A uniformly charge half ring of a radius ' $R$ ' has linear charge density ' $\sigma$ '. The electric…
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)
$\frac{\sigma}{6 \epsilon_0}$
$\frac{\sigma}{2 \epsilon_0}$
$\frac{\sigma}{\epsilon_0}$
$\frac{\sigma}{4 \epsilon_0}$
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 $\mathrm{q}=\sigma \times \pi \mathrm{R}$
$\therefore \mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma \pi \mathrm{R}}{\mathrm{R}}=\frac{\sigma}{4 \varepsilon_0}$