Question

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.

Answer 1

Answer: 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