# Path followed by a free test charge in an electric field region represent the electric lines of forces?

As we know that if we put any free test charge in an electric field which is moving horizontally then it will experience an repulsive force and move in that direction so by that we can say that for Straight lines the above statement is true that path followed by a free test charge in electric field represent the electric lines of forces which will be straight line. So we can say that it is TRUE.

But after that I read one exception case which states that the above statement is not true for Curved path.

Why is that so ? Means why it is for curved path and what is the exception ?Why it is not true?

Your statement implies that the Electric field $\boldsymbol E$ will be parallel, during the whole motion, to the instant velocity $\boldsymbol u(t)$, i.e: $$\boldsymbol E = \boldsymbol E_\parallel + \boldsymbol E_\perp= \boldsymbol E_\parallel.$$ If you have a curved trajectory there must be a component of the force which is perpendicular to the velocity in each instant of time, i.e. the centripetal force:

$$\boldsymbol F = \boldsymbol F_\parallel + \boldsymbol F_\perp \neq \boldsymbol F_\parallel$$ and if the only force is the effect of the electric field on the test charge, i.e. $\boldsymbol F = q\boldsymbol E$, we have that:

$$q \boldsymbol E= q(\boldsymbol E_\parallel + \boldsymbol E_\perp )\neq q \boldsymbol E_\parallel$$