Q.Consider the charges $q$, $q$, and $-q$ placed at the vertices of an equilateral triangle. What is the force on each charge?

The key idea is the Principle of Superposition of Electrostatic Forces, which states that the net force on any charge is the vector sum of the individual forces exerted on it by all other charges.

Let the side length of the equilateral triangle be $a$. The magnitude of the force between any two charges $q$ and $q$, or $q$ and $-q$, is given by Coulomb's Law:

1. Force on charge $q$ at vertex A:

   • Force from $q$ at B ($F_{AB}$): Repulsive, magnitude $F_0$, directed along BA.
   • Force from $-q$ at C ($F_{AC}$): Attractive, magnitude $F_0$, directed along AC.
   • The angle between $\vec{F}_{AB}$ and $\vec{F}_{AC}$ is $120^\circ$. The resultant force $\vec{F}_A$ has magnitude $F_A = \sqrt{F_0^2 + F_0^2 + 2F_0^2 \cos(120^\circ)} = \sqrt{2F_0^2 + 2F_0^2(-1/2)} = F_0$. Its direction is parallel to the side BC, pointing away from B and C.

2. Force on charge $q$ at vertex B:

   • Force from $q$ at A ($F_{BA}$): Repulsive, magnitude $F_0$, directed along AB.
   • Force from $-q$ at C ($F_{BC}$): Attractive, magnitude $F_0$, directed along BC.
   • The angle between $\vec{F}_{BA}$ and $\vec{F}_{BC}$ is $120^\circ$. The resultant force $\vec{F}_B$ has magnitude $F_B = \sqrt{F_0^2 + F_0^2 + 2F_0^2 \cos(120^\circ)} = F_0$. Its direction is parallel to the side AC, pointing away from A and C.

3. Force on charge $-q$ at vertex C:

   • Force from $q$ at A ($F_{CA}$): Attractive, magnitude $F_0$, directed along CA. …

Electric charges exert forces on each other as described by Coulomb's Law. When multiple charges are present, the net force on any single charge is the vector sum of the individual forces exerted on it by all other charges. This is known as the Principle of Superposition. Since force is a vector quantity, both its magnitude and direction must be considered when adding forces.

Let the charges be $q_A = q$ at vertex A (bottom-left), $q_B = q$ at vertex B (bottom-right), and $q_C = -q$ at vertex C (top). Let the side length of the equilateral triangle be $a$.

The distance between any two charges is $a$.

Let's define a reference force magnitude $F_0$ for convenience:

The magnitude of the force between $q$ and $-q$ will also be $F_0$.

We will now calculate the net force on each charge using vector addition.

1. Force on charge $q_A$ (at vertex A)

Charge $q_A$ experiences two forces:

• Force from $q_B$ ($\vec{F}_{AB}$): Since $q_A$ and $q_B$ are both positive ($q$), the force is repulsive. Its magnitude is $F_0$. The direction is away from B, along the line AB, pointing horizontally to the left.
• Force from $q_C$ ($\vec{F}_{AC}$): Since $q_A$ is positive ($q$) and $q_C$ is negative ($-q$), the force is attractive. Its magnitude is $F_0$. The direction is towards C, along the line AC, pointing upwards and to the right.

The angle between the line AB (horizontal) and AC is $60^\circ$. Therefore, the angle between $\vec{F}_{AB}$ (pointing left) and $\vec{F}_{AC}$ (pointing up-right) is $180^\circ - 60^\circ = 120^\circ$.

To find the net force $\vec{F}_A = \vec{F}_{AB} + \vec{F}_{AC}$, we use the parallelogram law of vector addition:

The resultant force $\vec{F}_A$ has a magnitude of $F_0$. Since the two component forces have equal magnitudes, the resultant force bisects the angle between them. The angle between $\vec{F}_{AB}$ (left) and $\vec{F}_{AC}$ (up-right) is $120^\circ$. The resultant will be $60^\circ$ from each. This means $\vec{F}_A$ points $60^\circ$ above the negative x-axis, which is parallel to the side BC.

2. Force on charge $q_B$ (at vertex B)

Charge $q_B$ experiences two forces:

• Force from $q_A$ ($\vec{F}_{BA}$): Since $q_B$ and $q_A$ are both positive ($q$), the force is repulsive. Its magnitude is $F_0$. The direction is away from A, along the line BA, pointing horizontally to the right.
• Force from $q_C$ ($\vec{F}_{BC}$): Since $q_B$ is positive ($q$) and $q_C$ is negative ($-q$), the force is attractive. Its magnitude is $F_0$. The direction is towards C, along the line BC, pointing upwards and to the left.

The angle between the line BA (horizontal) and BC is $60^\circ$. Therefore, the angle between $\vec{F}_{BA}$ (pointing right) and $\vec{F}_{BC}$ (pointing up-left) is $180^\circ - 60^\circ = 120^\circ$.

To find the net force $\vec{F}_B = \vec{F}_{BA} + \vec{F}_{BC}$:

$|\vec{F}_B| = \sqrt{F_0^2 + F_0^2 + 2 F_0 F_0 \left(-\frac{1}{2}\right)}$ …

Concept: Coulomb Force Superposition

The net force on any charge is the vector sum of forces from the other two charges, using Coulomb’s law.

Reasoning

1. Let side length be $a$. For a charge $q$ at vertex A, forces from the other $q$ (at B) and $-q$ (at C) are:

   • $F_{AB} = \frac{k q^2}{a^2}$ (repulsive, along AB)
   • $F_{AC} = \frac{k q^2}{a^2}$ (attractive, along AC)

2. These two forces are equal in magnitude and at $60^\circ$ to each other. Their resultant magnitude:

   $F_{\text{net}} = \sqrt{F_{AB}^2 + F_{AC}^2 + 2 F_{AB} F_{AC} \cos 60^\circ} = \sqrt{3} \cdot \frac{k q^2}{a^2}$ …

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.

--- …