A proton sits at coordinates $(x, y) = (0, 0)$, and an electron at $(d, h)$, where $d >> h$. At time $t = 0$, $a$ uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible, the $y$ coordinates of the two particles will be equal (at equal time)

An electron falls through a distance of $1.5\, cm$ in a uniform electric field of magnitude $2.0\times10^4\, N/C$ as shown in the figure. The time taken by electron to fall through this distance is ($m_e = 9.1\times10^{-31}\,kg$, Neglect gravity)

An electron enters a parallel plate capacitor with horizontal speed $u$ and is found to deflect by angle $\theta$ on leaving the capacitor as shown below. It is found that $\tan \theta=0.4$ and gravity is negligible. If the initial horizontal speed is doubled, then the value of $\tan \theta$ will be

A Charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E.$ Neglecting the effect of gravity, the kinetic energy of the charged particle after $'t'$ second is

An electron and a proton are in a uniform electric field, the ratio of their accelerations will be

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E.$ The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h.$ The time of fall of the electron, in comparison to the time of fall of the proton is