A thin ring of radius R = 10 cm carries a charge Q = +2.0 µC, spread evenly round it…

Step 1. The field on the axis (a). Each bit $dQ$ of the ring is $\sqrt{R^2 + z^2}$ away, and only the part of its field along the axis adds up:

At $z = R$: $E = \dfrac{kQ}{2\sqrt2\,R^2} = 6.36 \times 10^{5}\ \text{N/C}$.

Step 2. The greatest field (b, c). $\dfrac{dE}{dz} = kQ\,\dfrac{R^2 - 2z^2}{(R^2 + z^2)^{5/2}} = 0$ at $z = \dfrac{R}{\sqrt2} = 7.07\ \text{cm}$, where $E_{\max} = \dfrac{2kQ}{3\sqrt3\,R^2} = 6.93 \times 10^{5}\ \text{N/C}$.

Step 3. Speed at the centre (d). The bead is pulled towards $O$. The work done on it from $z = R$ to $O$ is

so $v = \sqrt{2W/m} = 7.26\ \text{m/s}$.

Step 4. Small oscillations (e). For $z \ll R$ the force is $F \approx -\dfrac{kQq}{R^3}z$. So $\omega = \sqrt{\dfrac{kQq}{mR^3}} = 94.87\ \text{rad/s}$ and $T = \dfrac{2\pi}{\omega} = 66.2\ \text{ms}$.

Step 5. A positive bead (f). Now the force pushes the bead away from $O$. All the way out it gains $\int_0^\infty qE\,dz = \dfrac{kQq}{R} = 0.090\ \text{J}$, so $v = \sqrt{\dfrac{2kQq}{mR}} = 13.42\ \text{m/s}$.