Significance of the negative sign in $V=-\int \vec E · \mathrm d\vec r$

Question

Why is there a negative sign in the relation $$V=-\int \vec E · \mathrm d\vec r$$ between the electric field and the electric potential?

Is this because of some derivation in vector calculus because I have not been taught that approach to this concept yet (as I am in high school)? Does this also represent the fact in the direction of increasing electric field the electric potential decreases? How should one interpret the negative sign and integral intuitively in more complex non linear cases?

What is a “non-linear case”? Electromagnetic fields in classical EM are linear. — G. Smith, Aug 12 '19 at 22:14
Possible duplicates: https://physics.stackexchange.com/q/249670/2451 , https://physics.stackexchange.com/q/356893/2451 , https://physics.stackexchange.com/q/315528/2451 and links therein. — Qmechanic, Aug 12 '19 at 22:22
By non-linear I mean electric fields dependent on higher powers of r — Schwarz Kugelblitz, Aug 13 '19 at 09:10

G. Smith · Accepted Answer · 2019-08-12T22:21:43.997

It is the integral form of

$$\vec{E}=-\vec{\nabla} \varphi,$$

which says that the field points from high potential to low and not vice versa. Just as with other concepts of potential or potential energy (say, gravitational), electrostatic potential is defined so that (positively charged) objects tend to move “down the potential hill”, not up it. It is also defined so that electrostatic potential energy is additive with other forms of energy such as kinetic energy.

Significance of the negative sign in $V=-\int \vec E · \mathrm d\vec r$

1 Answers1