Electrostatics: Diagram-Based Questions (5)

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Tags Electrostatics

Question

Three point charges +q+q, +q+q, and −q-q sit at the corners of an equilateral triangle of side aa. Find the magnitude of the net force on the −q-q charge.

Solution — Step by Step

Each +q+q attracts the −q-q charge with magnitude:

F0=14πϵ0q2a2=kq2a2F_0 = \frac{1}{4\pi\epsilon_0} \frac{q^2}{a^2} = \frac{kq^2}{a^2}

The two forces act along the two sides of the triangle meeting at −q-q, both pointing toward the respective +q+q.

The angle between the two sides of an equilateral triangle meeting at one vertex is 60°60°. So the two attractive forces have 60°60° between them.

Fnet=F02+F02+2F02cos⁡60°F_{\text{net}} = \sqrt{F_0^2 + F_0^2 + 2 F_0^2 \cos 60°}

Fnet=2F02+F02=F03F_{\text{net}} = \sqrt{2 F_0^2 + F_0^2} = F_0 \sqrt{3}

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

Final answer: Fnet=3kq2a2F_{\text{net}} = \frac{\sqrt{3} k q^2}{a^2}.

Why This Works

Coulomb’s law gives the magnitude and direction of the force between any pair. For multiple charges, the net force is the vector sum — superposition principle. The cosine rule (or parallelogram law) handles the angle between non-collinear forces.

Alternative Method

Resolve along the perpendicular bisector of the side joining the two +q+q charges. Components along that bisector add (2F0cos⁡30°2 F_0 \cos 30°), and components perpendicular cancel by symmetry. Either way, you get F03F_0 \sqrt{3}.

For three charges at vertices of a regular polygon, the force on the middle charge often has elegant symmetry. Always look for cancellations before brute-forcing components.

Common Mistake

Students sometimes use cos⁡120°\cos 120° instead of cos⁡60°\cos 60°, confusing the angle between sides with the exterior angle. Draw the diagram carefully: at the vertex where −q-q sits, the two sides going to +q+q charges make a 60°60° angle, not 120°120°.

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