A proton that is initially at rest is accelerated through an electric potential difference of magnitude $500 \mathrm {~V}$. What speed does the proton gain? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m \right.$ proton $\left. = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} \right)$

Question

Quizplus · Accepted Answer

The kinetic energy gained by the proton is equal to the work done on it by the electric field, which is given by $qV$, where $q$ is the charge of the proton and $V$ is the potential difference. The kinetic energy can also be expressed as $\frac{1}{2}mv^2$, where $m$ is the mass of the proton and $v$ is its velocity. Setting these equal and solving for $v$ gives $v = \sqrt{\frac{2qV}{m}}$. Substituting the given values, $q = 1.60 \times 10^{-19} \, \text{C}$, $V = 500 \, \text{V}$, and $m = 1.67 \times 10^{-27} \, \text{kg}$, we find $v \approx 3.1 \times 10^5 \, \text{m/s}$.

A Proton That Is Initially at Rest Is Accelerated Through 500 V500 \mathrm {~V}500 V

A Proton That Is Initially at Rest Is Accelerated Through $500 \mathrm {~V}$