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

Options

A. `q^(2)/(4sqrt2piepsilon_(0)l^(2))`
B. `(-q^(2))/(4piepsilon_(0)l^(2))`
C. `(q^(2))/(4piepsilon_(0)l^(2))`
D. zero

Explanation

To find the force on each charge placed at the vertices of an equilateral triangle with charges \( q, q, \) and \( -q \), we can follow these steps:

### Step 1: Identify the Charges and Their Positions
Consider an equilateral triangle with vertices A, B, and C. Let:
- Charge at A = \( q \)
- Charge at B = \( q \)
- Charge at C = \( -q \)

### Step 2: Calculate the Forces on Charge at A
We need to find the net force acting on charge \( q \) at point A due to the other two charges.

1. **Force on A due to B (F_AB)**:
- Since both charges A and B are positive, they will repel each other.
- The magnitude of the force can be calculated using Coulomb's law:
\[ F_{AB} = k \frac{q \cdot q}{l^2} = k \frac{q^2}{l^2} \]
- The direction of \( F_{AB} \) will be away from B towards A.

2. **Force on A due to C (F_AC)**:
- Charge C is negative, which means it will attract charge A.
- The magnitude of the force is:
\[ F_{AC} = k \frac{q \cdot (-q)}{l^2} = -k \frac{q^2}{l^2} \]
- The direction of \( F_{AC} \) will be towards C.

### Step 3: Determine the Angles
In an equilateral triangle, the angles between the lines connecting the charges are 60 degrees. The angle between the forces \( F_{AB} \) and \( F_{AC} \) is 120 degrees (since they are on the same vertex).

### Step 4: Calculate the Resultant Force on Charge A
To find the resultant force \( F_R \) on charge A, we can use the vector addition of the two forces. The formula for the resultant of two forces \( F_1 \) and \( F_2 \) at an angle \( \theta \) is:
\[ F_R = \sqrt{F_{AB}^2 + F_{AC}^2 + 2 F_{AB} F_{AC} \cos(120^\circ)} \]

Substituting \( F_{AB} = k \frac{q^2}{l^2} \) and \( F_{AC} = -k \frac{q^2}{l^2} \):
\[ F_R = \sqrt{\left(k \frac{q^2}{l^2}\right)^2 + \left(-k \frac{q^2}{l^2}\right)^2 + 2 \left(k \frac{q^2}{l^2}\right) \left(-k \frac{q^2}{l^2}\right) \left(-\frac{1}{2}\right)} \]

### Step 5: Simplify the Expression
Calculating each term:
- \( F_{AB}^2 = \left(k \frac{q^2}{l^2}\right)^2 \)
- \( F_{AC}^2 = \left(-k \frac{q^2}{l^2}\right)^2 \)
- The cross term becomes:
\[ 2 \left(k \frac{q^2}{l^2}\right) \left(-k \frac{q^2}{l^2}\right) \left(-\frac{1}{2}\right) = k^2 \frac{q^4}{l^4} \]

Putting it all together:
\[ F_R = \sqrt{2 \left(k \frac{q^2}{l^2}\right)^2 + k^2 \frac{q^4}{l^4}} = k \frac{q^2}{l^2} \sqrt{2 + 1} = k \frac{q^2}{l^2} \sqrt{3} \]

### Step 6: Direction of the Resultant Force
The resultant force will be directed at an angle of 30 degrees from the line connecting A and B towards C.

### Final Result
The net force on charge \( q \) at point A is:
\[ F_R = k \frac{q^2}{l^2} \sqrt{3} \]
The same process can be repeated for charges at B and C, leading to similar magnitudes but different directions based on their positions.