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

• A

  $3 \mathrm{E}$ along $\mathrm{KO}$
• Correct answer
  B

  E along $O K$
• C

  $E$ along $K O$
• D

  $3 E$ along $O K$

Solution

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 $O K$. lternative : 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 $A K B$.

Hence, the correct answer is option: 'B'.