$A$ block of mass $m$ and charge $q$ is connected to a point $O$ with an inextensible string. This system is on a horizontal table. An electric field $E$ is applied perpendicular to the string and in the plane of the horizontal table. The tension in the string when it becomes parallel to the electric field is

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

$A$ particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The kinetic energy attained by the particle after moving a distance $y$ is

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of $1 \text{ MV}$,then the ratio of their kinetic energy will be

What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)

$A$ particle with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be: