The intensity of the electric field required to balance a proton of mass $1.7 \times 10^{-27} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is nearly:

The potential difference between two metal plates is $800 \, V$ and they are separated by a horizontal distance of $0.02 \, m$. $A$ particle of mass $1.96 \times 10^{-15} \, kg$ is held in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is .......

Two charges each of magnitude $q$ are fixed at a distance $2d$ apart. $A$ third charge (proton) placed at the midpoint is displaced slightly by a distance $x$ $(x << d)$ perpendicular to the line joining the two fixed charges. The proton will execute simple harmonic motion having an angular frequency: ($m =$ mass of the charged particle)

$A$ 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 $q$ 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

$A$ 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