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

1. The net electric field at the center of a uniformly charged full ring is zero because fields from opposite segments cancel out.
$$E_{\text{total}}=E_{AKB}+E_{ACDB}=0$$
2. Let the field due to segment $AKB$ be ${\vec{E}}_{AKB}=E$ along $KO$ (assuming $K$ is on one side).3. Therefore, $E+E_{ACDB}=0$.4. This implies ${\vec{E}}_{ACDB}=-E$, which means it has magnitude $E$ and its direction is opposite to $KO$, i.e., along $OK$.