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Explain the superposition principle for static electric forces and write its general equation.
Solution

To find force acting on a charge by other charges, principle of superposition is also used with Coulomb's law.
"When more than one Coulombian forces are acting on a charge, the resultant coulombian force acting on it is equal to the vector sum of the individual forces."
Suppose $q_{1}, q_{2}$ and $q_{3}$ are charges of a system as shown in figure.
Let $\overrightarrow{r_{1}}, \overrightarrow{r_{2}}$ and $\overrightarrow{r_{3}}$ are their respective position vectors from origin ' $\mathrm{O}$ '.
If $\overrightarrow{\mathrm{F}_{12}}$ is force acting on $q_{1}$ by $q_{2}$, then
$\overrightarrow{\mathrm{F}_{12}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{12}^{2}} \cdot \hat{r}_{12}$
And $\overrightarrow{\mathrm{F}_{13}}$ is force acting on $q_{1}$ by $q_{3}$, then $\overrightarrow{\mathrm{F}_{13}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{13}^{2}} \cdot \hat{r}_{13} \quad \ldots(2)$ Where $\overrightarrow{r_{12}}$ is vector in direction along $q_{2}$ to $q_{1}$.
and $\overrightarrow{r_{13}}$ is vector in direction along $q_{3}$ to $q_{1}$.
$\therefore \overrightarrow{r_{13}}=\overrightarrow{r_{3}}-\overrightarrow{r_{1}}$
If $\overrightarrow{\mathrm{F}}$ is force on $q_{1}$ by $q_{2}$ and $q_{3}$, then $\vec{F}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}$
$=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{12}^{2}} \cdot \hat{r}_{12}+\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{3}}{r_{13}^{2}} \cdot \hat{r}_{13}$