Electrostatics: Tricky Questions Solved (11)

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Question

Three identical point charges qq are placed at the vertices of an equilateral triangle of side aa. Find the electric potential energy of the system. Then, a fourth charge −q-q is brought from infinity and placed at the centroid. Find the change in the system’s energy.

Solution — Step by Step

For nn charges, the total potential energy is the sum over all pairs:

U=14πε0∑i<jqiqjrijU = \tfrac{1}{4\pi\varepsilon_0}\sum_{i<j} \tfrac{q_iq_j}{r_{ij}}

With three identical charges and three pairs, each at separation aa:

U3=3×14πε0q2a=3kq2aU_3 = 3 \times \tfrac{1}{4\pi\varepsilon_0} \tfrac{q^2}{a} = \tfrac{3kq^2}{a}

where k=1/(4πε0)k = 1/(4\pi\varepsilon_0).

For an equilateral triangle of side aa, the circumradius is r=a3r = \tfrac{a}{\sqrt{3}}.

 Vcentroid=3×kqr=3kqa/3=33 kqaV_{\text{centroid}} = 3 \times \tfrac{kq}{r} = \tfrac{3kq}{a/\sqrt{3}} = \tfrac{3\sqrt{3}\,kq}{a} Wext=(−q)×Vcentroid=−33 kq2aW_{\text{ext}} = (-q) \times V_{\text{centroid}} = -\tfrac{3\sqrt{3}\,kq^2}{a}

This equals the change in potential energy of the system.

 U4=U3+ΔU=3kq2a−33 kq2a=3kq2a(1−3)U_4 = U_3 + \Delta U = \tfrac{3kq^2}{a} - \tfrac{3\sqrt{3}\,kq^2}{a} = \tfrac{3kq^2}{a}(1 - \sqrt{3})

Since 3≈1.732>1\sqrt{3} \approx 1.732 > 1, U4<0U_4 < 0 — the four-charge system is bound.

Final answer: U3=3kq2aU_3 = \tfrac{3kq^2}{a}, ΔU=−33 kq2a\Delta U = -\tfrac{3\sqrt{3}\,kq^2}{a}, U4=3kq2a(1−3)U_4 = \mathbf{\tfrac{3kq^2}{a}(1 - \sqrt{3})}.

Why This Works

The potential energy of a charge distribution is path-independent because Coulomb force is conservative. Adding a new charge changes the energy by qVqV, where VV is the potential at that location due to all the other charges.

Notice we only included pair (i,j)(i,j) once. Forgetting this and double-counting is the most common arithmetic mistake.

Alternative Method

We can compute U4U_4 directly from all six pairs: three vertex pairs at distance aa giving +3kq2a+\tfrac{3kq^2}{a}, plus three pairs (each vertex with centroid) at distance a/3a/\sqrt{3} giving 3×k(q)(−q)a/3=−33kq2a3 \times \tfrac{k(q)(-q)}{a/\sqrt{3}} = -\tfrac{3\sqrt{3}kq^2}{a}. Same answer.

Common Mistake

Using the side length aa as the centroid-to-vertex distance. The correct distance is a/3a/\sqrt{3}, not aa or a/2a/\sqrt{2}. For an equilateral triangle, draw a quick figure and use r=side/3r = \text{side}/\sqrt{3} — derive it by dropping a perpendicular if you forget.

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