Speed of a Proton After it Accelerates Through a Potential Difference

Problem Statement

A proton is accelerated over a potential difference of 100 V. What is the final speed of the proton?

I Need a Hint!

We are not given any distances over which the proton accelerates - just the potential difference. Thus, it might be difficult to use electric force or Newton’s laws since we cannot directly turn potential difference into force or field without more information. Instead, think about how we might use Conservation of Energy here.

I Need a Another Hint!

If a charged particle (q) travels over a potential difference ($\Delta V$) the change in energy of the system is $\Delta U=q\Delta V$.

Solution

We can use the law of Conservation of Energy to solve this problem. If a proton with charge $e$ travels over a potential difference ($\Delta V$) the gain in energy of the proton is $\Delta U=e\Delta V$. This gain in energy will become kinetic energy:

$$e\Delta V=\frac{1}{2}m{v}^2$$

Solving for speed (v) yields:

$$\frac{1}{2}m{v}^2=e\Delta V$$ $$v^2=\frac{2e\Delta V}{m}$$ $$v=\sqrt{\frac{2e\Delta V}{m}}$$

Plugging in values yields:

$$v=\sqrt{\frac{2(1.6\times 10^{-19}C)(100V)}{1.67\times 10^{-27}kg}}=138425.7m/s$$