1. **Problem statement:** We want to derive the formula for kinetic energy in terms of charge $q$, mass $m$, velocity $v$, and potential difference $U$. 2. **Given:** The kinetic energy $E_{kin}$ is related to the charge $q$ and potential difference $U$ by the work-energy principle: $$E_{kin} = qU$$ 3. **Formula for kinetic energy:** The kinetic energy of a particle with mass $m$ and velocity $v$ is given by: $$E_{kin} = \frac{1}{2}mv^2$$ 4. **Equate the two expressions for kinetic energy:** Since the kinetic energy gained by the charge $q$ accelerated through potential difference $U$ is $qU$, we have: $$qU = \frac{1}{2}mv^2$$ 5. **Solve for $v^2$:** Multiply both sides by 2: $$2qU = mv^2$$ 6. **Isolate $v^2$:** $$v^2 = \frac{2qU}{m}$$ 7. **If the charge is an electron with charge $e$ and $z$ is the charge number, then $q = ez$. Substituting:** $$v^2 = \frac{2ezU}{m}$$ 8. **Rearranging to express $m/z$:** $$\frac{m}{z} = \frac{2eU}{v^2}$$ **Summary:** Starting from the work-energy principle and kinetic energy formula, we derived that $$\frac{m}{z} = \frac{2eU}{v^2}$$ which relates mass-to-charge ratio to potential difference and velocity.