A charge \(q\) is uniformly distributed on a ring of radius \(r\). A sphere of an equal radius is constructed with its centre lying on the periphery of the ring. The flux of electric field through the surface of the sphere will be

• A

  \(\frac{q}{\varepsilon_0}\)
• B

  \(\frac{q}{2 \varepsilon_0}\)
• Correct answer
  C

  \(\frac{q}{3 \varepsilon_0}\)
• D

  \(\frac{q}{4 \varepsilon_0}\)

Solution

Charge on ring $q$, centre of ring $=O$ Centre of sphere $=O^{\prime}$ Linear charge density of ring, $\lambda=\frac{q}{2 \pi a}$ Charge on arc $A B$ of ring, $\begin{aligned} & q_{A B}=\lambda(\operatorname{arc} A B)=\frac{1}{2 \pi a} \cdot q \cdot \frac{2 \pi a}{3} \\ & q_{A B}=q / 3\end{aligned}$ i.e., charge enclosed by sphere $=q / 3$ $\therefore$ Flux coming out of sphere $=q / 3 \varepsilon_0$.

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