A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field `vecE` at the centre O is

Options

A. `- (q)/(4 pi^(2) in_(0) r^(2)) hat(j)`
B. `- (q)/(2 pi^(2) in_(0) r^(2)) hat(j)`
C. `(q)/(2 pi^(2) in_(0) r^(2)) hat(j)`
D. `(q)/(4 pi^(2) in_(0) r^(2)) hat(j)`

Answer

C

Explanation

Let `lambda` be the linear charge desntiy on the semicirular ring. Then `lambda = q//pi r`. Take an element of the ring of length `Delta l` as shown in Fig.

Charge on element, `dq =lambda Delta l = lambda (r Delta theta)`.
Elecatric field intensity at `O` due to charge on element is `dE = (1)/(4 pi in_(0)) (lambda r Delta theta)/(r^(2))` along `OA`
Two recentagular components of `vec(dE)`are `dE cos theta` along x-direction and `dE sin theta` along negative y-direction.If we find the electric field at `O`due to charge on all other elements of the ring and resolve them into two recentagular components, we note that the components along X-direction cancel out in pairs and components along negative y-direction add up. Therefore, total electric field intensity at `O` due to charge on the entire semicircular ring is
`vec(E) = int_(0)^(180^(@)) dE sin theta (- hat(j))`
`= int_(0)^(180^(@)) (1)/(4pi in_(0)) (lamnda r Delta theta sin theta)/(r^(2)) (- hat(j))`
`= (1)/(4pi in_(0)) (lambda)/(r ) [- cos theta]_(0)^(180^(@) 9- hat(j))`
`= (1)/(4pi in_(0)) (lambda)/( r) [cos 180^(@) - cos 0^(@)] hat(j)`
`= (1)/(4pi in_(0)) ((q// pi r))/(r ) [-1- 1] hat(j)`
`= (q)/(2 pi^(2) in_(0) r^(2)) hat(j)`