# Electric Potential of 3 point charges

1. Feb 5, 2010

### Kittlinljd

1. The arrangement of charges shown in the figure is called a linear electric quadrupole. The positive charges are located at +-s. Notice that the net charge is zero.
Find an expression for the electric potential on the x-axis at distances y>>s.

2. Relevant equations
V= $$\sum$$1/4pi $$\epsilon$$0 * qi/ri

3. The attempt at a solution
Im just not sure how to go about modeling this equation.. otherwise I don't know were to start.
I have tried: 1/4pi$$\epsilon$$0 *Q/y thinking that it was a simple question, and I now understand that I should pick a s-value and use that to come up with the radius. However, I am stuck on how.
Thanks

Last edited: Feb 5, 2010
2. Feb 5, 2010

### CompuChip

Let's start by finding expressions for the ri.

Can you express them in terms of y and/or s?

3. Feb 5, 2010

### Spinnor

You have the definition of the potential. Say I sit at x=0, y=10. Ignore the constants but not the charge and distance. You should be able to write down the potential at this point, (remember the distance from x=0, y=10 to each charge are 10 + s, 10, 10-s. Your general formula must take into account the fact that the charges are not at the same place.

V(y=10) = 1/(10 + s) -2/10 + 1/(10 - s)

In general V(y) = 1/(y + s) - 2/y + 1/(y - s).

4. Feb 6, 2010

### Kittlinljd

so I change the constants to -s or s depending on which direction on the direction on the y axis? that makes sense.
What I tried was V = K $$\sum$$Q/(y+s) -2Q/(y) + Q/(y-s) , and it said that the answer does not depend on the variable s. Would this have something to do with Q= 2qs^2, being the electric quadrupole moment.

Last edited: Feb 6, 2010
5. Feb 6, 2010

### CompuChip

You solved the problem for a point on the y-axis, like Spinnor said in his example.

However, you want to take a point on the x-axis.
Did you read my post (#2) ?

6. Feb 7, 2010

### Kittlinljd

ahhhh, okay so now I see so I get:
K* Q/(y^3)
which is the answer masteringphysics took!
Thanks So much!!!

7. Feb 7, 2010

### CompuChip

I assume that you worked this out using the Pythagorean theorem, and found that s cancels out