With three or more charges, each pair still obeys Coulomb's law, and the force from one charge does not change because others are nearby. So the net force is the sum of the separate forces, added as vectors, the way you learned in Adding Vectors by Components.
Why it works
This is the superposition principle, and it is an experimental fact: the force between two charges is the same whether or not other charges are around. So you can find each force on its own and then add. Forces are vectors, so they add tip to tail. On a line, that means adding with signs; at an angle, it means adding components.
Key idea
$\vec F_{\text{net}} = \vec F_1 + \vec F_2 + \cdots$, each from Coulomb's law. Find each force's size, decide its direction (toward or away from the charge that makes it), then add components: $F_x = \sum F_{ix}$, $F_y = \sum F_{iy}$, $F = \sqrt{F_x^2 + F_y^2}$.
Worked examples
Worked example: Three in a row
Charges on the $x$-axis: A $= +2.0\,\mu\text{C}$ at $x = 0$, B $= -3.0\,\mu\text{C}$ at $x = 0.20\,\text{m}$, C $= +4.0\,\mu\text{C}$ at $x = 0.50\,\text{m}$. Find the net force on B.
- From A ($0.20\,\text{m}$ away): $(8.99 \times 10^{9})\dfrac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.20)^2} = 1.35\,\text{N}$. Opposite signs attract, so B is pulled toward A: $-1.35\,\text{N}$.
- From C ($0.30\,\text{m}$ away): $(8.99 \times 10^{9})\dfrac{(4.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2} = 1.20\,\text{N}$, pulled toward C: $+1.20\,\text{N}$.
- Net: $-1.35 + 1.20 = -0.15\,\text{N}$, so $0.15\,\text{N}$ toward A (to the left).
Answer: $0.15\,\text{N}$ in the $-x$ direction
Worked example: At a right angle
A $+2.0\,\mu\text{C}$ charge is at the origin, a $+3.0\,\mu\text{C}$ charge at $(0.30\,\text{m}, 0)$ and a $+4.0\,\mu\text{C}$ charge at $(0, 0.40\,\text{m})$. Find the net force on the charge at the origin.
- From the charge on the $x$-axis: $$F_1 = (8.99 \times 10^{9})\dfrac{(3.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.30)^2} = 0.599\,\text{N},$$ pushing in the $-x$ direction.
- From the charge on the $y$-axis: $$F_2 = (8.99 \times 10^{9})\dfrac{(4.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.40)^2} = 0.450\,\text{N},$$ pushing in the $-y$ direction.
- $F = \sqrt{0.599^2 + 0.450^2} = 0.749\,\text{N}$, at $\tan^{-1}(0.450/0.599) = 36.9^\circ$ below the $-x$ axis.
Answer: $0.749\,\text{N}$, $36.9^\circ$ below the $-x$ direction
Common mistake
Adding the sizes of forces that point in different directions. In the second example, $0.599 + 0.450 = 1.05\,\text{N}$ is wrong; the forces are at right angles, so the net is $0.749\,\text{N}$. Even on a line, a force pointing left must be subtracted. Check: the net force can never be bigger than the sum of the sizes, and equals that sum only when all the forces point the same way.