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(D) The total electric field at the centre $O$ of a uniformly charged ring is zero. Let the ring be divided into four equal parts: $AK$,$KB$,$BD$,and

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$E$ along $OK$

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(D) The total electric field at the centre $O$ of a uniformly charged ring is zero. Let the ring be divided into four equal parts: $AK$,$KB$,$BD$,and $DC$. Each part carries a charge of $+Q/4$. The electric field at the centre $O$ due to the part $AKB$ is the vector sum of the fields due to $AK$ and $KB$. Let this resultant be $E$ directed along $KO$. The part $ACDB$ consists of the segments $AC$,$CD$,and $DB$. Due to symmetry,the electric field at $O$ due to the charge on part $AC$ is equal and opposite to the field due to the charge on part $BD$. Thus,they cancel each other out. Therefore,the net electric field at $O$ due to the part $ACDB$ is equal to the electric field due to the part $CD$ alone. Since the total field of the entire ring is zero,the field due to $AKB$ must be equal and opposite to the field due to $ACDB$. If the field due to $AKB$ is $E$ directed along $KO$,then the field due to $ACDB$ must be $E$ directed along $OK$ to ensure the vector sum is zero.

$A$ 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

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