$A$ particle with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be:

An electron with kinetic energy $K_{1}$ enters between parallel plates of a capacitor at an angle $\alpha$ with the plates. It leaves the plates at an angle $\beta$ with the plates. If the kinetic energy of the electron when leaving is $K_{2}$,then the ratio of kinetic energies $K_{1}:K_{2}$ is ....... .

$A$ particle of mass $2 \ g$ and charge $6 \ \mu C$ is accelerated from rest through a potential difference of $60 \ V$. The speed acquired by the particle is (in $ms^{-1}$)

$A$ 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:

$A$ ball of mass $1 \text{ g}$ having a charge of $20 \mu\text{C}$ is tied to one end of a string of length $0.9 \text{ m}$ and can rotate in a vertical plane in a uniform electric field of $100 \text{ NC}^{-1}$ directed upwards. The minimum horizontal velocity that must be given to the ball at the lowest position so that it completes the vertical circle is (Let $g = 10 \text{ ms}^{-2}$): (in $\text{ ms}^{-1}$)

$A$ particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = (\frac{v_0}{\sqrt{2}}\hat{i} + \frac{v_0}{\sqrt{2}}\hat{j})$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is . . . . . . .