Clarification on Work Done by Electrostatic Force Along Arbitrary Paths in Feynman's Lectures

Question

I am reading the section about electric potential in the book "The Feynman Lectures on Physics, Vol. II" (see the screenshot of the section below). The author first write

There is some distribution of charge, which produces an electric field

So far so good. But here is what i really do not understand. The author write that

We ask about how much work it would take to carry a small charge from one place to another.

My Problem of understading:

For my understanding here there is only one "natural" force namely the electrostatic Force $\vec{F}=q\cdot \vec{E}$. Thus, the work $W$ done by this "natural" force is $W=-\int_a^b\vec{F}\cdot d\vec{s}$. So, both vectors $\vec{F}$ and $d\vec{s}$ are pointing in the same direction. But the author is showing in Fig. 4-2 two arbritrary paths where both vectors $\vec{F}$ and $d\vec{s}$ are in general not pointing the same direction. How could that be? Where is the force coming from? Can somebody explain me the definition of the work better, since i did not fully understand this?

Ref: The Feynman Lectures on Physics, Vol. II The New Millennium Edition Mainly Electromagnetism and Matter, section 4-3 Electrical Potential:

Answer 1

You have written, So, both vectors $\vec F$ and $d\vec s$ are pointing in the same direction and that is not true in general.

Note that theere is a dot product inside the integral, $\vec F \,{\LARGE \cdot} \,d\vec s = F \,s \cos \theta$ where $\theta$ is the angle between the vectors which varies depending on the path taken.

Answer 2

Imagine that there is a train track along the path. Along that train track we can move a train which is powered by an electric motor and an accumulator. Being an electric motor, it can either be used as a motor, draining the accumulator to move the train against the force field present, or it can be used as a dynamo, charging the accumulator when the train "goes with the flow", so to speak.

In all this, we are assuming that:

• No energy is lost due to friction, inefficiencies of the accumulator, etc.
• The train moves so slowly that we can ignore kinetic energy

Now the work done to the train is essentially the amount of energy drained from the accumulator: positive if the accu is drained, negative if it is instead recharged, zero if the charge remains the same.

Now wether the accu is drained or charged depends on the direction of the movement. If we move in parallel in the direction of the force field, we don't need to expend any energy. Instead, since the force field helps our movement, we can charge our accumulator. Work is negative. If we move orthogonally to the force field, then the force field can't accelerate or decelerate us, so to move we just need one tiny bit of energy to get moving in the first place, and then no more energy expenditure at all, since our momentum will carry us forward. And since we're ignoring kinetic energy, we practically didn't drain or charge our accu, so work is zero. And if we move in parallel against the force field, we need to spend energy to do so, our work is positive (in the sense that the train itself gets energy from our accumulator). If we move at some angle to the field, we need to take the scalar product of the force with our direction vector. To determine the amount of work done. If our angle changes, we need the integral to add up all the different infinitesimal portions of work.

To answer your specific question: The paths are determined by the path we choose our object to take. Be that due to a train track or by a steering wheel (take a car going up a mountain via a serpentine road: the force field points straight down, but we choose to go in curves and enforce that using a steering wheel) or some other means. Of course, this does require forces other than those of the force field. A train track applies constraint forces to the train, a steering wheel applies normal forces to a car (technically speaking it's the road applying the forces, but the steering wheel determines how strong they are).

However, an object without any external forces applied, but with the accumulator (or a heat sink) draining all excess kinetic energy, will in fact always go parallel to the force field. Which is what you are probably imagining.