Question

A thin conducting ring of radius $R$ is given a charge $+Q$ uniformly distributed over its circumference. The electric field at the centre $O$ of the ring due to the charge on the part $\text{AKB}$ of the ring is $E$. The electric field at the centre due to the charge on the part $\text{ACDB}$ of the ring is

• A. $3E$ along $\text{KO}$
• B. $E$ along $\text{OK}$
• C. $E$ along $\text{KO}$
• D. $3E$ along $\text{OK}$

Answer 1

The correct option is B $E$ along $\text{OK}$

The electric field due to part $\text{AKB}$ at centre will be along $\text{KO}$ with magnitude $E$.


We know that net electric field at the centre of ring having uniformly distributed over its circumference will be zero.

${\overrightarrow{E}}_{net}=0$

$\Rightarrow {\overrightarrow{E}}_{ACDB}+{\overrightarrow{E}}_{AKB}=0$

$\Rightarrow {\overrightarrow{E}}_{ACDB}=-{\overrightarrow{E}}_{AKB}$

Thus, the electric field due to part $\text{ACDB}$ will be along $\text{OK}$ and of same magnitude as of part $\text{AKB}$.

$\therefore E_{ACDB}=E$ along $\text{OK}$