0
$\begingroup$

Do elecrons move from low to high potential or they move due to attraction by positive charge? Or is it something else, what actually happens in flow of electricity.....the concept is just so confusing.

$\endgroup$
1
$\begingroup$

You ask the same thing. Electrons move from a low to high potential and they move due to attraction by a positive charge. The electric potential field is created by charged particles. You cannot have one without the other.

We have the same situation with gravity. The Earth is a massive object, and it attracts things like us and the moon. However, you can also think of there being a "gravitational field" surrounding the Earth which applies a force inward. They are two ways to solve the problem.

Which tool you use depends on the problem. If there are many "point charges," charged objects that are very small with respect to the distances you are looking at, then thinking in terms of electrons being attracted to them is a very effective way of thinking about it. If, on the other hand, there are distributed charges, spread out over large areas, its often easier to think about the problems in terms of electric fields rather than summing large numbers of particles. For example, in a capacitor, its much easier to solve problems by thinking of there being an "electric field" between the plates of the capacitor rather than having to do some summations over millions of charged particles on the plate.

$\endgroup$
0
$\begingroup$

Those are just two ways of describing the same thing. The potential causing an electron to move is due to other charges. The potential (or its negative gradient, the field) provides a local explanation rather than an action-at-a-distance explanation.

$\endgroup$
  • $\begingroup$ What's that action-at-a-distance thing? $\endgroup$ – Sakshi Singh Aug 24 '19 at 16:39
  • $\begingroup$ “Action at a distance” is the idea that a particle at one location can exert a force on a particle at another location (as in $F=q_1q_1/4\pi\epsilon_0r^2$ for electrostatics or $F=Gm_1m_2/r^2$ for gravitation) without there being anything in between (like a field). $\endgroup$ – G. Smith Aug 24 '19 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.