Electric Potential Above a Charged Equilateral Triangle: 04-BS-9, December 2016, Q2

Question 2: Electric Potential Above a Charged Equilateral Triangle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Centre charge	$+3e$
Vertex charges (3, one per corner)	$-e$ each
Triangle side $a$	$1$ Å $=10^{-10}$ m
Field point height $h$ above centroid	$1$ Å $=10^{-10}$ m

Find. The electric potential $V$ (relative to infinity) at the point $1$ Å above the centroid.

Equilateral triangle (side $a$) with $-e$ at each vertex and $+3e$ at the centroid; $P$ sits height $h$ directly above the centroid.

Approach. Potential is scalar, so sum $kq/d$ over all four charges: the centre charge is a distance $h$ from $P$, and each vertex charge is a distance $\sqrt{R_v^2+h^2}$ from $P$, where $R_v=a/\sqrt3$ is the centroid-to-vertex distance.

Centroid-to-vertex distance. $$R_v=\frac{a}{\sqrt3}=\frac{10^{-10}}{1.7321}=\boxed{5.774\times10^{-11}\ \text{m}}$$
Distance from $P$ to each vertex. $$d=\sqrt{R_v^2+h^2}=\sqrt{(5.774\times10^{-11})^2+(10^{-10})^2}$$ $$d=\boxed{1.1547\times10^{-10}\ \text{m}}$$
Superpose the four point-charge potentials (scalar sum). Centre charge at distance $h$, three identical vertex charges at distance $d$: $$V=\frac{k(3e)}{h}+3\cdot\frac{k(-e)}{d}=\frac{(8.992\times10^9)(3)(1.6\times10^{-19})}{10^{-10}}-\frac{3(8.992\times10^9)(1.6\times10^{-19})}{1.1547\times10^{-10}}$$ $$V=\boxed{5.782\ \text{V}}$$

Quantity	Result
Centroid-to-vertex distance $R_v$	$5.774\times10^{-11}$ m
$P$-to-vertex distance $d$	$1.1547\times10^{-10}$ m
Potential $V$ at $P$	$5.782$ V

Concept & Theory

Topic. Electric potential (a scalar) as the superposition of point-charge potentials $V=kq/d$ over a discrete charge arrangement.

Key relations. $V=\sum_i kq_i/d_i$; equilateral-triangle centroid-to-vertex distance $R_v=a/\sqrt3$ (from the side length $a$ via the $30$-$60$-$90$ half-triangle).

Why this works. Because potential is scalar, no vector decomposition is needed — only each charge's distance to $P$ matters, unlike the field calculation of Question 1/7 which needs direction. The three vertex charges are equidistant from $P$ (by the triangle's symmetry about its centroid's vertical axis), so their contributions combine as a single term times 3.

Common pitfall. Using the SIDE length $a$ in place of the centroid-to-vertex distance $R_v=a/\sqrt3$ when computing $d$ — the vertex is closer to the centroid than a full side length, a common geometry slip.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (electric potential, superposition of point charges).

Question 2 of 8: Electric Potential Above a Charged Equilateral Triangle

Question 2: Electric Potential Above a Charged Equilateral Triangle (20 marks)