This is the simple thing I’m confused about.
As I understand it, the electrostatic force from a charge out to a certain distance is inversely proportional to the distance squared (and electric field strength is in the units Newtons per Coulomb or volts per meter). But electric potential, or voltage is inversely proportional to the distance (r).
Now I get the general idea that electromagnetic signals/forces reduce in strength or energy they can transmit by the distance squared. However, doesn’t voltage also produce force and power from a distance? In other words, can electrostatic/electromagnetic energy be transmitted inversely proportional to the distance and not the distance squared?
Best Answer
The difference between electric potential at two points is proportional to the work done by moving charged particle in a present field. The work is the force multiplied by the distance: $F \cdot s$, or more generally $\Delta U \propto \int_P \vec{F} \cdot d\vec{s}$.
Given that the force scales as $\frac{1}{r^2}$ it should be easy to see why multiplying it by distance in units of $r$ we expect to get a potential that scales as $\frac{1}{r}$.
This process works in reverse too, the field (which is proportional to the force on a particle) is basically a derivative of the potential: $\vec{E} = -\vec{\nabla} U$. So changing potential in a region will result in force acting on a charged particles in this region.
For energy transfer - take a look at Poynting vector. It describes the flux of energy through the area.