Q144 / 150 · MHT CET 2021 (20 Sep Shift 2) · Physics
A uniformly charge half ring of a radius ' $R$ ' has linear charge density ' $\sigma$ '. The electric potential at the centre of the half ring is $\left(\epsilon_0=\right.$ permittivity of free space)
- A$\frac{\sigma}{6 \epsilon_0}$
- B$\frac{\sigma}{2 \epsilon_0}$
- C$\frac{\sigma}{\epsilon_0}$
- Correct answerD$\frac{\sigma}{4 \epsilon_0}$
Solution
If $\mathrm{q}$ is charge on the ring, then the potential at the centre is given by
$$
\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{R}}
$$
But $\mathrm{q}=\sigma \times \pi \mathrm{R}$
$$
\therefore \mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma \pi \mathrm{R}}{\mathrm{R}}=\frac{\sigma}{4 \varepsilon_0}
$$
Hence, the correct answer is option: 'D'.