A thin metallic ring of radius R and area of cross section A carries an electric charge q that is uniformly distributed. A point charge Q is placed at the centre. The signs of charges on the ring and that of Q are same. Young's modulus for the material of the ring is Y. What is the tension developed in the ring?

Options

A. `(qQ)/(4piepsilon_0R^2)`
B. `(qQ)/(8pi^2 epsilon_0R^2)`
C. `(3qQ)/(20pi^2 epsilon_0R^2)`
D. None

Answer

B

Explanation

Charge on metallic ring and point charge at the centre repel each others, tension develops in the ring. The force of tension for the ring and acts along the tangent to the ring. Select one segment on ring making an angle `Deltatheta` at the centre as shown in the diagram.
Charge on the segment : `q/(2pi)Deltatheta`
Force between point charge Q at the centre and this segment can be written as follows :
`F=1/(4piepsilon_0)((Q/(2pi)Deltatheta)Q)/R^2 =(qQ)/(8pi^2 epsilon_0 R^2) Deltatheta` …(i)
The components of tension T cos `(Deltatheta)/2` are cancelled from each other and T sin `(Deltatheta)/2` are added to act against electric force F. For equilibrium :
2 T sin `(Deltatheta)/2=F`
2T sin `(Deltatheta)/2 = (qQ)/(8pi^2 epsilon_0 R^2)Deltatheta`
For small values of `Deltatheta` we can approximate sin `(Deltatheta//2)` as `Deltatheta//2` , hence we get the following :
`2T (Deltatheta)/2 = (qQ)/(8pi^2 epsilon_0 R^2) Deltatheta rArr T= (qQ)/(8pi^2 epsilon_0 R^2)`