An electric line of force in $X$, $Y-$ plane is given by $x^2+y^2 = 1$. A particle with unit positive charge, initially at rest at the point $x = 1, y = 0$ in the $X, Y-$ plane

A positive charge particle of $100 \,mg$ is thrown in opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC ^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms ^{-1}$, how much distance (in $m$) it will travel before coming to the rest momentarily

An electron enters in an electric field with its velocity in the direction of the electric lines of force. Then

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 

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 $...........mm$

An electron and a proton are in a uniform electric field, the ratio of their accelerations will be