Two identical conducting spheres each having a radius of 0.500 cm are connected by a light 1.90 m long conducting wire. A charge of 65.0 µC is placed on one of the conductors. Assume the surface distribution of charge on each sphere is uniform. Determine the tension in the wire.
To determine the tension in the wire connecting the two spheres, we can use the concept of electrostatic equilibrium. In electrostatic equilibrium, the net force on each individual charge in a conductor is zero.
Here's how we can calculate the tension in the wire:
Step 1: Start by determining the charge on each sphere. Since both spheres are identical and the surface distribution of charge is uniform, the charge on each sphere will be half of the total charge:
Charge on each sphere = (65.0 µC) / 2 = 32.5 µC
Step 2: Next, calculate the electric field at the surface of each sphere. The electric field at the surface of a conducting sphere is given by:
Electric field = (k * Q) / r^2
Where k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), Q is the charge on the sphere, and r is the radius of the sphere.
Electric field at the surface of each sphere = (9.0 x 10^9 N m^2/C^2 * 32.5 x 10^(-6) C) / (0.005 m)^2
Step 3: Now, calculate the force acting on each sphere due to the electric field. Since the surface distribution of charge is uniform, the force acting on each sphere will be the product of the charge on the sphere and the electric field at its surface:
Force on each sphere = (32.5 x 10^(-6) C) * (Electric field at the surface of each sphere)
Step 4: Since both spheres are identical and the net force on each sphere should be zero, the tension in the wire connecting the spheres should be equal to the force acting on each sphere. Therefore, the tension in the wire is the double of the force acting on each sphere:
Tension in the wire = 2 * (Force on each sphere)
By plugging in the values from the previous calculations, you will get the final answer for the tension in the wire connecting the two spheres.