Solution

Gauss theoremwulawula

$$\oint_{2\pi r}E\,dl=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{q}{2\pi\varepsilon_0 r}$$

Let's write down the small change of the electrostatic field and then integrate it

$$dU=E \,dr\Rightarrow \int dU=\int E \,dr$$

Integrate from $R_1$ to $R_2$

$$U_0=\frac{q}{2\pi\varepsilon_0 r}\int_{R_1}^{R_2}\frac{dr}{r}=\frac{q}{2\pi\varepsilon_0 r}\ln\frac{R_2}{R_1}$$

Law of conservation of energy

$$\frac{mv^2}{2}=eU$$

$$e\frac{q}{2\pi\varepsilon_0 \frac{R_1+R_2}{2}}=\frac{mv^2}{(\frac{R_1+R_2}{2})}$$

From where we get

$$\frac{q}{2\pi\varepsilon_0}=\frac{mv^2}{e}$$

Substituting the previously obtained values

$$\frac{U_0}{\ln\frac{R_2}{R_1}}=\frac{2eU}{e}$$

From where

$$U=\frac{U_0}{2\ln\frac{R_2}{R_1}}$$