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 $A K B$ of the ring is $E$. The electric field at the centre due to the charge on the part $A C D B$ of the ring is

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Aicharya · Accepted Answer

Key Idea: It is given that the ring is conducting. As the ring is conducting, so electric field at its centre is zero, ie,
$\overrightarrow{\mathbf{E}}_{\text {total }}=0$
$\begin{aligned}
\overrightarrow{\mathbf{E}}_{A K B}+\overrightarrow{\mathbf{E}}_{A C D B} & =0 \\
\overrightarrow{\mathbf{E}}_{A C D B} & =-\overrightarrow{\mathbf{E}}_{A K B} \\
\overrightarrow{\mathbf{E}}_{A C D B} & =-\overrightarrow{\mathbf{E}} \text { (along } K O \text {) }
\end{aligned}$
Therefore, the electric field at the centre due to the charge on the part $A C D B$ of the ring is $E$ along $O K$.
Alternative :

The fields at $O$ due to $A C$ and $B D$ cancel each other.
The field due to $C D$ is acting in the direction $O K$ and equal in magnitude to $E$ due to $A K B$.

A thin conducting ring of radius $R$ is given a charge $+Q$. The electric field at the centre $O$ of the…

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