$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

$A$ point charge $Q$ is fixed. $A$ small charge $q$ and mass $m$ is given a velocity $v_0$ from infinity at a perpendicular distance $r_0$ as shown. If the distance of closest approach is $r_0/2$,find the value of $q$. [Given $mv_0^2 = \frac{Q^2}{4\pi \epsilon_0 r_0}$]

$A$ proton is located at coordinates $(x, y) = (0, 0)$,and an electron is 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,at what $y$ coordinate will the two particles have the same $y$ position at the same time?

In the following diagram,which particle has the highest $e/m$ value?

$A$ charged particle $q$ is shot towards another charged particle $Q$ which is fixed,with a speed $v$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be

$A$ particle of charge $1\ \mu C$ and mass $1\ g$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ s$. Then its final speed cannot be.......$m/s$.