To solve this problem, we need to analyze the motion of both the electron and the proton in the electric field.$\\$Given:$\\$• An electron falls from rest through a distance $h$ in a uniform electric field $E. \\$• The electric field is reversed, and a proton falls from rest through the same distance $h. \\$The force on a charged particle in an electric field is given by $F = qE,$ where $q$ is the charge.$\\$For the electron:$\\$• Charge $q_e = -e$ (where $e$ is the elementary charge)$. \\$• Mass $m_e$ of the electron.$\\$• Force on the electron $F_e = -eE. \\$• Acceleration of the electron $a_e = \frac{F_e}{m_e} = \frac{-eE}{m_e}. \\$For the proton:$\\$• Charge $q_p = +e. \\$• Mass $m_p$ of the proton.$\\$• Force on the proton $F_p = eE. \\$• Acceleration of the proton $a_p = \frac{F_p}{m_p} = \frac{eE}{m_p}. \\$The time of fall for an object under constant acceleration is given by the equation:$\\ t = \sqrt{\frac{2h}{a}}. \\$Time of fall for the electron:$\\ t_e = \sqrt{\frac{2h}{a_e}} = \sqrt{\frac{2h}{\frac{-eE}{m_e}}} = \sqrt{\frac{2hm_e}{eE}}. \\$Time of fall for the proton:$\\ t_p = \sqrt{\frac{2h}{a_p}} = \sqrt{\frac{2h}{\frac{eE}{m_p}}} = \sqrt{\frac{2hm_p}{eE}}. \\$To compare $t_e$ and $t_p,$ we take the ratio:$\\ \frac{t_e}{t_p} = \frac{\sqrt{\frac{2hm_e}{eE}}}{\sqrt{\frac{2hm_p}{eE}}} = \sqrt{\frac{m_e}{m_p}}. \\$The mass of the electron $m_e$ is much smaller than the mass of the proton $m_p. \\$Therefore, $\sqrt{\frac{m_e}{m_p}}$ is much smaller than 1.$\\$Thus, the time of fall of the electron is smaller than the time of fall of the proton.$\\$This corresponds to Option 3: smaller.