Va, Vb=k(q2)/r2, Va=k(q1)/r1 are also provided for reference. The attempt at a solution involved trying to determine the points for v=0, v>0, and v<0, but was unsuccessful. The attempt also involved squaring the equations for r1 and r2 and manipulating them to try and find a general form
  • #1
rslewis96
5
0

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.
 
Physics news on Phys.org
  • #2
Suggest stating the problem in its original form.
 

What is potential difference?

Potential difference, also known as voltage, is the measure of the electric potential energy per unit charge between two points in an electric field.

How is potential difference calculated?

Potential difference is calculated by subtracting the electric potential at one point from the electric potential at another point. It is represented by the equation V = W/Q, where V is the potential difference, W is the work done in moving a charge, and Q is the magnitude of the charge.

What is the unit of potential difference?

The unit of potential difference is volts (V), named after the Italian physicist Alessandro Volta. It is equivalent to one joule per coulomb (J/C).

Can potential difference be negative?

Yes, potential difference can be negative. This occurs when the electric potential at the starting point is greater than the electric potential at the ending point. It indicates that the direction of the electric field is opposite to the direction of movement of the charge.

How is potential difference measured?

Potential difference can be measured using a voltmeter, which is a device that measures the difference in electric potential between two points. It is connected in parallel to the circuit and has a high resistance to avoid altering the potential difference being measured.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
174
  • Introductory Physics Homework Help
Replies
2
Views
728
  • Introductory Physics Homework Help
Replies
17
Views
839
  • Introductory Physics Homework Help
Replies
2
Views
883
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
838
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top