A half ring of radius `r` has a linear charge density `lambda`.The potential at the centre of the half ring is

Options

A. `(lamda)/(4 epsi_(0))`
B. `(lamda)/(4pi^(2) epsi_(0)r)`
C. `(lamda)/(4pi epsi_(0)r)`
D. `(lamda)/(4pi epsi_(0)r^(2))`

Answer

A

Explanation

Potential due to small element .p. at the centre
`v= int dv = int k(lamda)/(r ) dl = (1)/(4pi epsi_(0)) (lamda)/(r )int dl`
`dv = K .lamda (dl)/(r ), = (1)/(4pi epsi_(0)) (lamda)/(r ) pi r rArr (lamda)/(4epsi_(0))`