Equipotential Surface

How to understand which of the surfaces are equipotential??
This is a question related to this:
A circle is drawn with centre as a charge +q> Now a charge +q is taken from B to Calong the cicumference of the circle. What work will be done?

well concept of equipotential surface refers to the surface (in a electric field configuration)on which all the points remain at the same potential. That is so say that let points b and c be on a equipotential surface, then the points b and c have the same potential (value) in the given electric field. Since we define the work done to move a charge from a point to another point in terms of the difference of electric potential between these two points. that is W = q*delta(V). Since on a equipotential surface the potential difference is zero, so the work done is also zero. But how can we figure out the configuration of this equipotential surface?. Again it depends upon the configuration of the field we have. Well the original definition of work is defined as W = F.s. So it is the dot product of two vectors. This is zero when the angle between the two vectors is 90 degrees. That means work done is zero when F and s (vectors) are at right angles.

Now in a electric field configuration the direction of force at any point depends upon the direction of the electric field. Therefore for work to done along a path to be zero (the vectors ds along the path) have to be perpendicular to the direction of the electric field. Therefore for charge ( free) which has its field lines emanating in all directions symmetrically outward, the surface is sphere ( 3 dimensions) and circle (2 dimensional space). Therefore when we take a charge around a equipotential surface, which is surface perpendicular to the direction of the electric field, the work done is zero and thereby the potential difference of any two points is zero.

Hope this Helps
Pinnaka