A. Electrostatic Force

Attraction or repulsion force between two charges at certain distance.

Attraction or repulsion force of two charges in empty space is:

$$\begin{array}{rl}F& =\frac{1}{4\pi {\epsilon }_o}\frac{Q_1{Q}_2}{r^2}=k\frac{Q_1{Q}_2}{r^2}\\ \\ k& =\frac{1}{4\pi {\epsilon }_o}=9\times {10}^9\\ \\ {\epsilon }_o& =8,85\times {10}^{-12}\text{ (permitivitas hampa udara)}\end{array}$$

When charge in a certain medium, the force is:

$$\begin{array}{rl}{F}_{\text{bahan}}& =\frac{1}{4\pi {\epsilon }_r{\epsilon }_o}\frac{Q_1{Q}_2}{r^2}=\frac{k}{{\epsilon }_r}\frac{Q_1{Q}_2}{r^2}\\ \\ F_{\text{bahan}}& =\frac{F_{\text{hampa udara}}}{{\epsilon }_r}\\ \\ {\epsilon }_r& =\text{permitivitas relatif bahan }({\epsilon }_r>1)\end{array}$$

Ratio of force in medium and in vacuum space:

$$\begin{array}{rl}\frac{F_{\text{medium}}}{F}& =\frac{{\displaystyle \frac{\text{coret}[red]k}{{\epsilon }_r}}\text{coret}[red]{\displaystyle \frac{Q_1{Q}_2}{r^2}}}{\text{coret}[red]k\text{coret}[red]{\displaystyle \frac{Q_1{Q}_2}{r^2}}}\\ \\ \frac{F_{\text{medium}}}{F}& =\frac{1}{{\epsilon }_r}\end{array}$$

Because ε_r > 1, F in medium is lower than in vacuum space.

Electric Field

a region in which an electric charge (a test charge) experiences a force because of influence of another charge.

Electric field strength:

$E=k{\displaystyle \frac{Q}{r^2}}$

When a charge q is in an electric field E, the charge will experience a force:

$F = q \cdot E$