$\def \u#1{\,\mathrm{#1}}$ $\def \abs#1{\left|#1\right|}$ $\def \ast{*}$ $\def \deg{^{\circ}}$ $\def \ten#1{\times 10^{#1}}$ $\def \redcancel#1{{\color{red}\cancel{#1}}}$ $\def \BLUE#1{{\color{blue} #1}}$ $\def \RED#1{{\color{red} #1}}$ $\def \PURPLE#1{{\color{purple} #1}}$ $\def \th#1,#2{#1,\!#2}$ $\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}$ $\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}$ $\def \dotspot{{\color{lightgray}{\circ}}}$

Electric Fields and Force

Because electric fields point downhill and positive charges feel a force downhill, it follows that positive charges will feel a force in the direction of the electric field. Also, the steeper the slope of the potential, the stronger the field, and from the height analogy we might guess (correctly) that the force on the charge will be larger as well.

Thus the force on a target charge $q_T$ is proportional to the electric field at that point, and we can write

$$\vec F = q_T\vec E$$

Like the electric potential, the electric field is defined everywhere in space, a little arrow at each point whether or not any particle is there to feel it.

An electric field vector with two charges: the positive charge is moving with the field, the negative charge is moving in the opposite direction.

Notice that when $q_T$ is negative, the force and the electric field point in opposite directions. In other words,

Positive targets feel a force with the electric field.
Negative targets feel a force against the electric field.

We can also use this formula to find the electric field at a particular location: place a positive charge $q_T$ there and measure the force on the charge. The electric field at that spot is then $$\vec E = \frac1{q_T}\vec F$$