The Question Coulomb Cannot Answer
The mutual electric force between two charges is given completely by Coulomb's law. But the moment a third charge enters the picture, a new question appears: how do we calculate the force on a charge when there are not one but several charges around it? NCERT is explicit that Coulomb's law alone is not enough to answer this.
Consider a system of $n$ stationary charges $q_1, q_2, q_3, \ldots, q_n$ in vacuum. What is the force on $q_1$ due to all the others? Mechanics offers a hint: forces of mechanical origin add according to the parallelogram law of vector addition. The decisive question is whether forces of electrostatic origin behave the same way. The answer is not a derivation — it is an experimental fact, elevated to a principle.
Superposition is verified, not derived
That electrostatic forces add vectorially and that each pairwise force is independent of the others are statements established by experiment. They do not follow from Coulomb's law for two charges. NCERT frames the principle as "experimentally, it is verified," so questions phrased as "which of the following is an experimental fact" can list the independence and additivity of Coulomb forces.
All of electrostatics is a consequence of two inputs: Coulomb's law and the superposition principle.
Statement of Superposition
The principle of superposition states that in a system of charges $q_1, q_2, \ldots, q_n$, the force on $q_1$ due to $q_2$ is exactly the Coulomb force given by the two-charge law — it is unaffected by the presence of the other charges $q_3, q_4, \ldots, q_n$. The total force $\mathbf{F}_1$ on $q_1$ is then the vector sum of the individual pairwise forces:
$$\mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} + \cdots + \mathbf{F}_{1n} = \frac{1}{4\pi\varepsilon_0}\sum_{i=2}^{n} \frac{q_1 q_i}{r_{1i}^{2}}\,\hat{\mathbf{r}}_{1i}$$
Here $\mathbf{F}_{12}$ is the force on $q_1$ due to $q_2$ alone, $\mathbf{F}_{13}$ the force due to $q_3$ alone, and so on; $\hat{\mathbf{r}}_{1i}$ is the unit vector pointing from $q_i$ toward $q_1$. The structure of the principle is best read in two clauses, set out below.
| Clause | What it says | What it forbids |
|---|---|---|
| Independence | Each pairwise force $\mathbf{F}_{1i}$ has the value it would have if only $q_1$ and $q_i$ existed. | No "screening" or weakening of one force by a nearby third charge. |
| Additivity | The net force is the vector sum of those independent pairwise forces. | Never the scalar sum of magnitudes (except when all are collinear and aligned). |
Adding Forces: Vectors, Not Scalars
Each Coulomb force points along the line joining the two charges — toward the source for attraction, away from it for repulsion. Because the contributions generally point in different directions, the net force must be assembled by vector addition: either the parallelogram law for two forces, or resolution into perpendicular components for three or more. The magnitude of the resultant is almost always smaller than the arithmetic sum of the individual magnitudes.
Forces add as vectors — magnitudes do not simply sum
A frequent error is to compute three pairwise forces and add the numbers. Two forces of equal magnitude $F$ at $120^\circ$ to each other give a resultant of magnitude $F$, not $2F$; at $90^\circ$ they give $F\sqrt{2}$; only when they are parallel and same-sense do they give $2F$. Always resolve into components or use the parallelogram law before reporting a net force.
Two equal forces $F$ at angle $\theta$: resultant $= 2F\cos(\theta/2)$.
Three charges on a triangle — the pairwise forces on one corner.
For like charges at $q_2$ and $q_3$, both pairwise forces on $q_1$ point away from the base; their vector sum $\mathbf{F}_1$ runs along the perpendicular bisector. The resultant is the diagonal of the parallelogram built on $\mathbf{F}_{12}$ and $\mathbf{F}_{13}$, never their numerical total.
Worked Vector Sum: Charges on a Triangle
NCERT supplies two model problems on equilateral-triangle geometry. Both reward the same instinct: exploit symmetry before reaching for components. The first places a test charge at the centroid; the second asks for the force on each vertex charge.
Consider three charges $q_1, q_2, q_3$ 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 $O$ of the triangle?
In the equilateral triangle $ABC$ of side $l$, drop a perpendicular $AD$ to $BC$. Then $AD = AC\cos 30^\circ = (\sqrt{3}/2)\,l$, and the distance $AO$ of the centroid from $A$ is $(2/3)\,AD = (1/\sqrt{3})\,l$. By symmetry $AO = BO = CO$.
Each charge therefore pushes $Q$ with a force of the same magnitude $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{(l^2/3)} = \dfrac{3}{4\pi\varepsilon_0}\dfrac{Qq}{l^2}$, directed along $AO$, $BO$ and $CO$ respectively.
The resultant of the forces $\mathbf{F}_2$ and $\mathbf{F}_3$ is $\dfrac{3}{4\pi\varepsilon_0}\dfrac{Qq}{l^2}$ along $OA$, by the parallelogram law. The total force on $Q$ is therefore $\dfrac{3}{4\pi\varepsilon_0}\dfrac{Qq}{l^2}\,(\hat{\mathbf{r}} - \hat{\mathbf{r}}) = 0$, where $\hat{\mathbf{r}}$ is the unit vector along $OA$.
