Q122 / 199 · NEET 2008 (Mains) · Physics

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 Figure

  • A
    3 E along $\mathrm{OK}$
  • B
    3 E along $\mathrm{KO}$
  • Correct answer
    C
    E along $\mathrm{OK}$
  • D
    E along KO

Solution

$$ \begin{aligned} \vec{E}_0 & =0 \\ \Rightarrow \quad \vec{E}_{A K B}+\vec{E}_{A C D B} & =0 \\ \vec{E}_{A C D B} & ={ }^{(-)} \vec{E}_{A K B} \\ & =-E(\text { along } K O) \\ & =E(\text { along } O K) \end{aligned} $$

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

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