Three identical charges of 2.0 uC are placed at the vertices of an equilateral triangle, 0.15m on a side. Find the net force acting on each angle.
For each charge, take the vector sum of the forces due to the other two particles. Use Coulomb's law for the force due to each of the other two particles. The vector sum will be perpendicular to the opposite base. The magnitude will be 2cos30 = sqrt3 times the force due to a single particle.
F = sqrt3*k*q^2/d^2
k is the Coulomb constant
q = 2.0*10^-6 C
d = 0.15 m
2.7682N
To find the net force acting on each angle, we need to calculate the net force exerted by the other two charges on each charge.
First, let's label the charges A, B, and C, and then calculate the net force on each charge:
Charge A: The net force on charge A is the sum of the forces due to charges B and C. Since charges B and C are equidistant from A and have the same magnitude, the net force on A will be directed towards the center of the triangle. The magnitude of the net force on A can be calculated using Coulomb's Law:
F(A) = k * (q * q) / (r * r)
where F(A) is the net force on A, k is the electrostatic constant (9 × 10^9 N m^2/C^2), q is the magnitude of the charge (2.0 × 10^-6 C), and r is the distance between the charges (0.15 m).
F(A) = (9 × 10^9 N m^2/C^2) * (2.0 × 10^-6 C)^2 / (0.15 m)^2
Calculating this expression, we find that F(A) is approximately 3.2 N.
Charge B: The net force on charge B is the sum of the forces due to charges A and C. Since charges A and C are equidistant from B and have the same magnitude, the net force on B will also be directed towards the center of the triangle. Using the same formula as before, we find:
F(B) = (9 × 10^9 N m^2/C^2) * (2.0 × 10^-6 C)^2 / (0.15 m)^2
Calculating this expression, we find that F(B) is approximately 3.2 N.
Charge C: The net force on charge C is the sum of the forces due to charges A and B. Since charges A and B are equidistant from C and have the same magnitude, the net force on C will also be directed towards the center of the triangle. Using the same formula as before, we find:
F(C) = (9 × 10^9 N m^2/C^2) * (2.0 × 10^-6 C)^2 / (0.15 m)^2
Calculating this expression, we find that F(C) is approximately 3.2 N.
So, the net force acting on each angle of the equilateral triangle is approximately 3.2 N.
To find the net force acting on each angle of the equilateral triangle, we can use the principle of superposition. The net force on each charge is the vector sum of the forces due to the other two charges.
First, let's calculate the magnitude of the force between two charges. The formula for the electric force between two point charges is given by Coulomb's law:
F = (k * q1 * q2) / r^2,
where F is the magnitude of the force, k is the electrostatic constant (9 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
In this case, all charges are identical with a magnitude of 2.0 uC (2.0 x 10^-6 C), and the distance between them is 0.15 m.
Using Coulomb's law, we find the magnitude of the force between any two charges:
F = (9 x 10^9 N m^2/C^2) * (2.0 x 10^-6 C)^2 / (0.15 m)^2.
Calculating the above expression, we get:
F ≈ 7.2 x 10^-3 N.
Since charges are placed at the vertices of an equilateral triangle, the forces on all three charges will have the same magnitude.
Now, since the charges are at the vertices of an equilateral triangle, the net force on each charge is the vector sum of two forces. Since the triangle is symmetric, we can determine the net force acting on each angle.
Using simple vector addition, we can find that the resultant force acting on each angle is:
Fnet = 2F * cos(30°).
Substituting the value of F into the equation, we get:
Fnet ≈ 2 * 7.2 x 10^-3 N * cos(30°).
Calculating the above expression, we find:
Fnet ≈ 2.78 x 10^-3 N.
Therefore, the net force acting on each angle of the equilateral triangle is approximately 2.78 x 10^-3 N.