Problem — Field at the centre of an equilateral triangle

Problem

Three point charges $q = +1{,}0\ \mu\text{C}$ are placed at the vertices of an equilateral triangle of side $\ell = 0{,}10\ \text{m}$. Calculate the electric field at the centre of the triangle. How does the result change if one of the three charges is $-q$ instead of $+q$?

Solution

Centre–vertex distance. In the equilateral triangle the centre (centroid) is a distance $R = \ell/\sqrt{3}$ from each vertex:

$R = \frac{\ell}{\sqrt{3}} = \frac{0{,}10}{1{,}732} = 0{,}0577\ \text{m}$

Field of a single charge at the centre. With $k = 8{,}99\cdot 10^9$:

$E_1 = k\,\frac{q}{R^2} = 8{,}99\cdot 10^9 \cdot \frac{1{,}0\cdot 10^{-6}}{(0{,}0577)^2} = \frac{8{,}99\cdot 10^3}{3{,}33\cdot 10^{-3}} = 2{,}70\cdot 10^6\ \text{N/C}$

Three charges $+q$ (symmetric case). The three field vectors have equal magnitude $E_1$ and point from the centre outwards (positive charges) at $120°$ from one another. Their vector sum is zero by symmetry:

$\vv{E}_{\text{centre}} = \vv{E}_1 + \vv{E}_2 + \vv{E}_3 = \vv 0$

Replacing one charge with $-q$. Changing one vertex from $+q$ to $-q$ is equivalent to starting from the symmetric configuration (zero field) and adding a charge $-2q$ at that vertex. The residual field is therefore just that of a charge $-2q$ placed at that vertex:

$E = k\,\frac{2q}{R^2} = 2\,E_1 = 2\cdot 2{,}70\cdot 10^6 = 5{,}4\cdot 10^6\ \text{N/C}$

Direction: since the charge is negative, the field at the centre points towards the vertex occupied by $-q$.

$\ev{E_{\text{centre}}\approx 0;\ \text{with }-q:\ E\approx 5{,}4\cdot 10^6\ \text{N/C towards }-q}$

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Topics: Electric field and potential Concepts: Electric field Skills: Superposition principle