- #1

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## Homework Statement

Can someone double check that this is right? I only multiplied charge of a proton by given potential difference to make the units work, without actually knowing if this is the right way to do this problem.

A proton is accelerated from rest through a potential difference of 0.100 MV. What is the final speed of the proton?

## The Attempt at a Solution

[tex]

\begin{array}{l}

E_k = \frac{1}{2}mv^2 \,\,\,\, \Rightarrow \,\,\,\,v = \sqrt {\frac{{2E_k }}{m}} = \sqrt {\frac{{2 \cdot 100\,{\rm{MV}} \cdot 10^6 {\rm{V/MV}} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} \\

\\

= \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 {\rm{V}} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 \frac{{{\rm{kg}} \cdot m^2 }}{{\frac{C}{s}s^3 }} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} \\

= \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 \frac{{{\rm{ \times m}}^2 }}{{\frac{{}}{{}} \cdot {\rm{s}}^2 }} \cdot 1.602 \cdot 10^{ - 19} }}{{1.67 \times 10^{ - 27} {\rm{ }}}}} =1.39x10^{8}m/s \\

\end{array}

\[/tex]

The final speed is rather high. This is one problem with E&M physics as opposed to mechanics. If a ball was tossed in the air and landed with this speed, I could use my intuition to tell me I did something wrong.

This velocity is half the speed of light. Wouldn't relativity come into play in these types of problems?