Q.Consider three charges $q_1$, $q_2$, $q_3$ each equal to $q$ at the vertices of an equilateral triangle of side $l$. What is the force on a charge $Q$ (with the same sign as $q$) placed at the centroid of the triangle?

Concept: Coulomb Force Superposition — the net force on $Q$ is the vector sum of three individual repulsive forces from $q_1, q_2, q_3$.

1. Each side of the equilateral triangle is $l$. The distance from a vertex to the centroid is $\frac{l}{\sqrt{3}}$. So each repulsive force has magnitude:

2. The three forces lie along the medians, pointing away from the vertices. At the centroid, the medians are separated by $120^\circ$.

3. Three vectors of equal magnitude, spaced $120^\circ$ apart, sum to zero. This is true regardless of the sign of $Q$ and $q$ (as long as they are the same sign, all forces are either all repulsive or all attractive).

The key idea is Coulomb’s law with superposition. Each vertex charge $q$ exerts a repulsive force on $Q$ (since both have the same sign). Because the triangle is equilateral, the centroid is equidistant from all three vertices, so each force has the same magnitude. And because the three vertices are symmetrically placed around the centroid, the three force vectors point along the medians, $120^\circ$ apart. When three equal vectors are arranged at $120^\circ$ intervals, they cancel exactly.

Let’s work through it step by step.

1. Distance from centroid to each vertex. In an equilateral triangle of side $l$, the centroid is also the circumcenter. The distance from the centroid to any vertex is the circumradius:

(Derivation: the altitude is $\frac{\sqrt{3}}{2}l$, and the centroid divides each median in the ratio $2:1$, so the distance from centroid to vertex is $\frac{2}{3}$ of the altitude: $\frac{2}{3} \cdot \frac{\sqrt{3}}{2}l = \frac{l}{\sqrt{3}}$.)

2. Magnitude of each force. By Coulomb’s law, the force on $Q$ due to a single vertex charge $q$ is

Since $q$ and $Q$ have the same sign, the force is repulsive — it points directly away from that vertex.

3. Direction of each force.

   The centroid lies at the intersection of the medians. The line from a vertex to the centroid is exactly along the median. So the force from vertex $A$ points from $O$ away from $A$ (straight down in the textbook figure), from $B$ away from $B$ (up-right), and from $C$ away from $C$ (up-left). These three directions are separated by $120^\circ$.

4. Vector addition.

   Place the three force vectors tail-to-tail at $O$. They have equal magnitude $F$ and are spaced $120^\circ$ apart. Their resultant is zero.

   Tip

   A quick way to see this: the sum of three equal vectors at $120^\circ$ is zero because they form the sides of an equilateral triangle when placed head-to-tail. Alternatively, resolve each into components: the horizontal components cancel pairwise, and the vertical components also sum to zero.

   Explicitly, take the direction from $O$ toward $A$ as the negative $y$-axis. Then:

   • $\vec{F}_A = -F\,\hat{j}$
   • $\vec{F}_B = F\sin 60^\circ\,\hat{i} + F\cos 60^\circ\,\hat{j} = \frac{\sqrt{3}}{2}F\,\hat{i} + \frac{1}{2}F\,\hat{j}$
   • $\vec{F}_C = -F\sin 60^\circ\,\hat{i} + F\cos 60^\circ\,\hat{j} = -\frac{\sqrt{3}}{2}F\,\hat{i} + \frac{1}{2}F\,\hat{j}$

   Adding:

   $\vec{F}_\text{net} = \left( \frac{\sqrt{3}}{2}F - \frac{\sqrt{3}}{2}F \right)\hat{i} + \left( -F + \frac{1}{2}F + \frac{1}{2}F \right)\hat{j} = 0\,\hat{i} + 0\,\hat{j} = 0.$

Method: Rotational-Symmetry Argument

This problem can be solved without computing a single force magnitude, just by reasoning about symmetry — often faster and less error-prone than vector addition.

1. Set up the symmetry.

   The three charges $q_1 = q_2 = q_3 = q$ sit at the vertices of an equilateral triangle, with $Q$ at the centroid. Rotate the entire triangle by $120^\circ$ about the centroid: vertex 1 moves to where vertex 2 was, vertex 2 to where vertex 3 was, and vertex 3 to where vertex 1 was.

