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.
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.
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