Forces between Multiple Charges

• The mutual electric force between two charges is given by Coulomb's law.
• When there are not one but several charges around a given charge, Coulomb's law alone is not enough to calculate the net force.
• Consider a system of n stationary charges q₁, q₂, q₃,…, q_n in vacuum — the question is: what is the force on q₁ due to q₂, q₃,…, q_n?
• Forces of mechanical origin add according to the parallelogram law of addition, and the same is found to be true for forces of electrostatic origin.

• It is experimentally verified that the force on any charge due to a number of other charges is the vector sum of all the forces on that charge, taken one at a time.
• The individual forces are unaffected by the presence of other charges.
• This is termed the Principle of Superposition.

Mathematical Formulation:

• Consider a system of three charges q₁, q₂, and q₃.
• The force on q₁ due to q₂ and q₃ is obtained by performing a vector addition of the forces due to each one of these charges.
• The force on q₁ due to q₂, denoted F₁₂, is given by Coulomb's law even though other charges are present:
  F₁₂ = \[\frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1q_2}{r_{12}^2}\hat{\mathbf{r}}_{12}\]
• The force on q₁ due to q₃, denoted F₁₃, is again the Coulomb force even though q₂ is present:
  F₁₃ = \[\frac {1}{4πε_0}\] ⋅ \[\frac{q_1q_3}{r_{13}^2}\hat{\mathbf{r}}_{13}\]
• The total force F₁ on q₁ due to both q₂ and q₃ is (Equation 1.4):
  F₁ = F₁₂ + F₁₃ = \[\frac {1}{4πε_0}\]\[\begin{bmatrix} \frac{q_1q_2}{r_{12}^2}\hat{\mathbf{r}}_{12}+\frac{q_1q_3}{r_{13}^2}\hat{\mathbf{r}}_{13} \end{bmatrix}\]
• For a system of n charges, the force on q₁ due to all other charges is (Equation 1.5):
  F₁ = \[\frac{q_{1}}{4\pi\varepsilon_{0}}\sum_{i=2}^{n}\frac{q_{i}}{r_{1i}^{2}}\hat{\mathbf{r}}_{1i}\]
• The vector sum is obtained by the parallelogram law of addition of vectors.
• All of electrostatics is basically a consequence of Coulomb's law and the superposition principle.

Given: Three charges q₁ = q₂ = q₃ = q at vertices A, B, C of an equilateral triangle of side l. A charge Q (of the same sign as q) is placed at the centroid O.

• The perpendicular AD to side BC has length AD = AC cos ⁡30° = \[\frac {\sqrt 3}{2}\]l.
• The distance of centroid O from vertex A is AO = \[\frac {2}{3}\]AD = \[\frac {1}{\sqrt 3}\]l.
• By symmetry, AO = BO = CO.
• Force F₁ on Q due to charge q at A = \[\frac{3}{4\pi\varepsilon_0}\frac{Qq}{l^2}\] along AO.
• Force F₂ on Q due to charge q at B = \[\frac{3}{4\pi\varepsilon_0}\frac{Qq}{l^2}\] along BO.
• Force F₃ on Q due to charge q at C = \[\frac{3}{4\pi\varepsilon_0}\frac{Qq}{l^2}\] along CO.
• The resultant of F₂ and F₃ is \[\frac{3}{4\pi\varepsilon_0}\frac{Qq}{l^2}\] along OA, by the parallelogram law.
• Therefore, the total force on Q = 0, since F₁ and the resultant of F₂ + F₃ are equal and opposite.
• It is also clear by symmetry that the three forces will sum to zero — if the resultant were non-zero in some direction, rotating the system through 60° about O would produce a contradiction.

Given: Charges +q at A, +q at B, and −q at C at the vertices of an equilateral triangle of side l.

• The force of attraction or repulsion for each pair of charges has the same magnitude:
  F = \[\frac{q^2}{4\pi\varepsilon_0l^2}\]
• The forces on +q at A due to +q at B and −q at C are F₁₂ along BA and F₁₃ along AC, respectively.
• By the parallelogram law, the total force on q at A is F₁ = F\[\hat r_1\], where \[\hat r_1\] is a unit vector along BC.
• The total force on q at B is F₂ = F\[\hat r_2\], where \[\hat r_2\] is a unit vector along AC.
• The total force on −q at C is F₃ = \[\sqrt 3\] F \[\hat n\], where \[\hat n\] is the unit vector along the direction bisecting ∠BC A.
• The sum of all forces on the three charges is zero: F₁ + F₂ + F₃ = 0
• This result follows directly from the fact that Coulomb's law is consistent with Newton's Third Law.