A thin ring of radius 3.0 cm carries a charge of +14 nC, spread evenly…

Step 1. The centre. $kQ = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times 14 \times 10^{-9}\ \text{C} = 126\ \text{V m}$. So $V_c = \dfrac{kQ}{R} = \dfrac{126\ \text{V m}}{0.030\ \text{m}} = 4200\ \text{V}$.

Step 2. The release point. The distance to the ring is $\sqrt{(0.030)^2 + (0.040)^2}\ \text{m} = 0.050\ \text{m}$. $V_x = \dfrac{126\ \text{V m}}{0.050\ \text{m}} = 2520\ \text{V}$.

Step 3. The energy. $U = qV$. At the release point, $U_x = (-1.5 \times 10^{-6}\ \text{C}) \times 2520\ \text{V} = -3.78 \times 10^{-3}\ \text{J}$. At the centre, $U_c = (-1.5 \times 10^{-6}\ \text{C}) \times 4200\ \text{V} = -6.30 \times 10^{-3}\ \text{J}$. The potential energy falls by $2.52 \times 10^{-3}$ J, and this is the kinetic energy at the centre.

Step 4. The speed. From $\tfrac{1}{2}mv^2 = 2.52 \times 10^{-3}\ \text{J}$ we want $v$, so $v = \sqrt{\dfrac{2 \times 2.52 \times 10^{-3}\ \text{J}}{5.0 \times 10^{-4}\ \text{kg}}} = 3.17\ \text{m/s}$.