A ring of radius `R` is with a uniformly distributed charge `Q` on it .A charge `q` is now placed in the centre of the ring .Find the increment in tension in the ring.

Options

A. `(qQ)/(8pi^(2)epsilon_(0)r^(2))`
B. `(qQ)/(4pi^(2)epsilon_(0)r^(2))`
C. `[(qQ)/(4piepsilon_(0)r^(2))]`
D. `(qQ)/(2pi^(2)epsilon_(0)r^(2))`

Answer

A

Explanation

`2Tsin(d theta)/(2)=(q.dQ)/(4piepsilon_(0)r^(2))2T(d theta)/(2)=(qlambdard theta)/(4piepsilon_(0)r^(2))`
`T=(q.dQ)/(4piepsilon_(0)r^(2))T=(qQr)/((2pir)4piepsilon_(0)r^(2))T=(qQ)/(8pi^(2)epsilon_(0)r^(2))`