$A$ mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3\,s$ in this region is.......$m/s$.

In a uniform electric field,if a charge is fired in a direction different from the line of the electric field,then the trajectory of the charge will be a

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.

$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be:

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)

$A$ uniform electric field of $10\,N/C$ is created between two parallel charged plates (as shown in the figure). An electron enters the field symmetrically between the plates with a kinetic energy of $0.5\,eV$. The length of each plate is $10\,cm$. The angle $(\theta)$ of deviation of the path of the electron as it comes out of the field is $.........$ (in degrees).