An $\alpha$-particle is accelerated through a potential difference of $200\,V$. The increase in its kinetic energy is.......$eV$.

An electron of mass $m_e$ initially at rest moves through a certain distance in a uniform electric field in time $t_1$. $A$ proton of mass $m_p$ also initially at rest takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity,the ratio of $t_2/t_1$ is nearly equal to

$A$ particle of mass $m$ and charge $q$ is released from rest in a region of uniform electric field of intensity $E$. After traveling a distance $y$,the kinetic energy of the particle is:

$A$ charge of magnitude $2e$ and mass $4m$ is moving in an electric field $E$. The acceleration imparted to the above charge is

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

$A$ simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the final tension in the string changes. The final tension in the string, when the pendulum attains an equilibrium position, is . . . . . . . ($g$: acceleration due to gravity)