Forces between Multiple Charges

Coulomb’s law tells you the force between two charges. But real situations rarely involve just two charges. A dust particle near a circuit board, an ion inside a protein, a test charge surrounded by multiple source charges: in each case the charge of interest feels a pull or push from many neighbours at once. How do you find the combined effect?

The answer lies in a powerful and experimentally verified idea called the principle of superposition.

The Principle of Superposition

When you studied mechanics, you learned that forces of mechanical origin (pushes, pulls, tension, friction) add as vectors using the parallelogram law. The natural question is: do electric forces follow the same rule?

Experiments confirm that they do, and the result is called the principle of superposition:

The force on any charge due to several other charges is the vector sum of the individual forces that each of those charges would exert on it if acting alone. Crucially, each individual force follows Coulomb’s law exactly, and it does not change because other charges happen to be nearby.

Two things to notice here:

• Each pairwise force is independent. Placing a third charge $q_3$ next to $q_1$ and $q_2$ does not alter the force between $q_1$ and $q_2$ in the slightest. The third charge simply contributes its own separate force.
• The combination rule is vector addition. You cannot just add magnitudes. Direction matters, and the parallelogram law (or equivalently, component-wise addition) is how the individual forces combine.

All of electrostatics rests on just two foundations: Coulomb’s law and the superposition principle.

Working Out the Net Force: From Two to Three Charges

Consider three charges $q_1$, $q_2$, and $q_3$ positioned at points $\mathbf{r}_1$, $\mathbf{r}_2$, and $\mathbf{r}_3$ in vacuum (see Fig. 1.5(a)).

Fig 1.5: (a) A system of three charges, showing the individual forces on $q_1$. (b) A system of multiple charges, with the resultant force $\mathbf{F}_1$ on $q_1$ obtained by vector addition

Suppose you want the net force on $q_1$. By superposition, you calculate the force from $q_2$ alone, then the force from $q_3$ alone, and add them as vectors.

Step 1: Force on $q_1$ from $q_2$

Apply Coulomb’s law to just the pair ($q_1$, $q_2$), treating $q_3$ as if it did not exist:

$\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{\mathbf{r}}_{12}$

Here $r_{12}$ is the distance between $q_1$ and $q_2$, and $\hat{\mathbf{r}}_{12}$ is the unit vector pointing from $q_2$ toward $q_1$.

Step 2: Force on $q_1$ from $q_3$

Now apply Coulomb’s law to just the pair ($q_1$, $q_3$), treating $q_2$ as if it did not exist:

$\mathbf{F}_{13} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_3}{r_{13}^2} \hat{\mathbf{r}}_{13}$

Notice that this expression is exactly what Coulomb’s law gives for $q_1$ and $q_3$ in isolation. The presence of $q_2$ has no effect on $\mathbf{F}_{13}$.

Step 3: Add as vectors

The total force on $q_1$ is the vector sum:

$\mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{\mathbf{r}}_{12} + \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_3}{r_{13}^2} \hat{\mathbf{r}}_{13} \qquad \text{(1.4)}$

You carry out this addition using the parallelogram law (or by breaking each force into $x$ and $y$ components, summing components, and recombining).

The General Formula for $n$ Charges

The three-charge idea extends naturally to any number of charges. Consider $n$ stationary charges $q_1, q_2, q_3, \ldots, q_n$ arranged in vacuum (Fig. 1.5(b)).

By superposition, the force on $q_1$ from each of the other $n-1$ charges is given by Coulomb’s law for that pair alone. The net force on $q_1$ is the vector sum of all these individual contributions:

$\mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} + \ldots + \mathbf{F}_{1n}$

Writing each term out using Coulomb’s law and factoring out $q_1$:

$\mathbf{F}_1 = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1 q_2}{r_{12}^2} \hat{\mathbf{r}}_{12} + \frac{q_1 q_3}{r_{13}^2} \hat{\mathbf{r}}_{13} + \ldots + \frac{q_1 q_n}{r_{1n}^2} \hat{\mathbf{r}}_{1n} \right]$

In compact summation notation:

$\mathbf{F}_1 = \frac{q_1}{4\pi\varepsilon_0} \sum_{i=2}^{n} \frac{q_i}{r_{1i}^2} \hat{\mathbf{r}}_{1i} \qquad \text{(1.5)}$

This single formula is all you need. Given the magnitudes and positions of every charge, you can compute the force on any chosen charge by plugging into Equation (1.5) and carrying out the vector sum.

Solved Example 1.5: Equal Charges at the Corners of an Equilateral Triangle

Problem: Three charges, each equal to $q$, are placed at the three vertices A, B, C of an equilateral triangle with side length $l$. A charge $Q$ (same sign as $q$) is placed at the centroid O of the triangle. Find the net force on $Q$.

Fig 1.6: Three equal charges $q$ at the vertices of an equilateral triangle, with charge $Q$ at the centroid O

Setting Up: Finding the Vertex-to-Centroid Distance

Before computing forces, you need the distance from each vertex to the centroid.

Draw a perpendicular AD from vertex A to side BC. In an equilateral triangle, this perpendicular bisects BC, so D is the midpoint of BC.

The length of AD (the altitude) is:

$AD = AC \cos 30° = l \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\,l$

The centroid O lies along each altitude at a distance of $\frac{2}{3}$ of the altitude from the vertex. So:

$AO = \frac{2}{3} \times AD = \frac{2}{3} \times \frac{\sqrt{3}}{2}\,l = \frac{l}{\sqrt{3}}$

By the symmetry of the equilateral triangle, the centroid is equidistant from all three vertices:

$AO = BO = CO = \frac{l}{\sqrt{3}}$

Computing Individual Forces

Since $r = \frac{l}{\sqrt{3}}$, we get $r^2 = \frac{l^2}{3}$.

