The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around?

Consider a system of n stationary charges q1, q2, q3, ..., qn in vacuum. What is the force on q1 due to q2, q3, ..., qn? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin?

Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are not affected due to the presence of other charges. This is termed as the principle of superposition.

To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in Fig. 1.8(a).

FIGURE 1.8 A system of (a) three charges

“Image description by Dr. T. K. Bansal begins:
1. The figure shows 3 charges q1, q2, & q3, located at position vectors r1 vector, r2 vector, & r3 vector.
2. The direction of the line joining from q2 to q1 is represented as r12 cap and its length by r12.
3. The direction of the line joining q3 to q1 is represented by r13 cap and its length by r13.
4. Force acting on q1 due to q2 is represented by F12 vector and is directed outwards due to repulsion, likewise, the force on q1 due to charge q3 is represented by F13 and is directed outwards due to the mutual repulsion between the two charges.
5. The total force F1 acting on q1 due to both the charges q2 & q3 is given by the vector sum of the forces F12 vector and F13 vector.
End of description.”

The force on one charge, say q1, due to two other charges q2 and q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by F12 vector, F12 vector is given by Equation (1.3) even though other charges are present.

Thus, F12 vector = 1/(4πε0) × (q1q2)/r12^2 × r12 cap

In the same way, the force on q1 due to q3, denoted by F13 vector, is given by
F13 vector = 1/(4πε0) × (q1q3)/r13^2 × r13 cap
which again is the Coulomb force on q1 due to q3, even though other charge q2 is present.

Thus the total force F1 vector on q1 due to the two charges q2 and q3 is given as
F1 vector = F12 vector + F13 vector
= 1/(4πε0) × (q1q2)/r12^2 × r12 cap + 1/(4πε0) × (q1q3)/r13^2 × r13 cap …. (1.4)

The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.8(b).

FIGURE 1.8 (b) multiple charges.

The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The total force F1 vector on the charge q1, due to all other charges, is then given by the vector sum of the forces F12 vector, F13 vector, ..., F1n vector:
i.e., F1 vector = F12 vector + F13 vector + …… F1n vector
= 1/(4πε0) × [(q1q2)/r12^2 × r12 cap + (q1q3)/r13^2 × r13 cap + …… + (q1qn)/r1n^2 × r1n cap]
= q1/4πε0 × Σ i from 2 to n (q i)/r1i^2 × r1i cap (1.5)

The vector sum is obtained as usual, by the parallelogram law of vector addition.
All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle.

Example 1. 6

Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.9?

Fig. 1.9

Solution:
In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC,
AD = AC cos 30° = (√3 /2) l
and the distance AO of the centroid O from A is (2/3) AD = (1/√3) l.
By symmetry AO = BO = CO.

Thus,
Force F1 vector on Q due to charge q at A = 3/4πε0 × Qq/l^2 along AO
Force F2 vector on Q due to charge q at B = 3/4πε0 × Qq/l^2 along BO
Force F3 vector on Q due to charge q at C = 3/4πε0 × Qq/l^2 along CO

The resultant of forces F2 vector and F3 vector is 3/4πε0 × Qq/l^2 along OA, by the parallelogram law. Therefore, the total force on Q = 3/4πε0 × Qq/l^2 (r cap − r cap) = 0, where r cap is the unit vector along OA.

It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but is in some direction. Consider what would happen if the system was rotated through 60° about O.

Example 1.7

Consider the charges q, q, and −q placed at the vertices of an equilateral triangle, as shown in Fig. 1.10. What is the force on each charge?

Fig. 1.10

Solution:
The forces acting on charge q at A due to charges q at B and −q at C are F12 vector along BA and F13 vector along AC respectively, as shown in Fig. 1.10. By the parallelogram law, the total force F1 vector on the charge q at A is given by
F1 vector = F r1 cap where r1 cap is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the same magnitude F = q^2/4πε0l^2
The total force F2 vector on charge q at B is thus F2 vector = F r2 cap, where r2 cap is a unit vector along AC.
Similarly the total force on charge −q at C is F3 vector = √3 F n cap , where n cap is the unit vector along the direction bisecting the ∠BCA.
It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
F1 vector + F2 vector + F3 vector = 0
The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof is left to you as an exercise.