An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field $200 \, \frac{ N }{ C }$ 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. $\left[ g =9.8 \,m / s ^{2}, \sin 30^{\circ}=\frac{1}{2}\right.$; $\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]$

A particle of mass $m$ and charge $q$ is thrown in a region where uniform gravitational field and electric field are present. The path of particle

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E.$ The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h.$ The time of fall of the electron, in comparison to the time of fall of the proton is 

In an ink-jet printer, an ink droplet of mass $m$ is given a negative charge $q$ by a computer-controlled charging unit, and then enters at speed $v$ in the region between two deflecting parallel plates of length $L$ separated by distance $d$ (see figure below). All over this region exists a downward electric field which you can assume to be uniform. Neglecting the gravitational force on the droplet, the maximum charge that can be given so that it will not hit a plate is close to :

A stream of a positively charged particles having $\frac{ q }{ m }=2 \times 10^{11} \frac{ C }{ kg }$ and velocity $\overrightarrow{ v }_0=3 \times 10^7 \hat{ i ~ m} / s$ is deflected by an electric field $1.8 \hat{ j } kV / m$. The electric field exists in a region of $10 cm$ along $x$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is $………..mm$