Electric Field on an Electron in a Triangle of Electrons with Protons at the Centre: 04-BS-9, December 2017, Q1

Question 1: Electric Field on an Electron in a Triangle of Electrons with Protons at the Centre (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
Triangle side $a$ (equilateral)	$5\times10^{-11}$ m
Charge at each vertex	one electron, $-e$
Charge at the centroid	three protons, together $+3e$

Find. The magnitude and direction of $\vec E$ at one vertex (the location of one electron), due to the other five charges.

Equilateral triangle of electrons (vertices), three protons together at the centroid. Field at vertex A is the vector sum of the centroid's repulsion and the attraction toward the other two electrons.

Approach. Superpose the field at one vertex from: (1) the three centroid protons, treated as a single point charge $+3e$ at distance equal to the circumradius $R=a/\sqrt3$, directed radially outward along the centroid–vertex line; and (2) the two OTHER electrons, each $-e$ at distance $a$, each field pointing TOWARD that electron (attraction to a negative charge). By the triangle's symmetry the two off-axis components cancel sideways and add along the centroid–vertex line.

Field from the centroid charge $+3e$. Circumradius $R=a/\sqrt3$, so $R^2=a^2/3$: $$E_{\text{center}}=\frac{k(3e)}{R^2}=\frac{k(3e)}{a^2/3}=\frac{9ke}{a^2}$$ directed radially OUTWARD along the centroid–vertex line (away from the protons).
Field from the other two electrons. Each is at distance $a$ and pulls the field vector toward itself; resolving both onto the centroid–vertex line (the sideways components cancel by symmetry), each contributes a component $\tfrac{\sqrt3}{2}\cdot\tfrac{ke}{a^2}$ back TOWARD the centroid, so the two together give $$E_{\text{others}}=2\times\frac{\sqrt3}{2}\cdot\frac{ke}{a^2}=\frac{\sqrt3\,ke}{a^2}$$ directed back toward the centroid (opposing $E_{\text{center}}$).
Net field. Subtracting the inward contribution from the outward one: $$E=E_{\text{center}}-E_{\text{others}}=\frac{ke}{a^2}(9-\sqrt3)$$ $$E=\frac{(8.9918\times10^9)(1.6\times10^{-19})}{(5\times10^{-11})^2}(9-1.7321)$$ $$\boxed{E=4.183\times10^{12}\ \text{V/m}}$$ directed radially OUTWARD — along the extension of the centroid–vertex line, i.e. away from the three protons.

Quantity	Result
Field from centroid ($+3e$)	$5.179\times10^{12}$ V/m (outward)
Net field from other two electrons	$0.997\times10^{12}$ V/m (inward)
Net field at the electron	$4.183\times10^{12}$ V/m, directed radially outward (away from the centroid)

Concept & Theory

Topic. Superposition of Coulomb fields from discrete point charges arranged with the symmetry of an equilateral triangle.

Key relations. $E=kq/r^2$ per point charge; equilateral-triangle circumradius $R=a/\sqrt3$; vector superposition resolved along the axis of symmetry.

Why this works. The centroid–vertex line is an axis of symmetry for the remaining two electrons, so their sideways (tangential) components must cancel identically, leaving a purely radial net field that is simply the algebraic sum of the (outward) centroid repulsion and the (inward) attraction from the two flanking electrons.

Common pitfall. Using the side length $a$ instead of the circumradius $R=a/\sqrt3$ for the centroid distance, or forgetting the $\cos30^\circ=\sqrt3/2$ projection factor when resolving the two flanking-electron fields onto the symmetry axis.

Source. Sadiku, Elements of Electromagnetics, Ch. 2 (Coulomb's law, superposition of point charges).

Question 1 of 8: Electric Field on an Electron in a Triangle of Electrons with Protons at the Centre

Question 1: Electric Field on an Electron in a Triangle of Electrons with Protons at the Centre (20 marks)