A small element `l` cut from a circular ring of radius a and `lamda` charge per unit length. The net electric field at the centre of ring is

Options

A. `(lambda)/(2pi epsilon_(0)a^(2))`
B. `(lambda)/(4pi epsilon_(0)a^(2))`
C. `(lambda^(2))/(2pi epsilon_(0)a^(2))`
D. `(lambda)/(2pi epsilon_(0)a)`

Answer

D

Explanation

Electric intensity at centre O, due to small element dl of charged ring
`dE = (lambda dl)/(4pi epsilon_(0) a^(2)) = (lambda (ad theta))/(4pi epsilon_(0)a^(2))`
As is clear from figure, horizontal components of dE will cancel out in pairs and vertical components will add.
`:. E = underset(0)overset(pi)int dE sin theta = underset(0)overset(pi)int (lambda)/(4pi epsilon_(0)a) sin theta d theta = (lambda)/(4pi epsilon_(0)a)[-cos theta]_(0)^(pi)`
`E = (lambda)/(4pi epsilon_(0)a)(- cos pi + cos 0) = (lambda)/(2pi epsilon_(0)a)`