Principle of Superposition

What this figure shows. A central charge $q_1$ at point $A_1$, surrounded by four other point charges $q_2,q_3,q_4,q_5$ scattered at various positions and distances at points $A_2,A_3,A_4,A_5$. Four separate force vectors, $\vec{F}_{12},\vec{F}_{13},\vec{F}_{14},\vec{F}_{15}$, are drawn from $q_1$, each directed along the straight line joining $q_1$ to the respective other charge, representing the individual pairwise Coulomb force on $q_1$ due to each of the four surrounding charges considered one at a time. The figure establishes that the total/net force on $q_1$ is the VECTOR sum of these four individually-computed forces, never a simple scalar addi …

Worked out. Charges of $2\,\mu C$ (A), $3\,\mu C$ (B) and $4\,\mu C$ (C) are placed with AB = 4.0 cm and BC = 3.0 cm at right angles, so $AC=\sqrt{4^2+3^2}=5.0$ cm. The force on A due to B alone is $F_{AB}=\frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{AB^2}=9\times10^9\times\frac{(2\times10^{-6})(3\times10^{-6})}{(0.04)^2}\approx33.7$ N, directed along BA (away from B); the force on A due to C alone is $F_{AC}=9\times10^9\times\frac{(2\times10^{-6})(4\times10^{-6})}{(0.05)^2}\approx28.8$ N, directed along CA (away from C). Using the triangle's geometry, the angle between $\vec{F}_{AB}$ and $\vec{F}_{AC}$ at A works out to $\angle BAC\approx36.9^\circ$ (from $\cos(\angle BAC)=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=0.8$), so the resultant is $F=\sqrt{F_{AB}^2+F_{AC}^2+2F_{AB}F_{AC}\cos(36.9^\circ)}\approx59.4$ N, directed $16.9^\circ$ north of west -- a full worked example of …