Total Electrostatic Energy of a Charge Triangle with Protons at the Centroid: 04-BS-9, May 2018, Q2

Question 2: Total Electrostatic Energy of a Charge Triangle with Protons at the Centroid (20 marks)

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Given.

Quantity	Value
Triangle side $a$ (equilateral)	$1\times10^{-10}$ m
Charge at each vertex	one electron, $-e$ (three vertices)
Charge at the centroid	three protons together, $+3e$

Find. The total electrostatic potential energy of the four-charge assembly, relative to all four charges infinitely far apart.

Equilateral triangle of electrons (side a) with three protons together at the centroid, at circumradius R=a/√3 from every vertex. Total energy sums all six unique pairwise Coulomb interactions.

Approach. Potential energy is a pairwise sum, $U=\sum_{i<j}\dfrac{q_iq_j}{4\pi\varepsilon_0r_{ij}}$, over every unique pair of the four charges: three vertex–vertex pairs at separation $a$, and three vertex–centroid pairs at separation $R=a/\sqrt3$ (the equilateral triangle's circumradius).

Vertex–vertex contribution (3 pairs). Each pair is $(-e)(-e)=+e^2$ at separation $a$: $$U_{vv}=3\times\frac{e^2}{4\pi\varepsilon_0a}=\frac{3e^2}{4\pi\varepsilon_0a}$$
Vertex–centroid contribution (3 pairs). Each pair is $(-e)(+3e)=-3e^2$ at separation $R=a/\sqrt3$: $$U_{vc}=3\times\frac{-3e^2}{4\pi\varepsilon_0(a/\sqrt3)}=\frac{-9\sqrt3\,e^2}{4\pi\varepsilon_0a}$$
Total energy. Adding the two contributions, with $1/4\pi\varepsilon_0=8.99\times10^9$ N·m$^2$/C$^2$: $$U=U_{vv}+U_{vc}=\frac{e^2}{4\pi\varepsilon_0a}\left(3-9\sqrt3\right)$$ $$U=\frac{(8.99\times10^{9})(1.6\times10^{-19})^2}{1\times10^{-10}}(3-15.5885)$$ $$\boxed{U=-2.897\times10^{-17}\ \text{J}}$$

Quantity	Result
Vertex–vertex energy $U_{vv}$ (3 pairs)	$6.904\times10^{-18}$ J
Vertex–centroid energy $U_{vc}$ (3 pairs)	$-3.588\times10^{-17}$ J
Total electrostatic energy $U$	$-2.897\times10^{-17}$ J

Concept & Theory

Topic. Total electrostatic potential energy of a discrete multi-charge assembly, built by pairwise summation.

Key relations. $U=\sum_{i<j}q_iq_j/(4\pi\varepsilon_0r_{ij})$; equilateral-triangle circumradius $R=a/\sqrt3$.

Why this works. Potential energy is intrinsically a pairwise quantity, so a four-charge system needs all $\binom{4}{2}=6$ unique pair terms — not just "energy of the vertex charges in the centroid's field." The negative result here means the assembly is energetically bound relative to all four charges being infinitely far apart, because the large attractive centroid–vertex terms outweigh the smaller repulsive vertex–vertex terms.

Common pitfall. Omitting the three vertex–vertex terms (treating the problem as "three charges in the centroid's field" only), or double-counting a pair by summing both $i,j$ and $j,i$.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (electric potential energy of point-charge systems).

Question 2 of 8: Total Electrostatic Energy of a Charge Triangle with Protons at the Centroid

Question 2: Total Electrostatic Energy of a Charge Triangle with Protons at the Centroid (20 marks)