2. Observe that the configuration looks identical after rotation.

   Because all three vertex charges are equal ($q_1 = q_2 = q_3 = q$), swapping their positions this way leaves the physical charge distribution completely unchanged. An observer at the centroid cannot tell the triangle was rotated.

3. The force on $Q$ must obey the same symmetry.

   Since the source charges look identical before and after the rotation, the electric force they produce on $Q$ (sitting exactly at the centroid, the rotation axis) must also look identical before and after — i.e., the net force vector $\vec{F}$ must map onto itself when rotated by $120^\circ$.

4. Ask what vectors are invariant under a $120^\circ$ rotation.

   Rotating any nonzero vector by $120^\circ$ always produces a different vector (pointing in a different direction) — $120^\circ$ is neither $0^\circ$ nor a multiple of $360^\circ$. The only vector that is unchanged by such a rotation is the zero vector.

5. Conclude.

   Therefore $\vec{F}$ must equal the zero vector:

This symmetry method generalizes well: for any $n$ equal charges arranged symmetrically ($n \geq 3$) around a central point, the net force or field at the center is zero by the same rotational argument — no need to redo the component algebra for a square, pentagon, or hexagon of equal charges.

Here are the most common mistakes students make when solving this classic Coulomb force superposition problem, along with how to avoid each.

1. Forgetting the Vector Nature of Force

The Mistake:

Students often compute the magnitude of the force from each $q$ on $Q$ correctly, but then simply add them as scalars (e.g., $F_{\text{net}} = F_1 + F_2 + F_3$).

Why it’s wrong:

Coulomb force is a vector. Forces from different charges point in different directions. Adding magnitudes directly ignores direction and gives an incorrect (usually larger) result.

How to Avoid:

Always draw a clear diagram showing the direction of each force vector. Use vector addition (component method or symmetry) — never scalar addition.

2. Not Using Symmetry to Simplify

The Mistake:

Students calculate all three force vectors explicitly, resolve into components, and sum — a long, error-prone process.

Why it’s wrong:

It wastes time and increases the chance of algebraic mistakes. The problem has perfect symmetry.

How to Avoid:

Recognize that the three charges are identical and placed at vertices of an equilateral triangle. The centroid is equidistant from all vertices. By symmetry, the three force vectors are equal in magnitude and spaced $120^\circ$ apart. Their vector sum is zero.

Key result: The net force on $Q$ at the centroid is $\mathbf{F}_{\text{net}} = 0$.

3. Incorrect Distance Calculation

The Mistake:

Using $l$ (side length) as the distance between a vertex charge and the centroid.

Why it’s wrong:

The distance from a vertex to the centroid of an equilateral triangle is not $l$. It is $\frac{l}{\sqrt{3}}$.

How to Avoid:

Memorize or derive:

• Centroid divides the median in ratio $2:1$.
• Median length $= \frac{\sqrt{3}}{2} l$.
• Distance from vertex to centroid $= \frac{2}{3} \times \text{median} = \frac{2}{3} \cdot \frac{\sqrt{3}}{2} l = \frac{l}{\sqrt{3}}$.

Use $r = \frac{l}{\sqrt{3}}$ in Coulomb’s law.

4. Sign Confusion in Force Direction

The Mistake:

If $Q$ and $q$ have the same sign, students sometimes draw forces as attractive.

Why it’s wrong:

Like charges repel. All three forces on $Q$ are repulsive and point radially outward from each vertex.

How to Avoid:

Always check: same sign → repulsion (force away from the other charge). Opposite sign → attraction (force toward the other charge). Draw arrows accordingly.

5. Assuming the Net Force is Non-Zero Without Checking

The Mistake:

After computing magnitudes, students assume the forces don’t cancel and proceed to find a non-zero resultant.

Why it’s wrong:

Symmetry guarantees cancellation. The three equal-magnitude vectors at $120^\circ$ to each other always sum to zero.

How to Avoid:

Before doing heavy algebra, pause and check for symmetry. If the configuration is symmetric and all charges are identical, the net force at the center is zero.

Quick Summary Checklist

Final takeaway: For this exact problem, the answer is zero — but only if you handle vectors, distances, and directions correctly.