To determine the magnitude of the resultant force on the charged particle at vertex A of the equilateral triangle formed by the three charged particles, we need to analyze the forces acting on it due to the other two charges located at vertices B and C. Each charge is +20 microcoulombs, and since they are all positive, they will repel each other.
Understanding the Forces
In this scenario, we have three particles, each with a charge of +20 µC, positioned at the vertices of an equilateral triangle with sides measuring 1 meter. The forces acting on the charge at point A will be due to the charges at points B and C.
Calculating the Forces
First, we can calculate the force exerted on the charge at A by the charge at B and the charge at C using Coulomb's Law, which is given by:
F = k * |q1 * q2| / r²
Where:
- F is the force between the charges.
- k is Coulomb's constant, approximately 8.99 x 10^9 N m²/C².
- q1 and q2 are the magnitudes of the charges.
- r is the distance between the charges.
For our case:
- q1 = q2 = 20 µC = 20 x 10^-6 C
- r = 1 m
Substituting these values into Coulomb's Law:
F = (8.99 x 10^9 N m²/C²) * (20 x 10^-6 C)² / (1 m)²
Calculating this gives:
F = (8.99 x 10^9) * (400 x 10^-12) = 3.596 N
Direction of Forces
Now, we need to consider the direction of these forces. The force exerted by charge B on charge A (F_AB) will point away from B towards A, and similarly, the force exerted by charge C on charge A (F_AC) will point away from C towards A. Since the triangle is equilateral, the angle between the lines connecting A to B and A to C is 60 degrees.
Resultant Force Calculation
To find the resultant force on charge A, we can use vector addition. The forces F_AB and F_AC can be resolved into their components:
Let:
- F_AB = 3.596 N (acting along the line AB)
- F_AC = 3.596 N (acting along the line AC)
The x-component of F_AB is:
F_ABx = F_AB * cos(60°) = 3.596 * 0.5 = 1.798 N
The y-component of F_AB is:
F_ABy = F_AB * sin(60°) = 3.596 * (√3/2) ≈ 3.113 N
For F_AC, the x-component is:
F_ACx = F_AC * cos(60°) = 3.596 * 0.5 = 1.798 N
The y-component of F_AC is:
F_ACy = F_AC * sin(60°) = 3.596 * (√3/2) ≈ 3.113 N
Now, we can sum the components:
Total x-component = F_ABx + F_ACx = 1.798 + 1.798 = 3.596 N
Total y-component = F_ABy + F_ACy = 3.113 + 3.113 = 6.226 N
Magnitude of the Resultant Force
Finally, we can find the magnitude of the resultant force using the Pythagorean theorem:
F_resultant = √(Total x-component² + Total y-component²)
F_resultant = √(3.596² + 6.226²) ≈ √(12.93 + 38.76) ≈ √51.69 ≈ 7.19 N
Thus, the magnitude of the resultant force on the charged particle at A is approximately 7.19 N.
