When we say condition of a vector field F being conservative is curl F=0,does it mean that F=F(r)?.I know normally it does not look so.Please,then site an example where F is not a function of r,but still curl F=0.

\vec{F}=y\hat{i}-x\hat{j} is such an example.

Hi neelakash,

If the curl of a vector field is zero, it means the vector field can be represented as the gradient of some function. It does not mean that the vector field is conservative; for example, one can consider time dependent vector fields which nevertheless have zero curl at all times.

Furthermore, its straightforward to come up with examples of vector fields that have no curl and depend on more than just r. Simply take the gradient of any function that doesn't depend just on r. Example: f = x^4 + y^3 + z^2, \vec{F} = \vec{\nabla} f = 4 x^3 \vec{e}_1 + 3 y^2 \vec{e}_2 + 2 z \vec{e}_3 .

Hope this helps.

I was not here for a couple of days...So,I could not reply.However,I suppose from your post that even a time dependent field (having curl zero) can be represented as a gradient of a scalar field?Of course,this time varying field is not conservative.
What is then the condition of being conservative?

One can definitely write a time-dependent vector field with zero curl as the gradient of some (time-dependent) function.

In the context of mechanics, one usually calls a vector field conservative if it can be derived from a time independent potential. In other words, if it has zero curl and is time independent.

A word of warning is in order: in my experience people occasionally get a bit lazy with these definitions.