Coulomb's law and superposition of forces

Two charged pith balls pushed apart, a charged comb bending a thin stream of water, a lightning bolt finding its path to the ground — every one of these is the electrostatic force at work, and Coulomb's law is the equation that pins down exactly how strong that force is. It says the force between two point charges depends only on how much charge each carries and how far apart they sit, falling off as the square of the distance, just like gravity but enormously stronger. The moment a third charge enters the picture, a second idea takes over: the superposition principle, which says every pair of charges pushes or pulls independently of every other pair, and the net effect on any one charge is just the vector sum of all those individual pulls and pushes. Together these two ideas are the entire toolkit of electrostatics — from a two-charge textbook problem to the charge distributed over a charged sphere.

The core insight

The force between two point charges is $F = kq_1q_2/r^2$ along the line joining them, and when several charges act on one charge their forces add as vectors, not as numbers.

The core idea

Coulomb's law. For two point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum or air, the magnitude of the electrostatic force each exerts on the other is

F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = k\frac{q_1 q_2}{r^2}

where: $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ is the permittivity of free space, and $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is the Coulomb constant. The force lies along the straight line joining the two charges — repulsive when $q_1$ and $q_2$ share a sign, attractive when they are opposite — and, by Newton's third law, the charges pull or push on each other with equal and opposite forces. Inside a medium of relative permittivity (dielectric constant) $\varepsilon_r$ , the same law holds with $\varepsilon_0$ replaced by $\varepsilon_0\varepsilon_r$ , so the force is weaker by a factor of $\varepsilon_r$ than in air.

Superposition principle. When more than two charges are present, find the total force on any chosen charge by computing each pairwise Coulomb force as though the other charges were not there, then adding all of these forces as vectors:

\vec F_{\text{net}} = \vec F_1 + \vec F_2 + \vec F_3 + \cdots

Each $\vec F_i$ is worked out independently from Coulomb's law using only the chosen charge and one other charge at a time; only at the end are the vectors combined, using components or the parallelogram rule — never by adding magnitudes as plain numbers unless the forces happen to lie along the same straight line.

Continuous charge. A charge smeared over a body rather than sitting at a point is described by a charge density — linear ( $\lambda = dq/dl$ , C/m), surface ( $\sigma = dq/dA$ , C/m²), or volume ( $\rho = dq/dV$ , C/m³). Conceptually, the body is cut into infinitesimal point charges $dq$ , Coulomb's law is applied to each, and the contributions are summed (integrated); the syllabus asks only that this idea be understood qualitatively, not carried out.

Figure

Two force vectors on a charge at right angles combine by the parallelogram rule into a single diagonal resultant force

Derivation ANet force from two Coulomb forces at an angle

The goal is to derive a single formula for the magnitude of the net electrostatic force on a charge that is being pulled or pushed simultaneously by two other charges whose individual Coulomb forces $F_1$ and $F_2$ meet at some angle $\theta$ between them, so that the same formula automatically covers the collinear case, the perpendicular case, and every angle in between that a superposition problem might present.

Step 1. Compute $F_1$ and $F_2$ separately from Coulomb's law, each treating only one of the two source charges as present, exactly as the superposition principle requires: $F_1 = kq q_1/r_1^2$ and $F_2 = kq q_2/r_2^2$ , where $q$ is the charge experiencing both forces.

Step 2. Represent $\vec F_1$ and $\vec F_2$ as two vectors drawn from the charge $q$ , separated by the angle $\theta$ between their lines of action, and complete the parallelogram they define; the resultant $\vec F_{\text{net}}$ is the diagonal of that parallelogram starting at $q$ .

Step 3. Apply the parallelogram law (equivalently, the law of cosines on the triangle formed by $\vec F_1$ , $\vec F_2$ and $\vec F_{\text{net}}$ ) to get the magnitude of the resultant:

F_{\text{net}} = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta}

Step 4. Check the special cases the syllabus actually tests. When the two charges lie on the same side so the forces point the same way, $\theta = 0^\circ$ and $F_{\text{net}} = F_1 + F_2$ , plain addition. When they oppose, $\theta = 180^\circ$ and $F_{\text{net}} = |F_1 - F_2|$ . When the forces are perpendicular, $\theta = 90^\circ$ , $\cos\theta = 0$ , and the formula collapses to the familiar Pythagorean sum $F_{\text{net}} = \sqrt{F_1^2 + F_2^2}$ — the case used in the worked example below.

