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 41.0 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.

4.19 N

Explanation: In order to determinate the tension applied on the wire we have to calculate the electric force between the conductor spheres connected by the wire.

As the wire is a conductor the spheres are at same potential so we have:

V1=V2

V1=k*Q1/r1 and V2=k*Q2/r2

where r1=r2, then

Q1=Q2

so the electric force is given by:

F=k*Q^2/d^2  where d is the distance between the spheres.

Finally replacing the values,  we have

F=9*10^9(41*10^-6)^2/(1.9)^2= 4.19 N

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it