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

An electron enters a parallel plate capacitor with horizontal speed $u$ and is found to deflect by angle $\theta$ on leaving the capacitor as shown below. It is found that $\tan \theta = 0.4$ and gravity is negligible. If the initial horizontal speed is doubled, then the value of $\tan \theta$ will be

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of $1 \text{ MV}$,then the ratio of their kinetic energy will be

An electron with mass $9.1 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is placed in an electric field of $1 \times 10^6 \ V/m$. How much time will it take for the electron to reach a velocity equal to $1/10$th of the speed of light?

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).