The force on $Q$ due to the charge $q$ at vertex A is:

$|\mathbf{F}_1| = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{l^2/3} = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}$

This force acts along AO (directed away from A toward O and beyond, since $Q$ and $q$ are of the same sign and repel).

By symmetry, the forces from the charges at B and C have exactly the same magnitude:

$|\mathbf{F}_2| = |\mathbf{F}_3| = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}$

$\mathbf{F}_2$ acts along BO and $\mathbf{F}_3$ acts along CO.

Finding the Resultant

Now add the three force vectors. First combine $\mathbf{F}_2$ and $\mathbf{F}_3$. These two forces are equal in magnitude and their directions point outward from the centroid along BO and CO respectively. By the parallelogram law, their resultant points along OA (the direction opposite to $\mathbf{F}_1$) and has magnitude:

$|\mathbf{F}_2 + \mathbf{F}_3| = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}$

(You can verify this by resolving $\mathbf{F}_2$ and $\mathbf{F}_3$ into components: their components perpendicular to OA cancel, while their components along OA add up to exactly one unit of force magnitude, directed along OA.)

This resultant is equal in magnitude to $\mathbf{F}_1$ but points in the opposite direction (along OA, while $\mathbf{F}_1$ points along AO extended). Therefore:

$\mathbf{F}_1 + (\mathbf{F}_2 + \mathbf{F}_3) = \mathbf{0}$

The net force on $Q$ at the centroid is zero.

The Symmetry Shortcut

You can see this result more quickly through symmetry. The system has three-fold rotational symmetry: rotating the entire arrangement by 120° about the centroid maps every charge onto the position of another identical charge. The physical situation is unchanged. If there were a net force pointing in some direction, a 120° rotation would swing it to a different direction, but the physics has not changed, so the force cannot change either. The only vector that remains unchanged under a 120° rotation is the zero vector. Hence the net force must be zero.

Solved Example 1.6: Two Like Charges and One Unlike Charge

Problem: Charges $q$, $q$, and $-q$ are placed at vertices A, B, and C of an equilateral triangle with side $l$. Find the net force on each charge.

Fig 1.7: Charges $q$, $q$, $-q$ at the vertices of an equilateral triangle, with individual and resultant forces on each charge

Key Observation: All Pair Forces Have the Same Magnitude

Every pair of charges is separated by the same distance $l$, and each charge has magnitude $q$. Whether a particular pair attracts or repels, the magnitude of the Coulomb force between them is the same:

$F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{l^2}$

The sign of the charges determines only the direction (toward each other for unlike charges, away for like charges).

Force on the Charge $q$ at A

The charge at A feels two forces:

• $\mathbf{F}_{12}$ (from $q$ at B): Both charges are positive, so the force is repulsive. It points along the line from B to A, that is, along BA. Magnitude: $F$.
• $\mathbf{F}_{13}$ (from $-q$ at C): One positive, one negative, so the force is attractive. It pulls $q_1$ toward C, pointing along AC. Magnitude: $F$.

To find the angle between these two force vectors, think carefully about the directions. The interior angle of the equilateral triangle at A is $60°$, but the two force vectors are not simply along the two sides meeting at A. One force ($\mathbf{F}_{12}$) points away from B (along BA), while the other ($\mathbf{F}_{13}$) points toward C (along AC). When you place both vectors tail-to-tail at A, the angle between them is $180° - 60° = 120°$.

Applying the parallelogram law for two equal vectors of magnitude $F$ with an included angle of $120°$:

$|\mathbf{F}_1| = \sqrt{F^2 + F^2 + 2F^2 \cos 120°} = \sqrt{2F^2 + 2F^2\!\left(-\tfrac{1}{2}\right)} = \sqrt{F^2} = F$

The resultant $\mathbf{F}_1$ has magnitude $F$ and points along $\hat{\mathbf{r}}_1$, a unit vector parallel to BC (from B toward C). You can verify this by resolving the two forces into components: the components perpendicular to BC cancel, while the components along BC reinforce.

Force on the Charge $q$ at B

By the same reasoning (the arrangement at B is a mirror image of the arrangement at A):

$|\mathbf{F}_2| = F$

The direction of $\mathbf{F}_2$ is along $\hat{\mathbf{r}}_2$, a unit vector parallel to AC.

Force on the Charge $-q$ at C

The charge $-q$ at C is attracted toward both positive charges at A and B. Both attractive forces have magnitude $F$. The angle between the directions CA and CB at vertex C is $60°$ (the interior angle of the equilateral triangle at C).

Using the parallelogram law with two equal forces $F$ at an angle of $60°$:

$|\mathbf{F}_3| = \sqrt{F^2 + F^2 + 2F^2 \cos 60°} = \sqrt{2F^2 + F^2} = \sqrt{3}\,F$

The resultant points along $\hat{\mathbf{n}}$, the unit vector along the angle bisector of $\angle BCA$, directed from C toward the midpoint of AB.

A Satisfying Check: Newton’s Third Law

Add up the net forces on all three charges:

$\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$

This is not a coincidence. Coulomb’s law obeys Newton’s third law: the force on $q_1$ due to $q_2$ is exactly opposite to the force on $q_2$ due to $q_1$ ($\mathbf{F}_{12} = -\mathbf{F}_{21}$). When you sum all forces on all charges, every such pair cancels, and the total is always zero for any isolated system of charges, regardless of the arrangement or the signs of the charges.