Electrostatics: Speed-Solving Techniques (4)

easy • 2 min read
Tags Electrostatics

Question

Three identical point charges, each of magnitude +q+q, are placed at the vertices of an equilateral triangle of side aa. Find the magnitude of the net electrostatic force on any one charge. Solve in under 30 seconds using symmetry — no vector decomposition required.

Solution — Step by Step

Each pair of charges exerts a force of magnitude:

F=kq2a2F = \frac{kq^2}{a^2}

So the chosen charge feels two forces, each of magnitude FF, directed along the two sides meeting at that vertex.

The angle between the two sides of an equilateral triangle is 60°60°. Net force magnitude using the parallelogram law:

Fnet=F2+F2+2F2cos⁡60°=3F2=F3F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2\cos 60°} = \sqrt{3F^2} = F\sqrt{3}

Fnet=3⋅kq2a2F_{\text{net}} = \sqrt{3} \cdot \frac{kq^2}{a^2}

The net force on each charge is 3 kq2/a2\sqrt{3}\, k q^2 / a^2, directed outward along the perpendicular bisector from the opposite side.

Why This Works

When two equal forces meet at angle θ\theta, the resultant magnitude is 2Fcos⁡(θ/2)2F\cos(\theta/2). For equilateral triangle vertices, θ=60°\theta = 60°, so 2Fcos⁡30°=2F⋅32=F32F\cos 30° = 2F \cdot \frac{\sqrt{3}}{2} = F\sqrt{3}.

This shortcut saves the vector decomposition step entirely.

For nn identical charges at vertices of a regular polygon, the net force on each is F3F\sqrt{3} (triangle), F2F\sqrt{2} along the diagonal (square — but you must include the diagonal pair), and so on. Sketch out the geometry, count contributions, use symmetry.

Alternative Method

Vector method: place one charge at origin and use coordinates. Force from (a,0)(a, 0) has components (−F,0)(-F, 0). Force from (a/2,a3/2)(a/2, a\sqrt{3}/2) has components (−F/2,−F3/2)(-F/2, -F\sqrt{3}/2). Sum: (−3F/2,−F3/2)(-3F/2, -F\sqrt{3}/2). Magnitude: 9F2/4+3F2/4=3F2=F3\sqrt{9F^2/4 + 3F^2/4} = \sqrt{3F^2} = F\sqrt{3}. Same answer, longer.

Common Mistake

Students sometimes write Fnet=2Fcos⁡60°=FF_{\text{net}} = 2F\cos 60° = F, forgetting that the parallelogram law uses cos⁡(θ/2)\cos(\theta/2) in the bisector direction, not cos⁡θ\cos\theta. The correct shortcut: Fnet=2Fcos⁡30°=F3F_{\text{net}} = 2F\cos 30° = F\sqrt{3}.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Capacitor combinations — series and parallel with charge and voltage distribution

medium

Capacitors in Series and Parallel — Equivalent Capacitance

easy

Coulomb's Law vs Gravitational Force — Why Electrostatic Is Stronger

easy

Derive expression for electric field due to infinite charged plane using Gauss's law

medium

Electric Dipole — Field on Axial and Equatorial Points

medium