If I would displace a negative one-coulomb charge r meters from a positive charge, Q, the negative charge would gain electric potential energy that would be calculated by summing up all the values of the electric potentials along the line of its displacement, r. (I don't know calculus, but I have heard of integration.) Why then is the formula for electric potential KQ/r? Doesn't this come from multiplying the electric force at displacement r from Q with r? Why do we do that when the electric force is actually changing along the path of displacement and not constant? Is not this, for example, like multiplying the final acceleration of an object by the duration of the acceleration to find the final velocity, which is invalid because the acceleration is actually constantly changing along the path?

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Nugatory
Mentor
If you actually calculate out the integration, you'll find that if the force at every point goes as ##1/r^2## the potential at any point goes as ##1/r##.

The change in potential energy is the difference between the potential at the starting point and the potential at the ending point. It's also what you get by summing (as an integral) the change in the potential across each infinitesimal step along the path from starting point to ending point.

Do not confuse the change in the potential between two points separated by a distance ##r## and the potential at a point at a distance ##r## from the positive charge. The best way to keep it all straight is to use different letters for the different quantities in the problem:
##R_0##: the distance of the starting point from the central charge.
##R_1##: the distance of the ending point from the central charge
##\Delta{R}=R_1-R_0##: the distance the negative charge is displaced
##r##: a label for an arbitrary distance from the central charge. The potential energy at a point at a distance ##r## from the central charge is ##-KQ/r## and the attractive force is ##KQ/r^2##.

The potential energy gain when I move the charge a distance ##\Delta{R}## from the starting point at ##R_0## to the ending point at ##R_1## is ##KQ(1/R_1-1/R_0##.

Inspiron
If you actually calculate out the integration, you'll find that if the force at every point goes as ##1/r^2## the potential at any point goes as ##1/r##.
.

So should I take this for granted till I study calculus? Is not there any other way to understand it without calculus?

Dale
Mentor
2021 Award
I don't know of another way. I suppose that you could graphically approximate it to convince yourself that it is correct.

vanhees71