Recall Newton’s Law of Gravitation: $$ \begin{equation}\begin{aligned} F_{gravity}=\frac{Gm_1m_2}{r^2}\\ \end{aligned}\end{equation} $$
Gravity happens by virtue of objects having mass. Electric repulsion happens by virtue of objects having charge, not mass: $$ \begin{equation}\begin{aligned} F_{electric}=\frac{kQ_1Q_2}{r^2}\\ \end{aligned}\end{equation} $$
This inverse square law is known as Coulomb’s Law, with the proportionality constant $k=\frac{1}{4\pi\epsilon_0}$: $$ \begin{equation}\begin{aligned} F=\frac{Q_1Q_2}{4\pi\epsilon_0r^2}\\ \end{aligned}\end{equation} $$ where $\epsilon_0$ is the permittivity of free space, $8.85\times 10^{-12}Fm^{-1}$.
Electric field strength
Similar to how gravitational field strength is the force acting per unit mass, electric field strength is the force acting per unit charge. Let $q$ be the charge of the body experiencing the force and $Q$ be the charge of the body producing the electric field: $$ \begin{equation}\begin{aligned} \frac{F}{q}&=\frac{Qq}{q\times 4\pi\epsilon_0r^2}\\ E&=\frac{Q}{4\pi\epsilon_0r^2}\\ \end{aligned}\end{equation} $$
Notice that, like with gravitational field strength being dependent on the mass of the planet producing the field, $M$ but not the mass of the object in the field ($m$), the electric field strength is dependent on the charge of the body producing the electric field, $Q$ but not on that of the object in the field ($q$).