# How to figure out if the vector product of force and displacement is $-\mathbf fd\mathbf x$ or $\mathbf fd\mathbf x$?

Consider the force due to a point charged particle at the origin and the particle which we move to also be charged with the same charge.

When we move a particle from point $$\mathbf{x}$$(position vector) from origin to infinity(radially), if there is some force $$\mathbf{f}$$ at $$\mathbf{x}$$ (position vector) then some work is done on the particle to move from $$\mathbf{x}$$ to $$\mathbf{x}+d\mathbf{x}$$ (to ultimately reach infinity) this work is given as: dot product of $$\mathbf{f}$$ and $$d\mathbf{x}$$(displacement vector) which in this case, since displacement is parallel to the force vector, gives $$\mathbf{f}d\mathbf{x}$$, this much I have no trouble understanding.

But, If we move from infinity to $$\mathbf{x}$$ (everything is the same, but the path of the particle is reversed), if there is some force $$\mathbf{f}$$ at some intermediate point in the path then, the displacement by $$d\mathbf{r}$$(vector) towards the final position(x) requires work given as: dot product of $$\mathbf{f}$$ and $$d\mathbf{x}$$, but now, the direction of force and displacement are opposite to each other, thus we must get $$-\mathbf{f}d\mathbf{x}$$ but to my dismay, this is false.

Can someone please explain why I am wrong?

• I was confused about the same thing! Here are people's responses: physics.stackexchange.com/q/549629 – plasmaQ Nov 16 '20 at 18:11
• “Vector product” means “cross product”. You are talking about the “scalar product” or “dot product”. – G. Smith Nov 16 '20 at 18:34
• Have you considered how the sign of the differential changes based on the limits of the integral? – BioPhysicist Nov 16 '20 at 19:12

There is nothing wrong!

The force is radially outward. If the particle travels from some point to infinite, then the work turns out to be positive. If the particle travel from infinite to some finite point the work is done by force will be negative. This can be understood from direction of force and displacement as you did. You can equally understand it as when the force is against the motion of particle and when it with the motion of the particle.

In later cases, in which particle is coming from infinite, The positive force must be done by some external force, so that the work energy theorem remain valid.

Let f be the magnitude of the force exerted by the particle at the origin on the particle being moved.

If the force exerted by the particle at the origin on the particle being moved is attractive, then you need to exert a force in the positive radial direction to move it +dr, and the work you expend is +fdr (and the work the particle does on you is -fdr).

If the force exerted by the particle at the origin on the particle being moved is repulsive, then you need to exert a force in the negative radial direction while moving it +dr, and the work you expend is -fdr (and the work the particle does on you is +fdr).

If the force exerted by the particle at the origin on the particle being moved is repulsive, then you need to exert a force in the negative radial direction to move it -dr, and the work you expend is +fdr (and the work the particle does on you is -fdr).

If the force exerted by the particle at the origin on the particle being moved is attractive, then you need to exert a force in the positive radial direction while it is moving -dr, and the work expended is -fdr (and the work the particle does on you is +fdr).