Forces Between Multiple Charges: Superposition Principle

When more than two charges are present in a system, the force on any one charge is not simply Coulomb's law applied once — it is the combined effect of every other charge acting independently. This idea is formalized as the Principle of Superposition, one of the most examined concepts in Electrostatics.

The total electrostatic force on any charge in a system of multiple charges is the vector sum of the forces exerted on it individually by each of the other charges, with each pairwise force being unaffected by the presence of the remaining charges.

Statement

The principle of superposition states that the net electric force acting on a given charge due to a number of other charges is equal to the vector sum of the individual forces exerted on it by each charge taken separately, assuming the other charges are absent.

Explanation / Mathematical Form

Consider a system of nnn point charges q₁,q₂,q₃,…,q_n.

The force acting on charge q₁ due to the other charges is:

\[\vec{F}_1=\vec{F}_{12}+\vec{F}_{13}+\cdots+\vec{F}_{1n}\]​

where
\[\vec F_{12}\]​ is the force on q_1​ due to q₂​,
\[\vec F_{13}\]​​ is the force due to q₃​, and so on.

According to Coulomb’s law, the force on q₁​ due to q₂​ is:

\[\vec F_{12}\]​ = \[\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}^2}\hat{r}_{12}\]​

Similarly, forces due to other charges can be written, and their vector sum gives the resultant force on q₁​.

Thus, the force between any two charges is independent of the presence of other charges.

Conclusion

The principle of superposition shows that:

• Electric forces obey vector addition.
• Each pair of charges interacts independently.
• The net force on a charge in a multi-charge system is found by adding all individual Coulomb forces vectorially.

For two point charges q₁​ and q₂​ separated by distance r_12​:

where r¹² is the unit vector from charge 2 to charge 1, and \[\frac {1}{4πε_0}\] ≈ 9 × 10⁹ N m²C⁻².

(Right-Triangle Arrangement: 2 µC, 3 µC, 4 µC)

Step 1: Find geometry:

• AB = 4.0 cm, BC = 3.0 cm, and using Pythagoras
• AC = \[\sqrt{(4^2 + 3^2)}\] = 5.0 cm.

Step 2: Force on A due to B (Coulomb's law):

• Plug q_A = 2 μC, q_B = 3 μC, r = 4 × 10⁻² m into F = \[\frac {1}{4πε_0}\frac {q_1q_2}{r^2}\]​​, giving F_AB = 33.7 N, directed along \[\vec {BA}\] (pointing away from B, since both charges are positive → repulsion).

Step 3: Force on A due to C:

• Plug q_A = 2 μC, q_C = 4 μC, r = 5 × 10⁻² m into the same formula, giving F_AC = 28.8 N, directed along \[\vec {CA}\].

Step 4: Combine as vectors:

• Since \[\vec F\] = \[\vec F_{AB}\] + \[\vec F_{AC}\]​, and the two forces act at an angle θ to each other (determined by triangle geometry), apply the law of vector addition: F = \[\sqrt{F_{AB}^2+F_{AC}^2+2F_{AB}F_{AC}\cos\theta}.\]

Step 5: Compute magnitude:

• Substituting the values gives the resultant F = 59.3 N.

Step 6: Compute direction:

• Using tan⁡ α (from the standard vector-addition direction formula), α = 16.9°, so the resultant force points 16.9° north of west.