When a proton is accelerated through $1\,V$,then its kinetic energy will be.....$eV$.

$A$ 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).

The coordinates of $A, B, C$ and $D$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$ respectively. How much work is done in moving a charge $q$ from $A$ to $D$ in a uniform electric field $\vec{E}$ directed along the positive $x$-axis?

Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is:

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

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field of $200 \, N/C$ as shown in the figure. $A$ body of mass $1 \, kg$ and charge $5 \, mC$ is allowed to slide down from rest at a height of $1 \, m$. If the coefficient of friction is $0.2$,find the time (in $s$) taken by the body to reach the bottom. $\left[ g = 9.8 \, m/s^2, \sin 30^{\circ} = 0.5, \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \right]$