It is clear by symmetry that the three forces sum to zero. Suppose instead the resultant were non-zero, pointing in some direction. Consider what would happen if the system were rotated through $60^\circ$ about $O$: the configuration is unchanged, yet the resultant would have turned — a contradiction. Hence it must vanish.
Consider the charges $q$, $q$, and $-q$ placed at the vertices of an equilateral triangle. What is the force on each charge?
The forces acting on charge $q$ at $A$ due to charge $q$ at $B$ and charge $-q$ at $C$ are $F_{12}$ along $BA$ and $F_{13}$ along $AC$ respectively. By the parallelogram law, the total force $\mathbf{F}_1$ on the charge $q$ at $A$ is $\mathbf{F}_1 = F\,\hat{\mathbf{r}}_1$, where $\hat{\mathbf{r}}_1$ is a unit vector along $BC$.
The force of attraction or repulsion for each pair of charges has the same magnitude $F = \dfrac{q^2}{4\pi\varepsilon_0\, l^2}$.
The total force $\mathbf{F}_2$ on charge $q$ at $B$ is thus $\mathbf{F}_2 = F\,\hat{\mathbf{r}}_2$, where $\hat{\mathbf{r}}_2$ is a unit vector along $AC$. Similarly, the total force on charge $-q$ at $C$ is $\mathbf{F}_3 = \sqrt{3}\,F\,\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is the unit vector along the direction bisecting $\angle BCA$.
It is interesting that the sum of the forces on the three charges is zero: $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0$. This follows straight from the fact that Coulomb's law is consistent with Newton's third law — every internal force on the system is matched by its reaction.
Superposition for forces becomes superposition for fields the moment you divide out the test charge. See how in Electric Field and Field Lines.
Equilibrium of a Charge: The Null Point
A charge is in equilibrium where the net force on it vanishes — the vector sum of all the Coulomb pulls and pushes is zero. The classic NEET problem fixes two source charges and asks where a third charge can sit so that it feels no net force. The location is called the null point, and finding it is a pure exercise in superposition.
Null point of a test charge between two like charges on a line.
Two like charges $q_1$ and $q_2$ a distance $d$ apart push a test charge $q$ in opposite directions. Net force is zero only between them, where the magnitudes balance.
Two like charges $q_1$ and $q_2$ are separated by distance $d$. Where on the line joining them does a third charge experience zero net force?
Place the test charge $q$ at distance $x$ from $q_1$, so it is $d-x$ from $q_2$. Between two like charges, the two forces on $q$ point in opposite directions, so equilibrium requires equal magnitudes:
$$\frac{1}{4\pi\varepsilon_0}\frac{q_1 q}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{q_2 q}{(d-x)^2}$$
The factor $q/4\pi\varepsilon_0$ cancels from both sides, leaving $\dfrac{q_1}{x^2} = \dfrac{q_2}{(d-x)^2}$, i.e. $\dfrac{d-x}{x} = \sqrt{\dfrac{q_2}{q_1}}$. Solving,
$$x = \frac{\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}\;d.$$
The null point lies nearer the smaller charge, and it divides the separation in the ratio of the square roots of the charge magnitudes. For two unlike charges, no balance point exists between them; the null point lies outside, on the line, beyond the smaller charge.
The null point ignores the test charge entirely
In the balance equation, the test charge $q$ appears as a common factor on both sides and cancels. The equilibrium position depends only on the two source charges and their separation — not on the magnitude of the test charge, and not on its sign. Changing $q$'s sign flips both forces together, so the balance point is unchanged. Questions that vary the test charge to "shift" the null point are testing exactly this cancellation.
Like sources → null point lies between them. Unlike sources → null point lies outside, beyond the weaker charge.
A Working Method for Any Configuration
Whatever the geometry — line, triangle, square, or scatter — the same procedure resolves every superposition problem. The steps below convert the principle into a reliable routine for the exam hall.
| Step | Action | Why |
|---|---|---|
| 1 | Identify the charge of interest; ignore the rest momentarily. | Superposition treats each source one at a time. |
| 2 | Compute each pairwise Coulomb force magnitude $\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_i|}{r_{1i}^2}$. | Each force is independent of the others. |
| 3 | Fix each direction: away for like charges, toward for unlike. | Sign of the product sets attraction vs repulsion. |
| 4 | Resolve every force into $x$ and $y$ components. | Vector sum, not scalar sum. |
| 5 | Add components separately; recombine magnitude and direction. | $F = \sqrt{F_x^2 + F_y^2}$ gives the net force. |
| 6 | Check symmetry first — it can collapse the work to zero. | Symmetric like-charge layouts often cancel. |
Forces between multiple charges in one screen
- Principle of superposition: the net force on a charge is the vector sum of the pairwise Coulomb forces due to every other charge, taken one at a time.
- Independence: each pairwise force keeps the value Coulomb's law assigns it; the presence of other charges does not alter it.
- Vectors, not scalars: add forces by the parallelogram law or by resolving into components; magnitudes do not simply sum.
- Symmetry: at the centroid of an equilateral triangle of equal charges, the three forces cancel to zero (NCERT Example 1.5).
- Newton's third law: internal forces of a closed charge system sum to zero (NCERT Example 1.6).
- Null point: set opposing magnitudes equal; the position depends only on the source charges and separation, never on the test charge — lying $x = \frac{\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}\,d$ from $q_1$ for like charges.