Worked example

Charges $q_1 = +2\ \mu\text{C}$ sits at the origin and $q_2 = +3\ \mu\text{C}$ sits at $(0.20\ \text{m}, 0)$ , exactly as before, but now $q_3 = -4\ \mu\text{C}$ is placed at $(0.20\ \text{m}, 0.30\ \text{m})$ , directly above $q_2$ rather than further along the same line. Find the net force on $q_2$ .

Step 1. Force on $q_2$ due to $q_1$ : separation $r_{12} = 0.20\ \text{m}$ , both positive so repulsive, pushing $q_2$ away from $q_1$ along $+x$ :

F_{12} = k\frac{q_1 q_2}{r_{12}^2} = (9\times10^9)\frac{(2\times10^{-6})(3\times10^{-6})}{(0.20)^2} = 1.35\ \text{N along } +x

Step 2. Force on $q_2$ due to $q_3$ : since $q_3$ sits directly above $q_2$ , the separation is $r_{32} = 0.30\ \text{m}$ , and since the charges are opposite in sign the force is attractive, pulling $q_2$ straight up along $+y$ toward $q_3$ :

F_{32} = k\frac{|q_2 q_3|}{r_{32}^2} = (9\times10^9)\frac{(3\times10^{-6})(4\times10^{-6})}{(0.30)^2} = 1.2\ \text{N along } +y

Step 3. Because $F_{12}$ points along $+x$ and $F_{32}$ points along $+y$ , the angle between them is $\theta = 90^\circ$ , so the derivation's perpendicular case applies directly:

F_{\text{net}} = \sqrt{F_{12}^2 + F_{32}^2} = \sqrt{1.35^2 + 1.2^2} = \sqrt{1.8225 + 1.44} = \sqrt{3.2625} \approx 1.81\ \text{N}

Step 4. The direction, measured from the $+x$ axis toward $+y$ , is $\theta = \tan^{-1}(1.2/1.35) \approx 41.6^\circ$ . The net force on $q_2$ is about $1.81\ \text{N}$ , directed $41.6^\circ$ above the line joining $q_1$ and $q_2$ .

Common traps

Adding $F_1$ and $F_2$ as plain numbers when the source charges are not collinear with the target charge — superposition demands vector addition, using components or the parallelogram/law-of-cosines rule.
Forgetting that Coulomb's law with unsigned (magnitude) charge values gives only the size of the force; the direction (toward or away from the other charge) must be fixed separately from the signs of the two charges.
Leaving charge in µC or nC instead of converting to coulombs before substituting into $F = kq_1q_2/r^2$ , which throws the answer off by several orders of magnitude.
Applying the vacuum value of $k$ inside a dielectric medium without dividing by the relative permittivity $\varepsilon_r$ .

Ask the tutor: want to try a three-charge superposition problem where the angle between the two forces is neither 0, 90, nor 180 degrees, so you have to use the full law-of-cosines formula?

Why do we compute each pairwise force separately instead of somehow combining the charges first?

Coulomb's law only describes the interaction between exactly two point charges at a time; there is no version of the formula for three or more charges acting together. The superposition principle is the rule that lets us handle more charges anyway: treat every other charge as if it alone were present, work out that one pairwise force with Coulomb's law, repeat for each remaining charge, and only at the very end add all the resulting force vectors together to get the true net force.

If two of the three charges are collinear with the target charge, do I still need vectors?

You still need to think in terms of direction, but the arithmetic simplifies. If both forces point the same way along that line, add their magnitudes; if they point opposite ways, subtract the smaller from the larger. That is vector addition restricted to one dimension, so it looks like ordinary addition or subtraction — the trouble starts only once a third charge is off that line, forcing you to resolve forces into perpendicular components or use the law of cosines.

Does the distance in Coulomb's law have to be measured along a straight line even for the perpendicular case?

Yes — $r$ in $F = kq_1q_2/r^2$ is always the straight-line (shortest) distance between the two point charges, regardless of how they are arranged relative to any axes you have drawn. In the worked example, $r_{32} = 0.30\ \text{m}$ is the straight distance from $q_3$ to $q_2$ , found directly from the geometry given, not a distance measured along either coordinate direction separately.

Coulomb's law and superposition of forces

The core idea

Worked example

Common traps

Before this

Where this leads