A thin half ring of radius `R = 20 cm` is uniformly charged with a total charge `q = 0.70 nC`.Find the magnitude of the electric field strength at the curvaite centre of this half-ring.

Explanation

The situation is shown in figure Here the semicircular wire subtend an angle `pi` at the centre, we known that the electric field strength due to a circular arc subtending an angle `phi` at at it centre can be given as
`E=(2Kqsinphi//2)/(phiR^(2))=(2Kq)/(piR^(2))" " ["Here "phi=pi]`

`=q/(2pi^2in_(0)R^(2))`
Substituting the value, we get
`=(7xx10^(-10))/(2xx(3.14)^2xx(8.85xx10^(-12))xx(0.2)^(2))`
`=100V//m`