## Force on a Charge Due to a Collection of Charges

The distribution of electric charges gives rise to a force field. Every charge in the region exerts a force on every other charge. If a test charge is brought into the field, it will experience a force due to all the other 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{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}