$A$ 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 $mv_0^2 = \frac{Q^2}{4\pi \epsilon_0 r_0}$]

$A$ simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied,the final tension in the string changes. The final tension in the string,when the pendulum attains an equilibrium position,is . . . . . . . ($g$: acceleration due to gravity)

An electron is accelerated through a potential difference of $200 \ V$. If $e/m$ for the electron is $1.6 \times 10^{11} \ C/kg$,the velocity acquired by the electron will be:

An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

$A$ particle of mass $2 \times 10^{-5} \ kg$ and charge $4 \times 10^{-3} \ C$ starts from rest in a uniform electric field of $5 \ V/m$. What is its kinetic energy after $10 \ s$?

An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be: