- #1

- 61

- 2

$$F=G\frac{m_1.m_2}{r^2}$$

Gravitational field $$g$$ is derived from this formula

$$g=G\frac{m_1}{r^2}$$ This is named gravitational "field" strength.

If Newton knew nothing about "field concept" and formulated his formula in the form of "action at a distance", how did he interpret $$g$$?

My question is valid for Coulomb as well.

Wasn't this formula

$$

E = \frac{Q}{4\pi\varepsilon_{0}r^2}

$$

invented by Coulomb?

If so, what did he name for $$E$$?