AIIMS 2003 Physics Coulomb's Law MCQ Question

To solve the problem, we need to analyze the forces acting on the charge at vertex A of the equilateral triangle formed by the three charges. Let's denote the charges at the vertices A, B, and C as $Q_A$, $Q_B$, and $Q_C$, respectively. For simplicity, we can assume that all three charges are equal and have a value of $Q$.

Step 1: Identify the Forces Acting on Charge at A

The forces acting on the charge at vertex A due to the other two charges (at vertices B and C) can be calculated using Coulomb's law:

$$F = k \cdot \frac{|Q_1 Q_2|}{r^2}$$

where $k = \frac{1}{4 \pi \epsilon_0}$, $Q_1$ and $Q_2$ are the magnitudes of the charges, and $r$ is the distance between them.

Step 2: Calculate the Forces from B and C

1. Force from Charge B on Charge A:

   • The distance $AB = a$.
   • The force $F_{AB}$ is given by:
   $$F_{AB} = k \cdot \frac{Q^2}{a^2}$$
   • This force acts along the line connecting A and B.

2. Force from Charge C on Charge A:

   • The distance $AC = a$.
   • The force $F_{AC}$ is similarly:
   $$F_{AC} = k \cdot \frac{Q^2}{a^2}$$
   • This force acts along the line connecting A and C.

Step 3: Analyze Directions of Forces

For an equilateral triangle:

• The angle between the line connecting A to B and the line connecting A to C is $60^\circ$.
• The forces $F_{AB}$ and $F_{AC}$ will have components in the direction normal to line BC (let's denote this direction as the y-axis) and along line BC (the x-axis).

Step 4: Resolve the Forces into Components

1. Components of $F_{AB}$:

   • The x-component (along BC):
   $$F_{AB,x} = F_{AB} \cdot \cos(60^\circ) = k \cdot \frac{Q^2}{a^2} \cdot \frac{1}{2} = \frac{k Q^2}{2a^2}$$
   • The y-component (normal to BC):
   $$F_{AB,y} = F_{AB} \cdot \sin(60^\circ) = k \cdot \frac{Q^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{k Q^2 \sqrt{3}}{2a^2}$$

2. Components of $F_{AC}$ (similar calculations):

   • The x-component:
   $$F_{AC,x} = F_{AC} \cdot \cos(60^\circ) = \frac{k Q^2}{2a^2}$$
   • The y-component:
   $$F_{AC,y} = F_{AC} \cdot \sin(60^\circ) = \frac{k Q^2 \sqrt{3}}{2a^2}$$

Step 5: Sum the Forces

For the x-components:

$$F_{x} = F_{AB,x} + F_{AC,x} = \frac{k Q^2}{2a^2} + \frac{k Q^2}{2a^2} = \frac{k Q^2}{a^2}$$

For the y-components:

$$F_{y} = F_{AB,y} + F_{AC,y} = \frac{k Q^2 \sqrt{3}}{2a^2} + \frac{k Q^2 \sqrt{3}}{2a^2} = \frac{k Q^2 \sqrt{3}}{a^2}$$

Step 6: Determine the Resultant Force

The force in the direction normal to BC (the y-direction) is:

$$F_{y} = \frac{k Q^2 \sqrt{3}}{a^2}$$

However, it is important to note that the x-components cancel each other out due to symmetry (they are equal and opposite), resulting in no net force in the x-direction. Therefore, the net force experienced by charge A in