What is the Solution to the Equipotential Surface Problem?

[J Hann]

It might help if you use the definition of the electric potential at a point:
"The work done in bringing a unit positive test charge from infinity to that point".
What path or paths (if you choose to use a plane instead of a straight line) , in this case,
would the test charge need to travel so that no work is done in moving the test charge?

 

[gracy]

I think I have understood this question properly that's why I want to show my understanding here

Each equipotential surface is a surface in 3D space, which corresponds to a line on the 2D graph where line can be curved or straight(That we will have to figure out).All the points on it will have same potential .With the two points given,I found out that same potential is zero then I had to find out the points of the equipotenial surface.$\frac{1}{4\pi\epsilon_0}\bigg(\frac{1}{\sqrt{(x-1)^2+(y-1)^2}}-\frac{1}{\sqrt{(x-1)^2+(y-2)^2}}\bigg)$=PThis is the equation for an arbitrary equipotential surface - If we put in the value of P, we will get the equipotential surface whose value of potential on it is equal to P

By solving that I got $y$=$1.5$

in a 3D Cartesian coordinates it is a collection of all points (x,1.5,z) where the values of x and z are arbitrary, any possible values from −∞ to +∞.

Hence in this case the equipotential will be represented by straight line

 

[BvU]

By solving that I got $y=1.5$

By solving that for P = 0

in a 3D Cartesian coordinates it is a collection of all points (x,1.5,z) where the values of x and z are arbitrary, any possible values from −∞ to +∞. Hence in this case the equipotential will be represented by straight line

The intersection of this equipotential plane with the plane z = 0 is a straight line parallel to the x-axis

In short: your understanding is OK !