A thin conducting ring of radius $\mathrm{R}$ is given a charge $+ \mathrm{Q}.$ The electric field at the centre $\mathrm{O}$ of the ring due to the charge on the part $\mathrm{AKB}$ of the ring is $\mathrm{E}.$The electric field at the centre due to the charge on the part $\mathrm{ACDB}$ of the ring is 

 

By the symmetry of the figure, the electric fields at $\mathrm{O}$ due to the portions $\mathrm{AC}$ and $\mathrm{BO}$ are equal in magnitude and opposite in direction. So, they cancel each other. 

Similarly, the field at $\mathrm{O}$ due to $\mathrm{CD}$and $\mathrm{AKB}$ are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part $\mathrm{ACDB}$ is $\mathrm{E}$ along $\mathrm{OK}.$