Field lines near the equilibrium point

[scotshocker]

Homework Statement

Charges 4q and -q are located at the points (-2a,0,0) and (-a,0,0), respectively. Write down the potential θ(x,y) for points in the xy plane, and then use a Taylor expansion to find an approximate expression for θ near the origin, which you can quickly show is the equilibrium point. (You can set a=1 to make things simpler)

Homework Equations

θ(x,y,s)=∫(ρ(x',y',z')dx'dy'dz')/4πεr

The Attempt at a Solution

setting a-1, ignoring the factor of q/4πε₀, the potential due to the two charges, at locations in the xy plane is θ(x,y)=(1/(sqrt((x-2)²+y²)-(1/sqrt((x-1)²+y²)

Have I set this equation up correctly?

 

[Simon Bridge]

Welcome to PF;

Have I set this equation up correctly?

... how would you check?

i.e. is the equilibrium point at the origin as the problem says?
(Notice that one of the charges is much bigger than the other? Have you accounted for that?)

Have you checked that you have added the vectors properly - say, by sketching them out head-to-tail?

 

[TSny]

Hello, scotshocker. Welcome to PF!

ignoring the factor of q/4πε₀, the potential due to the two charges, at locations in the xy plane is
θ(x,y)=(1/(sqrt((x-2)²+y²)-(1/sqrt((x-1)²+y²)

Have I set this equation up correctly?

Looks good except for a couple of signs. At what values of x would you expect the potential to be undefined?
[EDIT: And as Simon points out, the charges are not of equal magnitude.]

 

[scotshocker]

The fact that the one charge is larger than the other is the part that I am having trouble with. I am also confusing myself because both charges are on the same side of the origin. I am severely out of practice and am just trying to figure this out. Any suggestions?

 

[Simon Bridge]

The charges are also opposite signs.
Perhaps you'd better write down the equation for the potential along just the x-axis to start with - don't leave off the q and the a this time: I think you removed them too soon, before you understood the situation.
Once you see that, you'll probably make the connections you need.

Actually draw the axes - mark out x=0, x=+a, x=+2a, x=-a, x=-2a, etc.