A proton falls through a distance of $1.5 ext{ cm}$ in a uniform electric field of magnitude $2.0 \

Question

A proton falls through a distance of $1.5	ext{ cm}$ in a uniform electric field of magnitude $2.0 	imes 10^4	ext{ N/C}$. What is the approximate time of fall? (Take $m_p = 1.67 	imes 10^{-27}	ext{ kg}$, $e = 1.6 	imes 10^{-19}	ext{ C}$)

Professor PC Thomas Classes · Answer

Acceleration of proton $a_p = eE/m_p$. Time $t = \sqrt{2s m_p / (eE)} = \sqrt{2 	imes 0.015 	imes 1.67 	imes 10^{-27} / (1.6 	imes 10^{-19} 	imes 2.0 	imes 10^4)} \approx 1.3 	imes 10^{-7}	ext{ s}$.