Three point charges, each with charge q and mass 'm', are connected by three strings of equal length `l` whose other end is connected to a fixed point P. They form an equilateral triangle of side 'a' in a horizontal plane. The tension in each string will be

Options

A. `sqrt((mg)^2-(q^2/(4piepsilon_0a^2)sqrt3)^2)`
B. `(mgl)/sqrt(l^2-a^2//3)`
C. `(3q^2)/(4piepsilon_0a^2)`
D. none of the above

Answer

To find the tension in each string connecting three point charges arranged in an equilateral triangle, we can follow these steps:

### Step 1: Understand the Setup
We have three point charges, each with charge \( q \) and mass \( m \), connected by strings of length \( l \) to a fixed point \( P \). The charges form an equilateral triangle of side \( a \) in a horizontal plane.

### Step 2: Analyze the Geometry
Since the strings are of equal length \( l \) and connect to a fixed point \( P \), the charges will be at equal heights above the horizontal plane. The angle \( \theta \) that each string makes with the vertical can be determined using trigonometric relationships.

### Step 3: Establish Relationships
Using trigonometry, we can express \( \cos \theta \) and \( \sin \theta \):
- The horizontal distance from \( P \) to the midpoint of the side \( AC \) is \( \frac{a}{2} \).
- Thus, \( \cos \theta = \frac{a/2}{l} \) and \( \sin \theta = \sqrt{1 - \left(\frac{a/2}{l}\right)^2} \).

### Step 4: Force Balance in the Vertical Direction
In the vertical direction, the forces acting on each charge are the tension \( T \) in the string and the weight \( mg \):
\[ T \sin \theta = mg \quad \text{(1)} \]

### Step 5: Force Balance in the Horizontal Direction
In the horizontal direction, we consider the electrostatic forces between the charges. The electrostatic force \( F_{AC} \) between charges \( A \) and \( C \) is given by:
\[ F_{AC} = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \]
The horizontal component of the tension \( T \) must balance the net electrostatic force. The angle \( \alpha \) between the string and the horizontal plane is \( 30^\circ \) (since it's an equilateral triangle):
\[ T \cos \theta \cos 30^\circ = F_{AC} + F_{AB} \sin 30^\circ \quad \text{(2)} \]
Since \( F_{AB} \) is equal to \( F_{AC} \), we can simplify this equation.

### Step 6: Substitute and Solve
From equation (2), we can express the tension in terms of the electrostatic force:
\[ T \cos \theta \cdot \frac{\sqrt{3}}{2} = 2 \cdot \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \cdot \frac{1}{2} \]
This simplifies to:
\[ T \cos \theta = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \cdot \frac{2}{\sqrt{3}} \quad \text{(3)} \]

### Step 7: Combine Equations
Now we have two equations:
1. \( T \sin \theta = mg \)
2. \( T \cos \theta = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \cdot \frac{2}{\sqrt{3}} \)

### Step 8: Square and Add
Squaring both equations and adding them gives:
\[ T^2 (\sin^2 \theta + \cos^2 \theta) = m^2 g^2 + \left( \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \cdot \frac{2}{\sqrt{3}} \right)^2 \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ T^2 = m^2 g^2 + \left( \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \cdot \frac{2}{\sqrt{3}} \right)^2 \]

### Step 9: Solve for Tension
Finally, we can express the tension \( T \):
\[ T = \sqrt{m^2 g^2 + \left( \frac{1}{4 \pi \epsilon_0} \frac{q^2}{a^2} \cdot \frac{2}{\sqrt{3}} \right)^2} \]