Three negatively charged spheres, each with a charge of 4*10^-6C are fixed at the vertices of an equilateral triangle wise sides 20cm. Calculate the size and direction of the net force on each sphere.

To calculate the net force on each sphere, let's consider one sphere at a time and calculate the force between this sphere and the other two spheres. Then we will vectorially add the two forces to get the net force on the sphere.

Let's label the three spheres as A, B, and C with charges qA = qB = qC = 4*10^-6 C. Let AB = BC = CA = side length of the equilateral triangle = 20 cm = 0.2 m.

1. Force on sphere A due to sphere B: Since spheres A and B have the same charge, they will repel each other. The force between them is given by Coulomb's law:

F_AB = (k * qA * qB) / r_AB^2

where k is the electrostatic constant = 8.99*10^9 Nm²/C², qA and qB are the charges of the spheres, and r_AB is the distance between spheres A and B.

F_AB = (8.99*10^9 * 4*10^-6 * 4*10^-6) / (0.2)^2 = 719.2 N

The force is along the line connecting spheres A and B and directed away from sphere B. Let this force make an angle θ = π/3 radians with the "downward" direction.

2. Force on sphere A due to sphere C: Similarly, the force between spheres A and C is given by:

F_AC = (k * qA * qC) / r_AC^2

Since the charges and distances are the same as before, we get F_AC = F_AB = 719.2 N. This force is along the line connecting spheres A and C and directed away from sphere C. Let this force make an angle θ = -π/3 radians with the "downward" direction.

3. Net force on sphere A: To find the net force on sphere A, we will add F_AB and F_AC vectorially. Since both the forces are at an angle θ to the downward direction, we can decompose them into their horizontal (x-direction) and vertical (y-direction) components and then add the components.

F_AB_x = F_AB * cos(π/3) = 719.2 * cos(π/3) = 359.6 N (to the left)
F_AB_y = F_AB * sin(π/3) = 719.2 * sin(π/3) = 622.7 N (downward)

F_AC_x = F_AC * cos(-π/3) = 719.2 * cos(-π/3) = 359.6 N (to the right)
F_AC_y = F_AC * sin(-π/3) = 719.2 * sin(-π/3) = -622.7 N (upward)

F_net_x = F_AB_x + F_AC_x = 359.6 - 359.6 = 0 N
F_net_y = F_AB_y + F_AC_y = 622.7 - 622.7 = 0 N

So the net force on sphere A is F_net = sqrt(F_net_x^2 + F_net_y^2) = 0 N.

Since all the spheres have the same charges, and the triangle is equilateral, the net force on all three spheres will also be the same, 0 N due to symmetry.

To calculate the net force on each sphere, we need to understand that like charges repel each other. The net force is the vector sum of individual forces on each sphere.

To find the size and direction of the net force on each sphere, follow these steps:

Step 1: Calculate the electrostatic force between two spheres.

The electrostatic force between two charged objects is given by Coulomb's law:

F = (k * |Q₁| * |Q₂|) / r²

where F is the force, k is the electrostatic constant (9 * 10^9 Nm²/C²), |Q₁| and |Q₂| are the magnitudes of the charges, and r is the distance between the charges.

For our case, let's calculate the force between sphere 1 and sphere 2.

|Q₁| = |Q₂| = 4 * 10^-6 C (charge of each sphere)
r = 20 cm = 0.2 m (distance between the spheres)

Plugging these values into Coulomb's law:

F₁₂ = (9 * 10^9 Nm²/C² * (4 * 10^-6 C)²) / (0.2 m)²

Step 2: Calculate the net force on each sphere.

Since the sides of the equilateral triangle are all equal, the net force on each sphere will have the same magnitude but with different directions.

The net force on sphere 1 is the sum of the forces from sphere 2 and sphere 3. Since these two forces act at 120 degrees to each other, we need to find their resultant.

The net force on sphere 2 is the sum of the forces from sphere 1 and sphere 3, and the net force on sphere 3 is the sum of the forces from sphere 1 and sphere 2.

Step 3: Determine the direction of the net force on each sphere.

The direction of the net force can be found by constructing a vector triangle where each force is represented as a vector. The magnitude of the net force is the length of the resulting vector, and the direction is given by the angle.

In our case, since the equilateral triangle has 3 equal angles of 60 degrees, we can conclude that the net force on each sphere will be along one of the line bisecting each angle.

Calculating the values, we get:

F₁₂ = 0.36 N
F₃₁ = 0.36 N
F₂₃ = 0.36 N

So, the size of the net force on each sphere is 0.36 Newtons, and the direction of the net force on each sphere is along one of the lines bisecting each angle of the equilateral triangle.