# 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?

Here is an analogy. Let's consider dropping a ball on the earth's surface. We all know that Earth's gravitational field is a vector towards its centre. Only if the ball is dropped from rest, or was a given a downward shove, it's path will be straight(at least till it hits the ground). On the other hand, if you lob the ball horizontally, it's path is curved. In this case you can't claim that the curve traced by the ball is the direction of earth's gravitational field.

Much the same way, a test charge in an electric field for not necessarily trace out the electric lines of force (also called the electric field). The reason for this is because the electric field is in the direction of acceleration of the test charge, not in the direction of it's velocity.

If the initial velocity is in any direction other than that of the field, the test charge follows a curved path.

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:

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