The electric field at a point is given by $\vec E = E_0 \hat i \, V/m$. $A$ particle of charge $+q_0$ moves from point $A(0, a)$ to $B(a, 0)$ along a circular path as shown in the figure. Find the work done by the electric field in this motion.

$A$ charged particle of mass $0.003 \, g$ is held stationary in space by placing it in a downward electric field of $6 \times 10^4 \, N/C$. The magnitude of the charge is:

$A$ problem of practical interest is to make a beam of electrons turn at a $90^{\circ}$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 \times 10^{-17} \ J$ enters through a small hole in the bottom plate. The strength of the electric field that is needed if the electron is to emerge from an exit hole $1.0 \ cm$ away from the entrance hole,traveling at right angles to its original direction,is $y \times 10^5 \ N/C$. The value of $y$ is

$A$ charged particle is free to move in an electric field. It will travel

$A$ particle of mass $m$ and charge $q$ is thrown perpendicular to an electric field of intensity $E$ with an initial velocity $v$. The particle moves a distance $x$ perpendicular to the field and a distance $y$ along the direction of the field. If $y = \alpha x^{2}$,then $\alpha$ is given by:

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)