1 \begin{problem*}{23.43}
2 An electron and a proton are each placed at rest in a uniform electric
3 field of magnitude $520\U{N/C}$. Calculate the speed of each
4 particle $48.0\U{ns}$ after being released.
8 The acceleration magnitudes are
10 a_p &= \frac{|F_p|}{m_p} = \frac{|q_e|}{m_p} \cdot E
11 = \frac{1.60\E{-19}\U{C}}{1.67\E{-27}\U{kg}} \cdot 520\U{N/C}
13 a_e &= \frac{|F_e|}{m_e} = \frac{|q_e|}{m_e} \cdot E
14 = \frac{1.60\E{-19}\U{C}}{9.11\E{-31}\U{kg}} \cdot 520\U{N/C}
15 = 91.3\U{Tm/s$^2$} \;.
18 Using our constant-acceleration formula, the speed after
19 $t=48.0\U{ns}$ is given by
21 v_p &= a_p t = \ans{2.39\U{km/s}} \\
22 v_e &= a_e t = \ans{4.38\U{Mm/s}} \;.