To determine the tension in one of the strings, we first need to analyze the forces acting on the charge at one end of the strings. Each string forms an angle of 120 degrees, which means that if we consider one charge, it experiences repulsive forces due to the other charges.
Let’s label the charges as Q1, Q2, and Q3, each with a charge of 2 µC (2 x 10-6 C). The distance between them can be calculated using the geometry of the problem. The angle between each string is 120 degrees, resulting in the strings forming an equilateral triangle with sides equal to 1.00 m.
Using Coulomb’s Law, the force between any two charges can be expressed as:
F = k * |Q1 * Q2| / r²
Where:
- k = 8.99 x 109 N m2/C2
- r = distance between the charges (1.00 m)
- Q1 and Q2 = 2 x 10-6 C
Calculating the force between two charges:
F = (8.99 x 109) * (2 x 10-6 * 2 x 10-6) / (1.00)2
F = (8.99 x 109) * (4 x 10-12) = 3.596 x 10-2 N
This force acts at an angle, so we should decompose it into components. Each string is subjected to a tension that balances the forces in the horizontal and vertical directions. In this configuration, due to symmetry and the 120-degree angle, we can focus on the equilibrium of forces horizontally.
Considering one side of the triangle, the angle between the tension and the horizontal component is 60 degrees (as 120 degrees divided by 2 equals 60 degrees). Using trigonometry, the horizontal component of the tension (T) can be expressed as:
2 * T * cos(60°) = F
Where F is the total horizontal force from two charges acting on one charge. Thus:
2 * T * (1/2) = 3.596 x 10-2
So, this simplifies to:
T = 3.596 x 10-2
Therefore, the tension in one of the strings is approximately 0.03596 N.