Consider a circular arc of radius `R` which subtends an angle `phi` at its centre .Let us calculate the electric field strength at `C`

Explanation

Consider a polar segment on arc of angular width `d theta` at an angle `theta` form the angular bisector `XY` as shown ,the length of element segment is `Rd theta` .The charge on this element `dq` is
`dq=(Q)/(phi)d theta`
Due to this `dq` ,electric field at a centre of arc `C` is given as
`dE=(dq)/(4piepsilon_(0)R^(2))`
The electric field components `dE` to this segment `dE sin theta` which is perpendicular to the angle bisector gets cancelled out integration.
the net electric field at centre wil be along angle bisector which can be calculated by intersecting `dEcostheta` which limits from `-phi//2` to `phi//2`.
Hence net elctric field strength at centre C is
`E_(c)=intdEcostheta`
` =int_(phi//2)^(-phi//2)(Q)/(4piepsilon_(0)phiR^(2))costhetad theta`
`(Q)/(4pi epsilon_(0)phiR^(2))costhetad theta E_(c)=intdEcostheta`
`=int_(phi//2)^(-phi//2)(Q)/(4piepsilon_(0)R^(2)phi)costhetad theta`
`=(Q)/(4piepsilon_(0)R^(2)phi) =int_(phi//2)^(-phi//2) cos theta d theta`
`(Q)/(4pi epsilon_(0)phiR^(2)) ["sin " theta]_(phi//2)^(phi//2)`
`=(Q)/(4piepsilon_(0)R^(2)phi)[sinphi//2+sinphi//2]`
`E_(c)=(2Qsin(phi//2))/(4piepsilon_(0)R^(2)phi)`
for a semi circualr ring `phi=pi` So at centre
`E_(c)=(2Qsin(phi//2))/(4piepsilon_(0)R^(2)phi)=(2Qsin(phi//2))/(4phiepsilon_(0)R^(2)pi)=(2Q)/(4piepsilon_(0)R^(2))`