An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.

$A$ uniform vertical electric field $E$ is established in the space between two large parallel plates. $A$ small conducting sphere of mass $m$ is suspended in the field from a string of length $L$. If the sphere is given a $+q$ charge and the lower plate is charged positively,the period of oscillation of this pendulum is:

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.

$A$ particle of mass $m$ and charge $q$ travelling with a velocity $v$ along the $x$-axis enters a uniform electric field $\overrightarrow{E}$ directed along the $y$-axis. What will be the trajectory of the particle?

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)

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the center of a uniformly charged spherical region of total charge $Q$ and radius $R$. $q$ and $Q$ have opposite signs. The spherically charged region is not free to move. The value of $K$ is such that the particle will just reach the boundary of the spherically charged region. How much time does it take for the particle to reach the boundary of the region?