Force on a Charge Due to a Collection of Charges
It is linear. If we have two charge distributions and bring them together the total force is the vector sum of the forces due to the two charge distributions.
The force on a charge
\[Q\]
at point A due to a point charge \[q_B\]
at a point B is \[\mathbf{F}_{BA}=\frac{1}{4 \pi \epsilon_0} \frac{q_B Q}{r^2_{AB}} \mathbf{e}_{BA}\]
where \[r_{AB}\]
is the distance from A to B, \[\epsilon_0 = 8.854 \times 10^{-12} F/m\]
and \[\mathbf{e}_{BA}\]
is the unit vector from B to A.Suppose then that we have five charges as shown.
Each gives rise to a force, but the overall force is
\[\begin{equation} \begin{aligned} \mathbf{F} &= \mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3+\mathbf{F}_4+\mathbf{F}_5 \\ &= \frac{1}{4 \pi \epsilon_0} \frac{q_1Q}{d^2_1}\mathbf{e}_{1A}+ \frac{1}{4 \pi \epsilon_0} \frac{q_2Q}{d^2_2}\mathbf{e}_{2A}+ \frac{1}{4 \pi \epsilon_0} \frac{q_3Q}{d^2_3}\mathbf{e}_{3A} + \frac{1}{4 \pi \epsilon_0} \frac{q_4Q}{d^2_4}\mathbf{e}_{4A} + \frac{1}{4 \pi \epsilon_0} \frac{q_5Q}{d^2_5}\mathbf{e}_{5A} \\ &= \frac{Q}{4 \pi \epsilon_0} (\frac{q_1}{d^2_1}\mathbf{e}_{1A} + \frac{q_2}{d^2_2}\mathbf{e}_{2A} + \frac{q_3}{d^2_3}\mathbf{e}_{3A} + \frac{q_4}{d^2_4} \mathbf{e}_{4A} + \frac{q_5}{d^2_5} \mathbf{e}_{5A}) \end{aligned} \end{equation} \]
