Total Force on Point Charge from 2 Point Charges

Theorem
Let $p_1$, $p_2$ and $p_3$ be charged particles.

Let $q_1$, $q_2$ and $q_3$ be the electric charges on $p_1$, $p_2$ and $p_3$ respectively.

Let $\mathbf F_{ij}$ denote the force exerted on $q_j$ by $q_i$.

Let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of the other two charged particles.

Then the force $\mathbf F_1$ exerted on $q_1$ by the combined action of $q_2$ and $q_3$ is given by:

where:
 * $\mathbf F_{21} + \mathbf F_{31}$ denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$
 * $\mathbf r_{ij}$ denotes the displacement from $p_i$ to $p_j$
 * $r_{ij}$ denotes the distance between $p_i$ and $p_j$
 * $\varepsilon_0$ denotes the vacuum permittivity.

Proof

 * Two-charges-on-another.png

By definition, the force $\mathbf F_{21}$ and $\mathbf F_{31}$ are vector quantities.

Hence their resultant can be found by using the Parallelogram Law.

The result follows from Coulomb's Law of Electrostatics.